Introduction.

§ 1. Ring Domain, Ideal, Finiteness Condition.

§ 2. Representation of an Ideal as the Least Common Multiple of Finitely Many Irreducible Ideals.

§ 3. Equality of the Number of Components in Two Different Decompositions into Irreducible Ideals.

§ 4. Primary Ideals. Uniqueness of the Associated Prime Ideals in Two Different Decompositions into Irreducible Ideals.

§ 5. Representation of an Ideal as the Least Common Multiple of Greatest Primary Ideals. Uniqueness of the Associated Prime Ideals.

§ 6. Unique Representation of an Ideal as the Least Common Multiple of Relatively Prime Irreducible Ideals.

§ 7. Uniqueness of the Isolated Ideals.

§ 8. Unique Representation of an Ideal as a Product of Coprime-Irreducible Ideals.

§ 9. Extension of the Investigation to Modules. Equality of the Number of Components in Decompositions into Irreducible Modules.

§ 10. Special Case of the Polynomial Domain.

§ 11. Examples from Number Theory and from the Theory of Differential Expressions.

§ 12. Example from Elementary-Divisor Theory.

The content of the present paper is the transfer of the decomposition theorems for the rational integers, respectively of the ideals in algebraic number fields, to ideals in arbitrary integral domains, and more generally ring domains. To understand this transfer, let us first state the decomposition theorems for the rational integers in a form somewhat different from the usual formulation.

If in \[a=p_1^{\varrho_1}p_2^{\varrho_2}\cdots p_\sigma^{\varrho_\sigma}=q_1q_2\cdots q_\sigma\] one regards the prime-power factors \(q_i\) as the components of the decomposition, then these components have the following characteristic properties:

1. They are pairwise coprime; but no \(q_i\) can be represented as a product of pairwise coprime numbers, so that in this sense irreducibility holds. From the pairwise coprimeness it further follows that the product \(q_1\cdots q_\sigma\) is equal to the least common multiple \([q_1\cdots q_\sigma]\).

2. Any two of the components, \(q_i\) and \(q_k\), are relatively prime; that is, if \(bq_i\) is divisible by \(q_k\), then \(b\) is divisible by \(q_k\). In this sense too irreducibility holds.

3. Every \(q\) is primary; that is, if a product \(b\cdot c\) is divisible by \(q\), but \(b\) is not divisible, then a power¹ of \(c\) is divisible. The representation is moreover one by greatest primary components, since the product of two distinct \(q\) is no longer primary. Also with respect to decomposition into greatest primary components the \(q\) are irreducible.

4. Every \(q\) is irreducible in the sense that it cannot be represented as the least common multiple of two proper divisors.

The connection of these primary numbers \(q\) with the prime numbers \(p\) consists in this: to each \(q\) there exists one, and apart from sign only one, \(p\) which is a divisor of \(q\) and a power of which is divisible by \(q\): the associated prime number. If \(p^\varrho\) is the lowest power of this kind – \(\varrho\) the exponent of \(q\) – then here, in particular, \(p^\varrho\) is equal to \(q\). The uniqueness theorem may now be stated as follows:

For two different decompositions of a rational integer into the irreducible, greatest primary components \(q\), the number of components, the associated prime numbers (up to sign), and the exponents coincide. Since \(p^\varrho=q\), it follows from this also that the \(q\) themselves coincide (up to sign).

The indeterminacy arising from the sign is, as is well known, removed when one considers, instead of the numbers, the ideals derived from them (all numbers divisible by \(a\)); then the formulation holds in exactly the same way for the unique decomposition of the ideals of finite algebraic number fields into powers of prime ideals.

In what follows (§ 1), the basis will be a general ring domain satisfying only the finiteness condition that each ideal of the domain possess a finite ideal basis. Without such a finiteness condition, irreducible and prime ideals need not exist at all, as is shown by the domain of all algebraic integers, in which there is no decomposition into prime ideals.

It turns out that – corresponding to the four characteristic properties of the components \(q\) – in general four separate decompositions exist, each arising from the preceding one by subdivision. The decomposition into coprime-irreducible ideals is a product representation; the other three decompositions are reduced (§ 2) representations as least common multiples. The connection between a primary ideal – the irreducible ideals too are primary – and the associated prime ideal also persists: to every primary ideal \(\mathfrak{Q}\) there is uniquely determined an associated prime ideal \(\mathfrak{P}\) which is a divisor of \(\mathfrak{Q}\) and a power of which is divisible by \(\mathfrak{Q}\). If \(\mathfrak{P}^\varrho\) is the lowest such power – \(\varrho\) the exponent of \(\mathfrak{Q}\) – then here \(\mathfrak{P}^\varrho\) need not coincide with \(\mathfrak{Q}\). The uniqueness theorem is expressed as follows:

Decompositions 1 and 2 are unique; in two different decompositions 3 or 4 the number of components and the associated prime ideals coincide²; the isolated ideals occurring among the components (§ 7) are uniquely determined.

For the proof of the decomposition theorems, the finiteness condition first yields the “theorem on the finite chain”, first stated by Dedekind for finite number modules; from it the representation 4 of each ideal as the least common multiple of finitely many irreducible ideals is derived. By transforming the concept of reducibility of a component, one obtains from this the fundamental uniqueness theorem for decomposition 4 into irreducible ideals. By combining finitely many components at a time one rises to the remaining decompositions, whose uniqueness theorems then follow from the uniqueness theorem 4.

Finally it is shown (§ 9) that the representation by finitely many irreducible components still holds under weaker hypotheses; commutativity of the ring domain is not required, and it is enough to consider, in place of an ideal, a module with respect to the domain. In this more general case the equality of the number of components in two different decompositions still holds, whereas the notions prime and primary are tied to commutativity and to the ideal concept; by contrast, the notion coprime remains valid for ideals in noncommutative domains.

The simplest ring domain for which the four separate decompositions actually occur is the domain of all polynomials in \(n\) variables with arbitrary complex coefficients. Here the individual decompositions can be interpreted, irrationally, by the behavior of algebraic configurations, and the uniqueness theorem for the associated prime ideals corresponds to the fundamental theorem of elimination theory on the unique decomposability of algebraic configurations into irreducible ones. Further examples are given by all finite integral domains made of polynomials (§ 10). But even the simple domain of all even numbers, more generally of all numbers divisible by a fixed number, already gives an example of partially separated decompositions (§ 11). An example of ideal theory in noncommutative domains is provided by elementary-divisor theory (§ 12), where unique decomposition into irreducible ideals, respectively classes, holds. These irreducible classes completely characterize the irreducible constituents of the elementary divisors, and may perhaps be viewed as their equivalent for domains in which the usual elementary-divisor theory fails.

On the existing literature the following is to be said. The decomposition into greatest primary ideals for the polynomial domain with arbitrary complex, respectively integral, coefficients was given by Lasker, and in individual points further developed by Macaulay.³ Both rely on elimination theory, and hence use the fact that a polynomial can be represented uniquely as a product of irreducible polynomials. In fact the decomposition theorems for ideals are independent of this prerequisite, as ideal theory in algebraic number fields suggests and as the present paper shows. The primary ideal is also defined by Lasker and Macaulay on the basis of concepts from elimination theory.

The decomposition into irreducible ideals and the decomposition into relatively-prime irreducible ideals do not seem to have been noticed in the literature even for the polynomial domain; only in Macaulay is there a remark on the uniqueness of isolated primary ideals.

The decomposition into coprime-irreducible ideals was given for the polynomial domain by Schmeidler,⁴ using elimination theory for the proof of finiteness. There, however, the uniqueness theorem is stated only for classes of ideals, not for the ideals themselves. This latter uniqueness theorem appears in a joint paper,⁵ where ideals in noncommutative polynomial domains are involved. Only the finite ideal basis is used there; hence theorems and methods remain valid for general ring domains, and are sharpened in the present paper with respect to equality of number (§ 11). The present investigations are a strong generalization and further development of the conceptual formations underlying those two papers. The essential feature of the two papers is the passage from representation as a least common multiple to an additive decomposition of the system of residue classes. Here, for simplicity of presentation, we remain with the least common multiple; the additive decomposition is then matched by the transformation of the notion of reducibility into a property of the complement (§ 3). Yet by considerations essentially corresponding to those of the joint paper, all the theorems stated here can also be understood as additive decomposition theorems for the system of residue classes and certain partial systems. This system of residue classes forms a ring of the same generality as the ring originally taken as basis; indeed every ring may be regarded as the system of residue classes of the ideal corresponding to the totality of identical relations among the ring elements; or also of a partial system of these relations, if the remaining relations are assumed to be satisfied already in the domain.

This remark also gives the place of Fraenkel’s papers.⁶ Fraenkel considers additive decompositions of rings that are subjected to such restrictive conditions (existence of regular elements, division by these, decomposability condition) that, for the corresponding ideal, the four decompositions coincide. Because of this coincidence, his finiteness condition, that the ideal should have only finitely many proper divisors – from which he in part departs – also means no stronger restriction than ours. Fraenkel’s starting point is conditioned by the different, essentially algebraic, aims of his papers; by algebraic extension he then arrives at more general rings with less restrictive conditions.

1. The domain \(\Sigma\) taken as basis is to be a (commutative) ring in the abstract definition;⁷ that is, \(\Sigma\) consists of a system of elements \(a,b,c,\ldots,f,g,h,\ldots\) in which a relation satisfying the usual conditions is defined as equality; and in which by two operations (modes of composition), addition and multiplication, from any two ring elements \(a\) and \(b\) a third is always uniquely obtained, respectively as sum \(a+b\) and as product \(a\cdot b\). The ring and the otherwise wholly arbitrary operations must satisfy the following laws:

1. The associative law of addition: \((a+b)+c=a+(b+c)\).

2. The commutative law of addition: \(a+b=b+a\).

3. The associative law of multiplication: \((a\cdot b)c=a(b\cdot c)\).

4. The commutative law of multiplication: \(a\cdot b=b\cdot a\).

5. The distributive law: \(a(b+c)=ab+ac\).

6. The law of unrestricted and unique subtraction. There is in \(\Sigma\) a single element \(x\) satisfying the equation \(a+x=b\). (One denotes \(x=b-a\).)

From these properties follows the existence of zero; but a ring need not possess a unit; and the product of two elements can vanish without one of the factors vanishing. Rings for which, from the vanishing of a product, the vanishing of a factor always follows, and which in addition possess a unit, will be called proper integral domains. For the finite sum \(a+a+\cdots+a\) we introduce the usual abbreviated notation \(na\), where the integers \(n\) are to be regarded merely as abbreviating signs, not as ring elements, and are defined recursively by \(a=1a\), \(na+a=(n+1)a\).

2. By an ideal \(\mathfrak{M}\)⁸ in \(\Sigma\) is meant a system of elements from \(\Sigma\) satisfying the two conditions:

1. \(\mathfrak{M}\) contains, together with \(f\), also \(a\cdot f\), where \(a\) is an arbitrary element of \(\Sigma\).

2. \(\mathfrak{M}\) contains, together with \(f\) and \(g\), also the difference \(f-g\); hence with \(f\) also \(nf\) for every integer \(n\).

If \(f\) is an element of \(\mathfrak{M}\), we express this as usual by \(f\equiv0(\mathfrak{M})\) and say that \(f\) is divisible by \(\mathfrak{M}\). If every element of \(\mathfrak{N}\) is at the same time an element of \(\mathfrak{M}\), and hence divisible by \(\mathfrak{M}\), we say: \(\mathfrak{N}\) is divisible by \(\mathfrak{M}\); in symbols, \(\mathfrak{N}\equiv0(\mathfrak{M})\). \(\mathfrak{M}\) is called a proper divisor of \(\mathfrak{N}\) when it contains elements different from those of \(\mathfrak{N}\), and hence is not conversely divisible by \(\mathfrak{N}\). From \(\mathfrak{N}\equiv0(\mathfrak{M})\) and \(\mathfrak{M}\equiv0(\mathfrak{N})\) follows \(\mathfrak{N}=\mathfrak{M}\).

The other familiar notions are also retained word for word. By the greatest common divisor of two ideals \(\mathfrak{A}\) and \(\mathfrak{B}\) – \(\mathfrak{D}=(\mathfrak{A},\mathfrak{B})\) – we mean the totality of elements that can be represented in the form \(a+b\), where \(a\) runs through all elements of \(\mathfrak{A}\) and \(b\) through all elements of \(\mathfrak{B}\); \(\mathfrak{D}\) is again an ideal. Likewise, the greatest common divisor of infinitely many ideals – \(\mathfrak{D}=(\mathfrak{A}_1,\mathfrak{A}_2,\ldots,\mathfrak{A}_\nu,\ldots)\) – is defined as the totality of elements \(d\) representable as sums of elements from finitely many of the ideals at a time: \(d=a_{i_1}+a_{i_2}+\cdots+a_{i_n}\); here too \(\mathfrak{D}\) is again an ideal.

If, in particular, the ideal \(\mathfrak{M}\) contains a finite number of elements \(f_1,f_2,\ldots,f_\varrho\) such that \[\mathfrak{M}=(f_1\ldots f_\varrho),\qquad f=a_1f_1+\cdots+a_\varrho f_\varrho+n_1f_1+\cdots+n_\varrho f_\varrho\] for every \(f\equiv0(\mathfrak{M})\), where the \(a_i\) are quantities of the ring domain and the \(n_i\) are integers, then \(\mathfrak{M}\) is called a finite ideal; \(f_1, \ldots,f_\varrho\) an ideal basis.

In what follows we take as basis only rings \(\Sigma\) satisfying the finiteness condition: Every ideal in \(\Sigma\) is finite, hence possesses an ideal basis.

3. From the finiteness condition there follows directly the result on which all subsequent considerations rest.

Theorem I (theorem on the finite chain).⁹ Let \(\mathfrak{M},\mathfrak{M}_1,\mathfrak{M}_2,\ldots,\mathfrak{M}_\nu,\ldots\) be a countably infinite system of ideals in \(\Sigma\), each of which is divisible by the following one. Then from some finite index \(n\) onward all ideals are identical, \(\mathfrak{M}_n=\mathfrak{M}_{n+1}=\cdots\). In other words: if \(\mathfrak{M},\mathfrak{M}_1,\mathfrak{M}_2,\ldots,\mathfrak{M}_s, \ldots\) form a simply ordered chain of ideals such that every ideal is a proper divisor of the one immediately preceding it, then the chain breaks off after finitely many steps.

Indeed, let \(\mathfrak{D}=(\mathfrak{M}_1,\mathfrak{M}_2,\ldots,\mathfrak{M}_\nu,\ldots)\) be the greatest common divisor of the system, and let \(f_1,\ldots,f_k\) be a basis of \(\mathfrak{D}\), which always exists by the finiteness condition. From the divisibility assumption it follows that every element of \(\mathfrak{D}\) is at the same time an element of one ideal of the chain; for from \[f=g+h,\qquad g\equiv0(\mathfrak{M}_r),\qquad h\equiv0(\mathfrak{M}_s),\qquad (r\le s)\] one gets \(g\equiv0(\mathfrak{M}_s)\) and hence \(f\equiv0(\mathfrak{M}_s)\). The same holds when \(f\) is a sum of several constituents. Hence there is a finite index \(n\) such that \[f_1\equiv0(\mathfrak{M}_n);\quad\ldots;\quad f_k\equiv0(\mathfrak{M}_n);\quad \mathfrak{D}=(f_1\ldots f_k)\equiv0(\mathfrak{M}_n).\] Since conversely \(\mathfrak{M}_n\equiv0(\mathfrak{D})\), we have \(\mathfrak{M}_n=\mathfrak{D}\); and since further \(\mathfrak{M}_n\equiv0(\mathfrak{D})\) and \(\mathfrak{D}=\mathfrak{M}_n\equiv0(\mathfrak{M}_r)\), we also have \(\mathfrak{M}_r=\mathfrak{D}=\mathfrak{M}_n\) for every \(r>n\), proving the theorem.

We note that conversely the existence of an ideal basis again follows from this theorem, so that the finiteness condition could also have been stated in this basis-free form.

The least common multiple \([\mathfrak{B}_1,\mathfrak{B}_2,\ldots,\mathfrak{B}_k]\) of the ideals \(\mathfrak{B}_1,\mathfrak{B}_2,\ldots,\mathfrak{B}_k\) is defined as usual as the totality of the elements divisible by \(\mathfrak{B}_1\), by \(\mathfrak{B}_2\), \(\ldots\), and by \(\mathfrak{B}_k\); in symbols, \[\text{from } f\equiv0(\mathfrak{B}_i),\quad (i=1,2,\ldots,k), \quad\text{follows}\quad f\equiv0([\mathfrak{B}_1,\mathfrak{B}_2,\ldots,\mathfrak{B}_k]),\] and conversely. The least common multiple is again an ideal; the ideals \(\mathfrak{B}_i\) are also called the components of the decomposition.

Definition I. A representation \(\mathfrak{M}=[\mathfrak{B}_1\cdots\mathfrak{B}_k]\) is called a reduced representation if no \(\mathfrak{B}_i\) is absorbed into the least common multiple \(\mathfrak{A}_i\) of the remaining ideals, and if no \(\mathfrak{B}_i\) can be replaced by a proper divisor.¹⁰ If the conditions are satisfied only for the ideal \(\mathfrak{B}_i\), the representation is called reduced with respect to \(\mathfrak{B}_i\). The least common multiple \(\mathfrak{A}_i=[\mathfrak{B}_1,\ldots,\mathfrak{B}_{i-1},\mathfrak{B}_{i+1},\ldots,\mathfrak{B}_k]\) is called the complement of \(\mathfrak{B}_i\). Representations in which only the first condition is satisfied are called shortest representations.

For representing an ideal as a least common multiple it therefore suffices to restrict to reduced representations, by the following result.

Lemma I. Every representation of an ideal as the least common multiple of finitely many ideals can be replaced, in at least one way, by a reduced representation; such a representation can in particular be attained by successive decomposition.

Indeed, let \(\mathfrak{M}=[\mathfrak{B}_1^*\cdots\mathfrak{B}_l^*]\) be an arbitrary representation of \(\mathfrak{M}\). We omit, in order, those \(\mathfrak{B}_i^*\) which are absorbed into the least common multiple of the ideals retained. Since the remaining ideals still yield \(\mathfrak{M}\), in the resulting representation \[\mathfrak{M}=[\mathfrak{B}_1\cdots\mathfrak{B}_k]=[\mathfrak{A}_i,\mathfrak{B}_i]\] the first condition is satisfied, hence it is a shortest representation; and this condition remains satisfied if any \(\mathfrak{B}_i\) is replaced by a proper divisor. The second condition, however, is always attainable by the theorem on the finite chain (Theorem I). For if \[\mathfrak{B}_i\supset \mathfrak{B}_i'\supset \mathfrak{B}_i''\supset\cdots\supset \mathfrak{B}_i^{(\nu)}\supset\cdots, \qquad \mathfrak{M}=[\mathfrak{A}_i,\mathfrak{B}_i]=[\mathfrak{A}_i,\mathfrak{B}_i']=\cdots=[\mathfrak{A}_i,\mathfrak{B}_i^{(\nu)}]=\cdots,\] where every \(\mathfrak{B}_i^{(\nu)}\) is a proper divisor of the one immediately preceding, then by that theorem the chain \(\mathfrak{B}_i,\mathfrak{B}_i',\ldots,\mathfrak{B}_i^{(\nu)},\ldots\) must terminate after finitely many steps; hence in the representation \(\mathfrak{M}=[\mathfrak{A}_i,\mathfrak{B}_i^{(\nu)}]\) the ideal \(\mathfrak{B}_i^{(\nu)}\) can no longer be replaced by a proper divisor; and this holds a fortiori if \(\mathfrak{A}_i\) is replaced by a proper divisor. Applying the procedure successively to each \(\mathfrak{B}_i\), always forming the complement with the already reduced \(\mathfrak{B}\), gives a reduced representation.¹¹

In order to obtain such a representation successively, it remains to show that from the individual reduced representations \[\mathfrak{M}=[\mathfrak{B}_1,\mathfrak{C}_1],\qquad \mathfrak{C}_1=[\mathfrak{B}_2,\mathfrak{C}_2],\quad\ldots,\quad \mathfrak{C}_{k-1}=[\mathfrak{B}_k,\mathfrak{C}_k]\] it follows that the resulting representation \(\mathfrak{M}=[\mathfrak{B}_1\cdots\mathfrak{B}_k,\mathfrak{C}_k]\) is reduced. For this it suffices to show that from reduced representations \[\mathfrak{M}=[\mathfrak{B},\mathfrak{C}],\qquad \mathfrak{C}=[\mathfrak{C}_1,\mathfrak{C}_2]\] one obtains a reduced representation \(\mathfrak{M}=[\mathfrak{B},\mathfrak{C}_1,\mathfrak{C}_2]\). In fact, by hypothesis no \(\mathfrak{B}\) lies in its complement; if this happened for a \(\mathfrak{C}_i\), then, contrary to the hypothesis in the first representation, \(\mathfrak{C}\) would be replaceable by a proper divisor, since by the second reduced representation \(\mathfrak{C}_1\) and \(\mathfrak{C}_2\) are proper divisors of \(\mathfrak{C}\); so the representation is shortest. Moreover, by hypothesis no \(\mathfrak{B}\) can be replaced by a proper divisor; if this were possible for a \(\mathfrak{C}_i\), it would, contrary to the hypothesis, correspond to replacing \(\mathfrak{C}\) by a proper divisor, since the representation for \(\mathfrak{C}\) is reduced. Thus the lemma is proved.

Definition II. An ideal \(\mathfrak{M}\) is called reducible if it can be represented as the least common multiple of two proper divisors; in the contrary case \(\mathfrak{M}\) is called irreducible.

We now prove, by means of Theorem I on the finite chain and using reduced representations:

Theorem II. Every ideal is representable as the least common multiple of finitely many irreducible ideals.¹²

For an arbitrary ideal \(\mathfrak{M}\) is either irreducible; then \(\mathfrak{M}=[\mathfrak{M}]\) is a representation of the kind required by Theorem II; or else \(\mathfrak{M}=[\mathfrak{B}_1,\mathfrak{C}_1]\), where \(\mathfrak{B}_1\) and \(\mathfrak{C}_1\) are proper divisors of \(\mathfrak{M}\), and the representation may, by Lemma I, be assumed reduced. For \(\mathfrak{C}_1\) the same alternative holds: either it is irreducible, or there is a reduced representation \(\mathfrak{C}_1=[\mathfrak{B}_2,\mathfrak{C}_2]\).

Continuing in this way, one obtains the series of reduced representations \[\begin{equation} \mathfrak{M}=[\mathfrak{B}_1,\mathfrak{C}_1]=[\mathfrak{B}_1,\mathfrak{B}_2,\mathfrak{C}_2]=\cdots=[\mathfrak{B}_1,\ldots,\mathfrak{B}_n,\mathfrak{C}_n]=\cdots . \tag{1} \end{equation}\] In the chain \(\mathfrak{C}_1,\mathfrak{C}_2,\ldots,\mathfrak{C}_n,\ldots\) each \(\mathfrak{C}_n\) is a proper divisor of the one immediately preceding, and therefore the chain terminates after finitely many steps; there is an index \(n\) such that \(\mathfrak{C}_n\) is irreducible. By Lemma I, moreover, the representation \(\mathfrak{M}=[\mathfrak{A}_n,\mathfrak{C}_n]\) is reduced; \(\mathfrak{C}_n\) therefore cannot lie in its complement \(\mathfrak{A}_n\), and in the representation \(\mathfrak{M}=[\mathfrak{A}_n,\mathfrak{C}_n]\) the ideal \(\mathfrak{A}_n\) cannot be replaced by a proper divisor. Replacing, if necessary, \(\mathfrak{A}_n\) by a proper divisor,¹³ so that the representation becomes reduced, shows that every reducible ideal admits a reduced representation as the least common multiple of an irreducible ideal and a complementary ideal. In the series (1), therefore, all \(\mathfrak{B}_i\) may be assumed irreducible without loss of generality; repeating the preceding argument gives the existence of an irreducible \(\mathfrak{C}_n\), proving Theorem II.

To prove equality of number, the reducibility or irreducibility of an ideal must first be expressed in terms of properties of its complement, as follows.

Theorem III.¹⁴ Let the shortest representation \(\mathfrak{M}=[\mathfrak{A},\mathfrak{C}]\) be reduced with respect to \(\mathfrak{C}\). Then the necessary and sufficient condition for \(\mathfrak{C}\) to be reducible is the existence of two ideals \(\mathfrak{N}_1\) and \(\mathfrak{N}_2\) that are proper divisors of \(\mathfrak{M}\), such that

It follows further: if conditions (2) are satisfied and \(\mathfrak{C}\) is irreducible, then at least one \(\mathfrak{N}_i\) is not a proper divisor of \(\mathfrak{M}\); \(\mathfrak{N}_i=\mathfrak{M}\).

Let \(\mathfrak{C}=[\mathfrak{C}_1,\mathfrak{C}_2]\), where \(\mathfrak{C}_1,\mathfrak{C}_2\) are proper divisors of \(\mathfrak{C}\). Then \[\mathfrak{M}=[\mathfrak{A},\mathfrak{C}]=[\mathfrak{A},\mathfrak{C}_1,\mathfrak{C}_2] =[[\mathfrak{A},\mathfrak{C}_1],[\mathfrak{A},\mathfrak{C}_2]].\] Here the ideals \([\mathfrak{A},\mathfrak{C}_i]\) are proper divisors of \(\mathfrak{M}\), since otherwise \([\mathfrak{A},\mathfrak{C}]\) would not be reduced with respect to \(\mathfrak{C}\). Since divisibility by \(\mathfrak{A}\) is also satisfied, condition (2) has been proved necessary. (The representation (2) is not reduced, since one \([\mathfrak{A},\mathfrak{C}_i]\) can be replaced by \(\mathfrak{C}_i\).)

Conversely, suppose (2) is satisfied. We form the ideals \[\mathfrak{C}_1=(\mathfrak{C},\mathfrak{N}_1),\qquad \mathfrak{C}_2=(\mathfrak{C},\mathfrak{N}_2),\qquad \mathfrak{C}^*=[\mathfrak{C}_1,\mathfrak{C}_2].\] Then \(\mathfrak{C}\) is divisible both by \(\mathfrak{C}_1\) and by \(\mathfrak{C}_2\), and hence also by their least common multiple \(\mathfrak{C}^*\). To show that \(\mathfrak{C}^*\) is divisible by \(\mathfrak{C}\), let \[f\equiv0(\mathfrak{C}^*);\quad f\equiv0(\mathfrak{C}_1);\quad f\equiv0(\mathfrak{C}_2), \quad\text{or equivalently}\quad f=c+n_1=\bar c+n_2,\] where \(c,\bar c\) are elements of \(\mathfrak{C}\) and \(n_1,n_2\) are elements of \(\mathfrak{N}_1,\mathfrak{N}_2\), respectively; in particular, \(n_i\) is divisible by \(\mathfrak{N}_i\). Thus the difference \[g=c-\bar c=n_2-n_1\] is divisible both by \(\mathfrak{C}\) and by \(\mathfrak{A}\), hence by \(\mathfrak{M}\). Since \(n_1=n_2+m\), moreover, \(n_1\) (and similarly \(n_2\)) is divisible both by \(\mathfrak{N}_1\) and by \(\mathfrak{N}_2\), hence by \(\mathfrak{M}\). Therefore \[f=c+m,\qquad f\equiv0(\mathfrak{C}),\qquad \mathfrak{C}^*=\mathfrak{C}.\] The ideals \(\mathfrak{C}_1\) and \(\mathfrak{C}_2\) are proper divisors of \(\mathfrak{C}\); for if \(\mathfrak{C}_1=(\mathfrak{C},\mathfrak{N}_1)=\mathfrak{C}\), then \(\mathfrak{N}_1\) would be divisible by \(\mathfrak{C}\) and hence, because it is also divisible by \(\mathfrak{A}\), would equal \(\mathfrak{M}\), contrary to the hypothesis. Thus \(\mathfrak{C}=[\mathfrak{C}_1,\mathfrak{C}_2]\) has been shown to be reducible; Theorem III is proved.

It may be noted that almost the same argument also proves the following.

Lemma II. If, in a shortest representation \(\mathfrak{M}=[\mathfrak{A},\mathfrak{C}]\), the ideal \(\mathfrak{C}\) can be replaced by a proper divisor, then \(\mathfrak{C}\) is reducible.

Indeed, suppose \[\mathfrak{M}=[\mathfrak{A},\mathfrak{C}]=[\mathfrak{A},\mathfrak{C}_1],\] and put \[\mathfrak{C}^*=[\mathfrak{C}_1,(\mathfrak{A},\mathfrak{C})].\] Again \(\mathfrak{C}\) is divisible by \(\mathfrak{C}^*\). From \(f\equiv0(\mathfrak{C}^*)\) it follows that \[f=c_1=a+c.\] The difference \(a=c_1-c\) is therefore divisible both by \(\mathfrak{A}\) and by \(\mathfrak{C}_1\), hence by \(\mathfrak{M}\); thus \(c_1=c+m\), \(f\equiv0(\mathfrak{C})\), and \(\mathfrak{C}^*=\mathfrak{C}\). Since, by hypothesis, both \(\mathfrak{C}_1\) and \((\mathfrak{A},\mathfrak{C})\) are proper divisors of \(\mathfrak{C}\), the ideal \(\mathfrak{C}=\mathfrak{C}^*\) has thereby been shown to be reducible.

An irreducible \(\mathfrak{C}\) therefore cannot be replaced by a proper divisor.

Now let two different shortest representations of \(\mathfrak{M}\) as the least common multiple of finitely many irreducible ideals be given: \[\mathfrak{M}=[\mathfrak{B}_1\cdots\mathfrak{B}_k]=[\mathfrak{D}_1\cdots\mathfrak{D}_l].\] By the remark following Lemma II, these representations are at the same time reduced. We first prove:

Lemma III. For every complement \(\mathfrak{A}_i=[\mathfrak{B}_1\cdots\mathfrak{B}_{i-1}\mathfrak{B}_{i+1}\cdots\mathfrak{B}_k]\) there is an ideal \(\mathfrak{D}_j\) such that \(\mathfrak{M}=[\mathfrak{A}_i,\mathfrak{D}_j]\).

Indeed, put \(\mathfrak{M}=[\mathfrak{D}_1,\mathfrak{C}_1]\), \(\mathfrak{C}_1=[\mathfrak{D}_2,\mathfrak{C}_{12}]\), and so on. Then \[\mathfrak{M}=[\mathfrak{A}_i,\mathfrak{M}]=[\mathfrak{A}_i,\mathfrak{D}_1,\mathfrak{C}_1] =[[\mathfrak{A}_i,\mathfrak{D}_1],[\mathfrak{A}_i,\mathfrak{C}_1]].\] Here, for \(\mathfrak{N}_1=[\mathfrak{A}_i,\mathfrak{D}_1]\) and \(\mathfrak{N}_2=[\mathfrak{A}_i,\mathfrak{C}_1]\), the conditions (2) of Theorem III are satisfied, since \(\mathfrak{M}=[\mathfrak{A}_i,\mathfrak{B}_i]\) is reduced with respect to \(\mathfrak{B}_i\) and the representation is a shortest one. Since \(\mathfrak{B}_i\) was assumed irreducible, one \(\mathfrak{N}_i\) must necessarily equal \(\mathfrak{M}\).

If \(\mathfrak{N}_1=\mathfrak{M}\), the lemma is proved. If \(\mathfrak{N}_2=\mathfrak{M}\), then correspondingly \(\mathfrak{M}=[[\mathfrak{A}_i,\mathfrak{D}_2],[\mathfrak{A}_i,\mathfrak{C}_{12}]]\), where the same argument again shows that one component must equal \(\mathfrak{M}\). Continuing in this way, either \(\mathfrak{M}=[\mathfrak{A}_i,\mathfrak{D}_j]\) with \(j<l\), or \(\mathfrak{M}=[\mathfrak{A}_i,\mathfrak{C}_{1\ldots l-1}]\); since \(\mathfrak{C}_{1\ldots l-1}=\mathfrak{D}_l\), the lemma is proved.

It now follows that:

Theorem IV. In two different shortest representations of an ideal as the least common multiple of irreducible ideals, the number of components is the same.

Indeed, the lemma gives, for \(i=1\), \[\mathfrak{M}=[\mathfrak{A}_1,\mathfrak{B}_1]=[\mathfrak{A}_1,\mathfrak{D}_{j_1}] = [\mathfrak{D}_{j_1},\mathfrak{B}_2,\ldots,\mathfrak{B}_k].\] Now consider the two decompositions \[\mathfrak{M}=[\mathfrak{D}_{j_1},\mathfrak{B}_2,\ldots,\mathfrak{B}_k] =[\mathfrak{D}_1,\mathfrak{D}_2,\ldots,\mathfrak{D}_l],\] and repeat the preceding argument with respect to the complement \(\bar{\mathfrak{A}}_2=[\mathfrak{D}_{j_1},\mathfrak{B}_3,\ldots,\mathfrak{B}_k]\) of \(\mathfrak{B}_2\). This gives \[\mathfrak{M}=[\bar{\mathfrak{A}}_2,\mathfrak{B}_2]=[\bar{\mathfrak{A}}_2,\mathfrak{D}_{j_2}] =[\mathfrak{D}_{j_1},\mathfrak{D}_{j_2},\mathfrak{B}_3,\ldots,\mathfrak{B}_k],\] and, by continuing the procedure, \[\mathfrak{M}=[\mathfrak{D}_{j_1},\mathfrak{D}_{j_2},\ldots,\mathfrak{D}_{j_k}].\] Since by hypothesis the representation \(\mathfrak{M}=[\mathfrak{D}_1\cdots\mathfrak{D}_l]\) is a shortest one, so that no \(\mathfrak{D}\) can be omitted, the distinct ideals among the \(\mathfrak{D}_{j_i}\) must exhaust all the \(\mathfrak{D}\); hence \(k\ge l\). Interchanging the \(\mathfrak{B}\) and \(\mathfrak{D}\) throughout the lemma and the subsequent arguments gives correspondingly \(l\ge k\), and therefore \(k=l\), proving equality of number. It follows further that the ideals \(\mathfrak{D}_{j_i}\) are all mutually distinct, since otherwise a shortest representation by the \(\mathfrak{D}_{j_i}\) would have fewer than \(k\) components; the notation may therefore be chosen so that \(\mathfrak{D}_{j_i}=\mathfrak{D}_i\). For the same reason, all the intermediate representations \(\mathfrak{M}=[\mathfrak{D}_{j_1}\cdots\mathfrak{D}_{j_r},\mathfrak{B}_{r+1},\ldots,\mathfrak{B}_k]\) are shortest and hence, by the remark following Lemma II, reduced.

Equality of number yields a converse of Lemma I:

Lemma IV. If, in a reduced representation, the components are combined into groups and their least common multiples are formed, then the resulting representation is reduced. In other words: from a reduced representation \[\mathfrak{M}=[\mathfrak{C}_{11}\cdots\mathfrak{C}_{1\mu_1};\;\ldots;\; \mathfrak{C}_{\sigma1}\cdots\mathfrak{C}_{\sigma\mu_\sigma}]\] it follows that \(\mathfrak{M}=[\mathfrak{N}_1\cdots\mathfrak{N}_\sigma]=[\mathfrak{N}_i,\mathfrak{L}_i]\) is also reduced, if \(\mathfrak{N}_i=[\mathfrak{C}_{i1}\cdots\mathfrak{C}_{i\mu_i}]\).

First, \(\mathfrak{N}_i\) cannot be absorbed into its complement \(\mathfrak{L}_i\), since this is true of none of its divisors \(\mathfrak{C}_{ij}\); the representation is therefore a shortest one. To show that \(\mathfrak{N}_i\) cannot be replaced by a proper divisor, decompose the \(\mathfrak{C}\) into their irreducible ideals \(\mathfrak{B}\),¹⁵ so that the representations, reduced by Lemma I, become \[\mathfrak{M}=[\mathfrak{B}_{11}\cdots\mathfrak{B}_{1\lambda_1};\;\ldots;\; \mathfrak{B}_{\sigma1}\cdots\mathfrak{B}_{\sigma\lambda_\sigma}]; \qquad \mathfrak{N}_i=[\mathfrak{B}_{i1}\cdots\mathfrak{B}_{i\lambda_i}].\] Now let \(\mathfrak{M}=[\mathfrak{N}_i^*,\mathfrak{L}_i]\) be reduced with respect to \(\mathfrak{N}_i^*\), and let \(\mathfrak{N}_i^*\) be a proper divisor of \(\mathfrak{N}_i\). By Lemma II, \[\mathfrak{N}_i=[\mathfrak{N}_i^*,(\mathfrak{N}_i,\mathfrak{L}_i)],\] and this representation is necessarily reduced with respect to \(\mathfrak{N}_i^*\), since otherwise \(\mathfrak{N}_i^*\) could also be replaced in \(\mathfrak{M}\) by a proper divisor. If necessary, also replace \((\mathfrak{N}_i,\mathfrak{L}_i)\) by a proper divisor so that a reduced representation of \(\mathfrak{N}_i\) results. If the two components of \(\mathfrak{N}_i\) are now decomposed into irreducible ideals, the number \(\lambda_i\) of irreducible ideals of \(\mathfrak{N}_i\) is the sum of the numbers belonging to the components; the number of irreducible ideals corresponding to the proper divisor \(\mathfrak{N}_i^*\) is therefore necessarily smaller than \(\lambda_i\). But then the decomposition of \(\mathfrak{M}=[\mathfrak{N}_i^*,\mathfrak{L}_i]\) into irreducible ideals would also yield fewer than \(\sum_i\lambda_i\) ideals, contradicting equality of number. As the special case \(\sigma=2\), it follows further that the representation \(\mathfrak{M}=[\mathfrak{N}_i,\mathfrak{L}_i]\) is also reduced with respect to the complement \(\mathfrak{L}_i\).

What follows concerns the connection between primary and irreducible ideals.

Definition III. An ideal \(\mathfrak{Q}\) is called primary if \(a\cdot b\equiv0(\mathfrak{Q})\) and \(a\not\equiv0(\mathfrak{Q})\) necessarily imply \(b^\varkappa\equiv0(\mathfrak{Q})\), where the exponent \(\varkappa\) is finite.

The definition may also be stated as follows: if a product \(a\cdot b\) is divisible by \(\mathfrak{Q}\), then either one factor is divisible by \(\mathfrak{Q}\), or a power of each factor is. In particular, if \(\varkappa\) is always equal to 1, the ideal is called a prime ideal.

By virtue of the existence of bases, the definition of a primary (respectively prime) ideal yields the following formulation, which involves only products of ideals¹⁶:

Definition IIIa. An ideal \(\mathfrak{Q}\) is called primary if \(\mathfrak{A}\cdot\mathfrak{B}\equiv0(\mathfrak{Q})\) and \(\mathfrak{A}\not\equiv0(\mathfrak{Q})\) necessarily imply \(\mathfrak{B}^\lambda\equiv0(\mathfrak{Q})\). If \(\lambda\) is always equal to 1, the ideal is called a prime ideal. Thus, for a prime ideal \(\mathfrak{P}\), \(\mathfrak{A}\cdot\mathfrak{B}\equiv0(\mathfrak{P})\) and \(\mathfrak{A}\not\equiv0(\mathfrak{P})\) always imply \(\mathfrak{B}\equiv0(\mathfrak{P})\).

Indeed, since Definition III is contained in IIIa as the special case \(\mathfrak{A}=(a)\), \(\mathfrak{B}=(b)\), every ideal primary according to IIIa is also primary according to III. Conversely, let \(\mathfrak{Q}\) be primary according to III and suppose the hypothesis of IIIa is satisfied: \(\mathfrak{A}\cdot\mathfrak{B}\equiv0(\mathfrak{Q})\). Then either \(\mathfrak{A}\equiv0(\mathfrak{Q})\), or there is at least one element \(a\equiv0(\mathfrak{A})\) such that \(a\cdot\mathfrak{B}\equiv0(\mathfrak{Q})\) and \(a\not\equiv0(\mathfrak{Q})\). If \(b_1,\ldots,b_r\) is an ideal basis of \(\mathfrak{B}\), then Definition III, since \(a\cdot b_i\equiv0(\mathfrak{Q})\), gives \[b_1^{\varkappa_1}\equiv0(\mathfrak{Q});\quad\ldots;\quad b_r^{\varkappa_r}\equiv0(\mathfrak{Q}).\] Since \[b=f_1b_1+\cdots+f_r b_r+n_1b_1+\cdots+n_r b_r,\] the product of any \(\lambda=\varkappa_1+\cdots+\varkappa_r\) elements \(b\) is therefore divisible by \(\mathfrak{Q}\), proving that ideals primary according to III satisfy Definition IIIa. In the special case of prime ideals \(\mathfrak{P}\), however, \(a\cdot\mathfrak{B}\equiv0(\mathfrak{P})\), and hence \(a\cdot b\equiv0(\mathfrak{P})\) for every \(b\equiv0(\mathfrak{B})\), together with \(a\not\equiv0(\mathfrak{P})\), imply \(b\equiv0(\mathfrak{P})\) and thus \(\mathfrak{B}\equiv0(\mathfrak{P})\). This proves the equivalence of the two definitions.

The connection between primary and prime ideals is established by observing that the totality \(\mathfrak{P}\) of all elements \(p\) having the property that some power of \(p\) is divisible by \(\mathfrak{Q}\) forms a prime ideal. First, it is clear that \(\mathfrak{P}\) is an ideal, since, along with \(p_1\) and \(p_2\), the elements \(ap_1\) and \(p_1-p_2\) also have the stated property. By the same basis argument used in Definition IIIa, there is furthermore a number \(\lambda\) such that \(\mathfrak{P}^\lambda\equiv0(\mathfrak{Q})\).

Now let \[ab\equiv0(\mathfrak{P}),\qquad a\not\equiv0(\mathfrak{P}).\] Then, by the definition of \(\mathfrak{P}\), \[a^\lambda b^\lambda\equiv0(\mathfrak{Q}),\qquad a^\lambda\not\equiv0(\mathfrak{Q}),\] and therefore, by the definition of \(\mathfrak{Q}\), \[b^{\lambda\varkappa}\equiv0(\mathfrak{Q}),\qquad\text{and consequently}\qquad b\equiv0(\mathfrak{P}),\] which proves that \(\mathfrak{P}\) is a prime ideal. The ideal \(\mathfrak{P}\) is also characterized as the greatest common divisor of all ideals \(\mathfrak{B}\) having the property that some power of \(\mathfrak{B}\) is divisible by \(\mathfrak{Q}\). For every such \(\mathfrak{B}\) is divisible by \(\mathfrak{P}\), and hence so is their greatest common divisor \(\mathfrak{D}\). Conversely, \(\mathfrak{P}\) itself is one of these ideals \(\mathfrak{B}\), and hence is divisible by \(\mathfrak{D}\), proving \(\mathfrak{P}=\mathfrak{D}\). Thus \(\mathfrak{P}\) is a prime ideal that is a divisor of \(\mathfrak{Q}\) and some power of which is divisible by \(\mathfrak{Q}\); this property determines it uniquely. For from \[\mathfrak{Q}\equiv0(\mathfrak{P}),\qquad \mathfrak{P}^\lambda\equiv0(\mathfrak{Q}), \qquad \mathfrak{Q}\equiv0(\bar{\mathfrak{P}}),\qquad \bar{\mathfrak{P}}^{\mu}\equiv0(\mathfrak{Q})\] it follows that \[\mathfrak{P}^\lambda\equiv0(\bar{\mathfrak{P}}),\qquad \bar{\mathfrak{P}}^{\mu}\equiv0(\mathfrak{P}),\] and hence, by the property of prime ideals, \[\mathfrak{P}\equiv0(\bar{\mathfrak{P}}),\qquad \bar{\mathfrak{P}}\equiv0(\mathfrak{P}),\qquad \mathfrak{P}=\bar{\mathfrak{P}}.\] In summary, we have:

Theorem V. For every primary ideal \(\mathfrak{Q}\) there exists one and only one prime ideal \(\mathfrak{P}\) that is a divisor of \(\mathfrak{Q}\) and some power of which is divisible by \(\mathfrak{Q}\); \(\mathfrak{P}\) will be called the “associated prime ideal”.¹⁷ \(\mathfrak{P}\) is characterized as the greatest common divisor of all ideals \(\mathfrak{B}\) having the property that some power of \(\mathfrak{B}\) is divisible by \(\mathfrak{Q}\). If \(\varrho\) is the least number such that \(\mathfrak{P}^\varrho\equiv0(\mathfrak{Q})\), then \(\varrho\) will be called the exponent of \(\mathfrak{Q}\).¹⁸

We now prove, as the connection between primary and irreducible ideals:

Theorem VI. Every nonprimary ideal is reducible; in other words, every irreducible ideal is primary.¹⁹

Let \(\mathfrak{K}\) be a nonprimary ideal, so that by Definition III there is at least one pair of elements \(a,b\) such that

We now form the two ideals \[\mathfrak{L}_0=(\mathfrak{K},a);\qquad \mathfrak{N}_0=(\mathfrak{K},b),\] which by (3) are proper divisors of \(\mathfrak{K}\) and for which, again by (3),

For the elements \(f\) of the least common multiple \(\mathfrak{K}_0=[\mathfrak{L}_0,\mathfrak{N}_0]\), the following alternative holds.

Either from \[f\equiv0(\mathfrak{L}_0);\qquad f\equiv0(\mathfrak{N}_0), \qquad\text{i.e.}\qquad f\equiv a_1\cdot b\;(\mathfrak{K})\] there always follows a representation \[f\equiv l_0\cdot b\;(\mathfrak{K});\qquad l_0\equiv0(\mathfrak{L}_0);\] then by (4), \(f\equiv0(\mathfrak{K})\), hence \(\mathfrak{K}_0\equiv0(\mathfrak{K})\), and, since also \(\mathfrak{K}\equiv0(\mathfrak{K}_0)\), \(\mathfrak{K}=\mathfrak{K}_0\), which shows that \(\mathfrak{K}\) is reducible.

Or there is at least one \(f\equiv0(\mathfrak{K}_0)\) for which no such \(l_0\) exists. With the \(a_1\) belonging to this \(f\), we then form \[\mathfrak{L}_1=(\mathfrak{L}_0,a_1)=(\mathfrak{K},a,a_1);\qquad \mathfrak{N}_1=(\mathfrak{K},b^2).\] Then, since \(a_1\cdot b\equiv0(\mathfrak{L}_0)\), (4) also gives

and \(\mathfrak{L}_1\) is a proper divisor of \(\mathfrak{L}_0\).

The same alternative holds for the elements \(f\) of \(\mathfrak{K}_1=[\mathfrak{L}_1,\mathfrak{N}_1]\).

Either from \[f\equiv0(\mathfrak{L}_1);\qquad f\equiv0(\mathfrak{N}_1);\] \[\text{i.e.}\qquad f\equiv a_2\cdot b^2\;(\mathfrak{K}) \quad\text{there always follows}\quad f\equiv l_1\cdot b^2;\qquad l_1\equiv0(\mathfrak{L}_1);\] and therefore by (4’), \(\mathfrak{K}=\mathfrak{K}_1\).

Or for at least one \(f\) there is no such \(l_1\), which leads to the formation of \(\mathfrak{L}_2=(\mathfrak{L}_1,a_2)\) and \(\mathfrak{N}_2=(\mathfrak{K},b^{2^2})\), with \(\mathfrak{L}_2\cdot\mathfrak{N}_2\equiv0(\mathfrak{K})\), where \(\mathfrak{L}_2\) is a proper divisor of \(\mathfrak{L}_1\). Continuing in this way, we define in general \[\mathfrak{L}_0=(\mathfrak{K},a);\quad \mathfrak{L}_1=(\mathfrak{L}_0,a_1);\quad\ldots;\quad \mathfrak{L}_\nu=(\mathfrak{L}_{\nu-1},a_\nu);\quad\ldots,\] \[\mathfrak{N}_0=(\mathfrak{K},b);\quad \mathfrak{N}_1=(\mathfrak{K},b^2);\quad\ldots;\quad \mathfrak{N}_\nu=(\mathfrak{K},b^{2^\nu});\quad\ldots,\] where the \(a_i\) are defined by the existence of an \(f\) such that \[f\equiv0(\mathfrak{L}_{i-1});\qquad f\equiv0(\mathfrak{N}_{i-1}),\] \[\text{i.e.}\qquad f\equiv a_i\cdot b^{2^{i-1}}\;(\mathfrak{K}); \qquad\text{but}\qquad a_i\not\equiv0(\mathfrak{L}_{i-1}).\] It then follows in general that \(\mathfrak{L}_i\cdot\mathfrak{N}_i\equiv0(\mathfrak{K})\), that \(\mathfrak{N}_i\) is a proper divisor of \(\mathfrak{K}\) by (3), and that \(\mathfrak{L}_i\) is a proper divisor of \(\mathfrak{L}_{i-1}\). By Theorem I on the finite chain, the chain of the \(\mathfrak{L}_i\) must therefore terminate after finitely many steps, say with \(\mathfrak{L}_n\). Thus, for every \(f\equiv0(\mathfrak{L}_n)\) and \(f\equiv0(\mathfrak{N}_n)\), one has \(f\equiv l_n\cdot b^{2^n}\;(\mathfrak{K})\) with \(l_n\equiv0(\mathfrak{L}_n)\), and consequently the preceding argument gives \(\mathfrak{K}=[\mathfrak{L}_n,\mathfrak{N}_n]\), which shows that \(\mathfrak{K}\) is reducible.

The uniqueness of the associated prime ideals follows from what has just been proved, as follows.

Let \[\mathfrak{M}=[\mathfrak{B}_1\cdots\mathfrak{B}_k]=[\mathfrak{D}_1\cdots\mathfrak{D}_k]\] be two shortest, hence reduced, representations of \(\mathfrak{M}\) as the least common multiple of irreducible ideals, whose numbers of components agree by Theorem IV. By that theorem the intermediate representations occurring there (where, as noted there, the index \(j_i=i\) may be put) \[\mathfrak{M}=[\mathfrak{D}_1\cdots\mathfrak{D}_{i-1}\mathfrak{B}_i\mathfrak{B}_{i+1}\cdots\mathfrak{B}_k] =[\mathfrak{D}_1\cdots\mathfrak{D}_{i-1}\mathfrak{D}_i\mathfrak{B}_{i+1}\cdots\mathfrak{B}_k] =[\bar{\mathfrak{U}}_i,\mathfrak{B}_i]=[\bar{\mathfrak{U}}_i,\mathfrak{D}_i]\] are also shortest representations. Hence \[\bar{\mathfrak{U}}_i\cdot\mathfrak{B}_i\equiv0(\mathfrak{D}_i),\qquad \bar{\mathfrak{U}}_i\not\equiv0(\mathfrak{D}_i), \qquad \bar{\mathfrak{U}}_i\cdot\mathfrak{D}_i\equiv0(\mathfrak{B}_i),\qquad \bar{\mathfrak{U}}_i\not\equiv0(\mathfrak{B}_i).\] Since by Theorem VI the irreducible ideals \(\mathfrak{B}_i\) and \(\mathfrak{D}_i\) are primary, it follows that there are two numbers \(\lambda_i\) and \(\mu_i\) such that

Let \(\mathfrak{P}_i\) and \(\bar{\mathfrak{P}}_i\) denote the associated prime ideals of \(\mathfrak{B}_i\) and \(\mathfrak{D}_i\), respectively; thus \(\mathfrak{P}_i^{\varrho_i}\equiv0(\mathfrak{B}_i)\) and \(\bar{\mathfrak{P}}_i^{\sigma_i}\equiv0(\mathfrak{D}_i)\). Then (5) gives \[\mathfrak{P}_i^{\lambda_i\varrho_i}\equiv0(\bar{\mathfrak{P}}_i), \qquad \bar{\mathfrak{P}}_i^{\mu_i\sigma_i}\equiv0(\mathfrak{P}_i),\] and hence, by the property of prime ideals, \[\mathfrak{P}_i\equiv0(\bar{\mathfrak{P}}_i), \qquad \bar{\mathfrak{P}}_i\equiv0(\mathfrak{P}_i), \qquad \mathfrak{P}_i=\bar{\mathfrak{P}}_i.\] This proves:

Theorem VII. In two different shortest representations of an ideal as the least common multiple of irreducible ideals, the associated prime ideals agree; equal prime ideals may occur among them,²⁰ and occur equally often in every decomposition. Consequently, the ideals themselves can be paired in at least one way so that, in each pair, a power of one ideal \(\mathfrak{B}_i\) is divisible by the assigned ideal \(\mathfrak{D}_i\), and conversely. Their numbers agree by Theorem IV.²¹

Definition IV. A shortest representation \(\mathfrak{M}=[\mathfrak{Q}_1\cdots\mathfrak{Q}_\alpha]\) will be called a least common multiple of greatest primary ideals if all \(\mathfrak{Q}\) are primary, but the least common multiple of two \(\mathfrak{Q}\) is no longer primary.

That at least one such representation always exists follows from the representation of \(\mathfrak{M}\) as a least common multiple of irreducible ideals. These ideals are primary; either there is already a representation by greatest primary ideals, or the least common multiple of some two ideals is again primary. Since the number of ideals has thereby decreased by one, repetition of the procedure leads after finitely many steps to the desired representation.

This representation is reduced by Lemma IV. Conversely, every reduced representation by greatest primary ideals arises in this way, as the decomposition of the \(\mathfrak{Q}\) into irreducible ideals shows.

In order to derive here from Theorem VII a corresponding uniqueness theorem, the connection with the associated prime ideals has to be investigated.

Theorem VIII. If the primary ideals \(\mathfrak{N}_1,\mathfrak{N}_2,\ldots,\mathfrak{N}_\lambda\) all possess the same associated prime ideal \(\mathfrak{P}\), then their least common multiple \(\mathfrak{Q}=[\mathfrak{N}_1,\mathfrak{N}_2,\ldots,\mathfrak{N}_\lambda]\) is also primary and has \(\mathfrak{P}\) as associated prime ideal. Conversely, if \(\mathfrak{Q}=[\mathfrak{N}_1\cdots\mathfrak{N}_\lambda]\) is a reduced representation of the primary ideal \(\mathfrak{Q}\), then all \(\mathfrak{N}_i\) are primary and have, as associated prime ideal, the associated prime ideal \(\mathfrak{P}\) of \(\mathfrak{Q}\).

For the proof of the first part, first observe that from \(\mathfrak{P}^{\varrho_i}\equiv0(\mathfrak{N}_i)\) for every \(i\) it also follows that \(\mathfrak{P}^\tau\equiv0(\mathfrak{Q})\), where \(\tau\) denotes the largest of the indices \(\varrho_i\). Since \(\mathfrak{P}\) is also a divisor of \(\mathfrak{Q}\), \(\mathfrak{P}\) is necessarily the associated prime ideal if \(\mathfrak{Q}\) is primary. From \[\mathfrak{A}\mathfrak{B}\equiv0(\mathfrak{Q}), \qquad \mathfrak{B}^k\not\equiv0(\mathfrak{Q})\quad\text{(for every }k)\] it follows that \[\mathfrak{B}\not\equiv0(\mathfrak{P}), \quad\text{hence}\quad \mathfrak{B}^k\not\equiv0(\mathfrak{N}_i)\quad\text{(for every }k),\] and consequently \(\mathfrak{A}\equiv0(\mathfrak{N}_i)\); hence \(\mathfrak{A}\equiv0(\mathfrak{Q})\). Thus \(\mathfrak{Q}\) is proved primary, and \(\mathfrak{P}\) the associated prime ideal.

Conversely, let first \[\mathfrak{Q}=[\mathfrak{N}_1\cdots\mathfrak{N}_\lambda]=[\mathfrak{N}_i,\mathfrak{L}_i]\] be a shortest representation of \(\mathfrak{Q}\) by primary ideals \(\mathfrak{N}_i\), and let \(\mathfrak{P}_i\) denote the corresponding associated prime ideals. From \[\mathfrak{L}_i\cdot\mathfrak{N}_i\equiv0(\mathfrak{Q}),\qquad \mathfrak{L}_i\not\equiv0(\mathfrak{Q})\] (because the representation is shortest) it follows that \[\mathfrak{N}_i^{\sigma_i}\equiv0(\mathfrak{Q}),\qquad \text{or}\qquad \mathfrak{P}_i^{\varrho_i\sigma_i}\equiv0(\mathfrak{Q}).\] Since \(\mathfrak{P}_i\) is at the same time a divisor of \(\mathfrak{Q}\), \(\mathfrak{P}_i\) therefore agrees, for every \(i\), with the associated prime ideal \(\mathfrak{P}\) of \(\mathfrak{Q}\).

It remains to show that in every reduced representation \(\mathfrak{Q}=[\mathfrak{N}_1\cdots\mathfrak{N}_\lambda]\) the \(\mathfrak{N}_i\) are primary.²² To this end, decompose the \(\mathfrak{N}_i\) into their irreducible ideals \(\mathfrak{B}\); in the resulting shortest representation \(\mathfrak{Q}=[\mathfrak{B}_1\cdots\mathfrak{B}_s]\), every ideal is primary and, by what has just been proved, has \(\mathfrak{P}\) as associated prime ideal. Then by the first part of the theorem proved above the same holds for each \(\mathfrak{N}_i\), completing the proof of Theorem VIII.

Addendum. It also follows that a prime ideal is necessarily irreducible. For from the reduced representation \(\mathfrak{P}=[\mathfrak{N},\mathfrak{K}]\) one obtains by Theorem VIII \(\mathfrak{N}\equiv0(\mathfrak{P})\); hence, since \(\mathfrak{P}\equiv0(\mathfrak{N})\), also \(\mathfrak{P}=\mathfrak{N}\), and similarly \(\mathfrak{P}=\mathfrak{K}\). The irreducibility of \(\mathfrak{P}\) also follows directly: for from \(\mathfrak{P}=[\mathfrak{N},\mathfrak{K}]\) follows \(\mathfrak{N}\mathfrak{K}\equiv0(\mathfrak{P})\), \(\mathfrak{N}\not\equiv0(\mathfrak{P})\), \(\mathfrak{K}\not\equiv0(\mathfrak{P})\), contrary to the defining property of a prime ideal.

Now let \[\mathfrak{M}=[\mathfrak{Q}_1\cdots\mathfrak{Q}_\alpha]=[\bar{\mathfrak{Q}}_1\cdots\bar{\mathfrak{Q}}_\beta]\] be two reduced representations of \(\mathfrak{M}\) as the least common multiple of greatest primary ideals. Decomposing the \(\mathfrak{Q}\) into irreducible ideals \(\mathfrak{B}\), and the \(\bar{\mathfrak{Q}}\) correspondingly into irreducible ideals \(\mathfrak{D}\), gives two reduced representations of \(\mathfrak{M}\) as the least common multiple of irreducible ideals; by Theorem VII, both the number of components and the associated prime ideals coincide. By Theorem VIII, all irreducible ideals \(\mathfrak{B}\) occurring for a fixed \(\mathfrak{Q}_i\) have the same associated prime ideal \(\mathfrak{P}_i\); whereas the \(\mathfrak{P}_j\) associated to \(\mathfrak{Q}_j\) is necessarily different from this, since otherwise, by Theorem VIII, there would be no representation by greatest primary ideals. Hence the number \(\alpha\) of the \(\mathfrak{Q}\) is equal to the number of different associated prime ideals \(\mathfrak{P}\) of the \(\mathfrak{B}\); these different \(\mathfrak{P}\) form the associated prime ideals of the \(\mathfrak{Q}\). The same holds for the \(\bar{\mathfrak{Q}}\) with respect to their decomposition into the \(\mathfrak{D}\). Theorem VII therefore gives equality of number for the \(\mathfrak{Q}\) and \(\bar{\mathfrak{Q}}\), and agreement of their associated prime ideals. At the same time it is seen that the grouping of the irreducible ideals into greatest primary ideals described at the beginning of the paragraph consists in combining all, and only those, with the same associated prime ideal. Theorem VIII further shows the irreducibility property of the greatest primary ideals: they admit no reduced representation as the least common multiple of greatest primary ideals.

In summary:

Theorem IX. In two reduced representations of an ideal as the least common multiple of greatest primary ideals, the number of components and the associated prime ideals, all of which are distinct from each other, coincide. In other words, to each \(\mathfrak{Q}\) there can be assigned one and only one \(\bar{\mathfrak{Q}}\) such that a power of \(\mathfrak{Q}\) is divisible by \(\bar{\mathfrak{Q}}\), and conversely.²³ The \(\mathfrak{Q}\) and \(\bar{\mathfrak{Q}}\) have the irreducibility property with respect to decomposition into greatest primary ideals.

Addendum. It should be noted that Theorem IX remains essentially valid if one assumes only a shortest representation instead of a reduced one. If, say, \[\mathfrak{M}=[\mathfrak{Q}_1\cdots\mathfrak{Q}_i^*\cdots\mathfrak{Q}_\alpha]\] is reduced with respect to \(\mathfrak{Q}_i^*\), and \(\mathfrak{Q}_i^*\) is a proper divisor of \(\mathfrak{Q}_i\), while \(\mathfrak{L}_i\) is the complement of \(\mathfrak{Q}_i\), then by Lemma IV \[\mathfrak{Q}_i=[\mathfrak{Q}_i^*;(\mathfrak{L}_i,\mathfrak{Q}_i)],\] and this representation is reduced with respect to \(\mathfrak{Q}_i^*\). By Theorem VIII – replacing \((\mathfrak{L}_i,\mathfrak{Q}_i)\) by a proper divisor if necessary when applying the theorem – \(\mathfrak{Q}_i^*\) is therefore primary and has the same associated prime ideal \(\mathfrak{P}_i\) as \(\mathfrak{Q}_i\). Continuing the procedure shows that to every such representation there can be assigned a reduced representation by greatest primary ideals in such a way that the number of components and the associated prime ideals coincide. Thus even for shortest representations it holds that in two different representations the number of components and the associated prime ideals coincide.

The associated, mutually distinct prime ideals thereby uniquely defined will be called simply “the associated prime ideals of \(\mathfrak{M}\).”

Definition V. An ideal \(\mathfrak{R}\) is called relatively prime to \(\mathfrak{S}\) if \(\mathfrak{T}\cdot\mathfrak{R}\equiv0(\mathfrak{S})\) necessarily implies \(\mathfrak{T}\equiv0(\mathfrak{S})\). If both \(\mathfrak{R}\) is relatively prime to \(\mathfrak{S}\) and \(\mathfrak{S}\) is relatively prime to \(\mathfrak{R}\), then \(\mathfrak{R}\) and \(\mathfrak{S}\) are called mutually relatively prime.²⁴

An ideal is called relatively-prime irreducible if it cannot be represented as the least common multiple of mutually relatively prime proper divisors.

In particular, if instead of \(\mathfrak{T}\) one takes the greatest common divisor \(\mathfrak{T}_0\) of all \(\mathfrak{T}\) for which \(\mathfrak{T}\mathfrak{R}\equiv0(\mathfrak{S})\), then also \(\mathfrak{T}_0\mathfrak{R}\equiv0(\mathfrak{S})\) and \(\mathfrak{S}\equiv0(\mathfrak{T}_0)\). Thus \(\mathfrak{T}_0=\mathfrak{S}\) when \(\mathfrak{R}\) is relatively prime to \(\mathfrak{S}\); and \(\mathfrak{T}_0\) is a proper divisor of \(\mathfrak{S}\) when \(\mathfrak{R}\) is not relatively prime to \(\mathfrak{S}\).²⁵

The uniqueness proof rests on:

Theorem X. 1. If \(\mathfrak{R}\) is relatively prime to the ideals \(\mathfrak{S}_1, \ldots,\mathfrak{S}_\lambda\), then \(\mathfrak{R}\) is also relatively prime to their least common multiple \(\mathfrak{S}\).

2. If the ideals \(\mathfrak{S}_1, \ldots,\mathfrak{S}_\lambda\) are relatively prime to \(\mathfrak{R}\), then their least common multiple \(\mathfrak{S}\) is also relatively prime to \(\mathfrak{R}\).

3. If \(\mathfrak{R}\) is relatively prime to \(\mathfrak{S}\) and \(\mathfrak{S}=[\mathfrak{S}_1\cdots\mathfrak{S}_\lambda]\) is a reduced representation for \(\mathfrak{S}\), then \(\mathfrak{R}\) is also relatively prime to each \(\mathfrak{S}_i\).

4. If \(\mathfrak{S}\) is relatively prime to \(\mathfrak{R}\), then every divisor \(\mathfrak{S}_i\) of \(\mathfrak{S}\) is also relatively prime to \(\mathfrak{R}\).

1. Since \(\mathfrak{T}\mathfrak{R}\equiv0(\mathfrak{S})\) necessarily gives \(\mathfrak{T}\mathfrak{R}\equiv0(\mathfrak{S}_i)\), the hypothesis gives \(\mathfrak{T}\equiv0(\mathfrak{S}_i)\), and hence \(\mathfrak{T}\equiv0(\mathfrak{S})\).

2. Put \[\mathfrak{C}_1=[\mathfrak{S}_2\cdots\mathfrak{S}_\lambda],\quad \mathfrak{C}_{12}=[\mathfrak{S}_3\cdots\mathfrak{S}_\lambda],\quad \ldots,\quad \mathfrak{C}_{12\ldots\lambda-1}=\mathfrak{S}_\lambda.\] From \(\mathfrak{T}\cdot\mathfrak{S}\equiv0(\mathfrak{R})\) it follows that \(\mathfrak{T}\cdot\mathfrak{S}_1\cdot\mathfrak{C}_1\equiv0(\mathfrak{R})\); hence, by the hypothesis, \(\mathfrak{T}\cdot\mathfrak{C}_1\equiv0(\mathfrak{R})\). It follows in turn that \(\mathfrak{T}\cdot\mathfrak{S}_2\cdot\mathfrak{C}_{12}\equiv0(\mathfrak{R})\), hence \(\mathfrak{T}\cdot\mathfrak{C}_{12}\equiv0(\mathfrak{R})\), and finally \(\mathfrak{T}\cdot\mathfrak{C}_{12\ldots\lambda-1}=\mathfrak{T}\cdot\mathfrak{S}_\lambda\equiv0(\mathfrak{R})\); hence \(\mathfrak{T}\equiv0(\mathfrak{R})\).

3. Let \(\mathfrak{C}_i\) denote the complement of \(\mathfrak{S}_i\). From \(\mathfrak{T}_0\cdot\mathfrak{R}\equiv0(\mathfrak{S}_i)\), together with \(\mathfrak{C}_i\cdot\mathfrak{R}\equiv0(\mathfrak{S}_i)\), it follows that \([\mathfrak{T}_0,\mathfrak{C}_i]\cdot\mathfrak{R}\equiv0(\mathfrak{S})\). If \(\mathfrak{T}_0\) were a proper divisor of \(\mathfrak{S}_i\), then, since \(\mathfrak{S}=[\mathfrak{S}_i,\mathfrak{C}_i]\) is reduced, \([\mathfrak{T}_0,\mathfrak{C}_i]\) would be a proper divisor of \(\mathfrak{S}\), contrary to the hypothesis.

4. From \(\mathfrak{T}\mathfrak{S}_i\equiv0(\mathfrak{R})\), where \(\mathfrak{T}\) is a proper divisor of \(\mathfrak{R}\), it follows also that \(\mathfrak{T}\mathfrak{S}\equiv0(\mathfrak{R})\); hence \(\mathfrak{T}\) would be a proper divisor of \(\mathfrak{R}\), contrary to the hypothesis.

Definition V shows in particular that every ideal is relatively prime to the unit ideal \(\mathfrak{D}\) consisting of all elements of \(\Sigma\);²⁶ the following theorems, however, apply only to ideals distinct from \(\mathfrak{D}\).

From Theorem X one first obtains:

Lemma V. Every representation of an ideal as the least common multiple of mutually relatively prime ideals distinct from \(\mathfrak{D}\) is reduced.

Let \[\mathfrak{M}=[\mathfrak{R}_1\mathfrak{R}_2\cdots\mathfrak{R}_\sigma]=[\mathfrak{R}_i, \mathfrak{L}_i]\] be such a representation. By Theorem X, 2, \(\mathfrak{L}_i\) is also relatively prime to \(\mathfrak{R}_i\); hence \(\mathfrak{R}_i\) cannot be contained in \(\mathfrak{L}_i\), and the representation is shortest. For from \(\mathfrak{L}_i\equiv0(\mathfrak{R}_i)\), with \(\mathfrak{L}_i\) relatively prime to \(\mathfrak{R}_i\), it would follow from \(\mathfrak{D}\mathfrak{L}_i\equiv0(\mathfrak{R}_i)\) that \(\mathfrak{D}\equiv0(\mathfrak{R}_i)\), hence \(\mathfrak{R}_i=\mathfrak{D}\), contrary to the hypothesis.

If now \(\mathfrak{R}_i\) could be replaced by a proper divisor \(\mathfrak{R}_i^*\), then by Lemma II \(\mathfrak{R}_i\) would be reducible and \[\mathfrak{R}_i=[\mathfrak{R}_i^*,(\mathfrak{R}_i, \mathfrak{L}_i)].\] If necessary, replace \((\mathfrak{R}_i, \mathfrak{L}_i)\) by a proper divisor \((\mathfrak{R}_i, \mathfrak{L}_i)^*\) and likewise replace \(\mathfrak{R}_i^*\) by a proper divisor so that a reduced representation of \(\mathfrak{R}_i\) is obtained. Then Theorem X, 3 shows that \(\mathfrak{L}_i\) is also relatively prime to \((\mathfrak{R}_i, \mathfrak{L}_i)^*\), and this ideal is distinct from \(\mathfrak{D}\) because the original representation was shortest. But \((\mathfrak{R}_i, \mathfrak{L}_i)^*\) is contained in \((\mathfrak{R}_i, \mathfrak{L}_i)\) and \((\mathfrak{R}_i, \mathfrak{L}_i)\) is contained in \(\mathfrak{L}_i\), giving the contradiction.

Theorem X further gives the following theorem, which mediates the connection with the associated prime ideals.

Theorem XI. If \(\mathfrak{R}\) is relatively prime to \(\mathfrak{S}\) and \(\mathfrak{S}\) is distinct from \(\mathfrak{D}\), then no associated prime ideal of \(\mathfrak{R}\)²⁷ is divisible by an associated prime ideal of \(\mathfrak{S}\). Conversely, if no such divisibility holds, then \(\mathfrak{R}\) is relatively prime to \(\mathfrak{S}\), and of course \(\mathfrak{S}\) is distinct from \(\mathfrak{D}\).

Let \[\mathfrak{R}=[\mathfrak{Q}_1\cdots\mathfrak{Q}_\alpha], \qquad \mathfrak{S}=[\mathfrak{Q}_1^*\cdots\mathfrak{Q}_\beta^*]\] be reduced representations of \(\mathfrak{R}\) and \(\mathfrak{S}\) by greatest primary ideals, and let \(\mathfrak{P}_1, \ldots,\mathfrak{P}_\alpha\) and \(\mathfrak{P}_1^*, \ldots,\mathfrak{P}_\beta^*\) be their associated prime ideals. We prove the assertion in the form: if some \(\mathfrak{P}\) is divisible by some \(\mathfrak{P}^*\), then \(\mathfrak{R}\) cannot be relatively prime to \(\mathfrak{S}\), and conversely.

Suppose then that \(\mathfrak{P}_\mu\equiv0(\mathfrak{P}_\nu^*)\), and consequently also \(\mathfrak{Q}_\mu\equiv0(\mathfrak{P}_\nu^*)\). By the definition of \(\mathfrak{P}_\nu^*\) this gives \(\mathfrak{Q}_\mu^{\sigma_\nu}\equiv0(\mathfrak{Q}_\nu^*)\), hence also \(\mathfrak{R}^{\sigma_\nu}\equiv0(\mathfrak{Q}_\nu^*)\). Let \(\mathfrak{R}^\tau\) be the lowest power of \(\mathfrak{R}\) divisible by \(\mathfrak{Q}_\nu^*\). If \(\tau=1\), then \(\mathfrak{R}\) is divisible by \(\mathfrak{Q}_\nu^*\); but since \(\mathfrak{S}\ne\mathfrak{D}\), \(\mathfrak{R}\) is not relatively prime to \(\mathfrak{Q}_\nu^*\). If \(\tau\ge2\), then \(\mathfrak{R}^{\tau-1}\mathfrak{R}\equiv0(\mathfrak{Q}_\nu^*)\) while \(\mathfrak{R}^{\tau-1}\not\equiv0(\mathfrak{Q}_\nu^*)\), so again \(\mathfrak{R}\) is not relatively prime to \(\mathfrak{Q}_\nu^*\);²⁸ in both cases Theorem X, 3 then implies that \(\mathfrak{R}\) is not relatively prime to \(\mathfrak{S}\).

Conversely, if \(\mathfrak{R}\) is not relatively prime to \(\mathfrak{S}\), then by Theorem X, 1 it is not relatively prime to at least one \(\mathfrak{Q}_\nu^*\). Thus \[\mathfrak{T}_0\mathfrak{R}\equiv0(\mathfrak{Q}_\nu^*), \qquad \mathfrak{T}_0\not\equiv0(\mathfrak{Q}_\nu^*),\] and since \(\mathfrak{Q}_\nu^*\) is primary, \[\mathfrak{R}^\tau\equiv0(\mathfrak{Q}_\nu^*),\qquad \text{hence also}\quad \mathfrak{Q}_1^\tau\cdots\mathfrak{Q}_\alpha^\tau\equiv0(\mathfrak{Q}_\nu^*).\] For the associated prime ideals this implies \[\mathfrak{P}_1^{\tau\varrho_1}\cdots\mathfrak{P}_\alpha^{\tau\varrho_\alpha}\equiv0(\mathfrak{P}_\nu^*),\] and therefore, by the property of prime ideals, \(\mathfrak{P}_\nu^*\) is contained in at least one of the \(\mathfrak{P}\). This proves Theorem XI.

Theorems X and XI now give the existence²⁹ and uniqueness of decomposition into relatively-prime irreducible ideals as follows.

Let \(\mathfrak{M}=[\mathfrak{Q}_1\cdots\mathfrak{Q}_\alpha]\) be a reduced (or at least shortest) representation of \(\mathfrak{M}\) by greatest primary ideals, and let \(\mathfrak{P}_1, \ldots,\mathfrak{P}_\alpha\) be the associated prime ideals. We collect the \(\mathfrak{P}\) into groups so that no ideal of one group is divisible by an ideal of a different group, while a single group cannot be split into two subgroups both having this property.

To construct such a grouping, note that by definition an individual group \(G\) must contain, together with each ideal \(\mathfrak{P}\), all its divisors and multiples occurring among \(\mathfrak{P}_1,\ldots,\mathfrak{P}_\alpha\) (that is, all ideals divisible by \(\mathfrak{P}\)). Let \(\mathfrak{P}^{(i_1)}\) denote all multiples of \(\mathfrak{P}\), \(\mathfrak{P}_{j_1}^{(i_1)}\) all divisors of \(\mathfrak{P}^{(i_1)}\), \(\mathfrak{P}_{j_1}^{(i_1i_2)}\) all multiples of \(\mathfrak{P}_{j_1}^{(i_1)}\), and so on; in general, let \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\) denote all multiples of \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_{\lambda-1})}\), and \(\mathfrak{P}_{j_1\ldots j_\lambda}^{(i_1\ldots i_\lambda)}\) all divisors of \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\). Since only finitely many ideals \(\mathfrak{P}\) are involved altogether, the procedure must terminate after finitely many steps, that is, must cease to produce an ideal different from all preceding ones. The totality of ideals \(\mathfrak{P}\) so obtained does indeed form a group \(G\) with the required properties.

For by definition \(G\) contains, together with each \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_{\lambda-1})}\), all multiples \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\), and, together with each \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\), all divisors \(\mathfrak{P}_{j_1\ldots j_\lambda}^{(i_1\ldots i_\lambda)}\). These include all divisors of \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_{\lambda-1})}\), while the multiples of \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\) are themselves again among the \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\). Thus the ideals not contained in \(G\) can be neither divisors nor multiples of those contained in \(G\).

The group \(G\) also satisfies the irreducibility condition. For if a division into two subgroups \(G^{(1)}\) and \(G^{(2)}\) were possible, and \(G^{(1)}\) contained, say, \(\mathfrak{P}_{j_1\ldots j_\lambda}^{(i_1\ldots i_\lambda)}\) (respectively \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\)), it would also contain all preceding ideals, since these are alternately multiples and divisors (respectively divisors and multiples), hence also \(\mathfrak{P}\) and therefore the entire group \(G\). Applying the same procedure to the ideals not contained in \(G\) yields a grouping \(G_1,\ldots,G_\sigma\) of all the \(\mathfrak{P}\) with the required properties. Such a grouping is unique: for if \(G'_1,\ldots,G'_\tau\) were a second grouping and \(\mathfrak{P}_{j_1\ldots j_\lambda}^{(i_1\ldots i_\lambda)}\) (respectively \(\mathfrak{P}_{j_1\ldots j_{\lambda-1}}^{(i_1\ldots i_\lambda)}\)) were an element of \(G'_i\), then by the preceding argument \(G'_i\) would contain the entire group \(G\) and hence, by the irreducibility property, would be identical with \(G\).

Let \(\mathfrak{Q}_{i\mu}\), where \(\mu\) runs from \(1\) to \(\lambda_i\), denote the ideals \(\mathfrak{Q}\) whose associated prime ideals are collected in a group \(G_i\). Since the \(\mathfrak{P}\) are all distinct, the primary ideals \(\mathfrak{Q}_{i\mu}\) belonging to a shortest representation are uniquely determined. Put \[\mathfrak{R}_i=[\mathfrak{Q}_{i1}\cdots\mathfrak{Q}_{i\lambda_i}],\qquad \text{then}\quad \mathfrak{M}=[\mathfrak{R}_1\cdots\mathfrak{R}_\sigma].\] Theorem XI remains applicable even if \(\mathfrak{R}_i\) is only a shortest representation, since by the addendum to Theorem IX its associated prime ideals are already well-defined. Because no associated prime ideal of \(\mathfrak{R}_i\) is divisible by one of \(\mathfrak{R}_j\), and conversely, Theorem XI shows that \(\mathfrak{R}_i\) and \(\mathfrak{R}_j\) are mutually relatively prime; and each \(\mathfrak{R}_i\) is relatively-prime irreducible by the irreducibility property of the group \(G_i\). By Lemma V the representation is reduced. Conversely, every decomposition into relatively-prime irreducible ideals leads, by Theorem XI, to the grouping just described.

Now let \[\mathfrak{M}=[\bar{\mathfrak{R}}_1\cdots\bar{\mathfrak{R}}_\tau]\] be a second representation of \(\mathfrak{M}\) by relatively-prime irreducible ideals. It is reduced by Lemma V. Decomposing the \(\mathfrak{R}\) into greatest primary ideals shows that the associated prime ideals agree, and hence the groupings of these prime ideals agree. Thus \(\tau=\sigma\), and the notation may be chosen so that \(\mathfrak{R}_i\) and \(\bar{\mathfrak{R}}_i\) belong to the same group. If \[\mathfrak{M}=[\mathfrak{R}_i, \mathfrak{L}_i]=[\bar{\mathfrak{R}}_i, \bar{\mathfrak{L}}_i]\] is the representation by ideal and complement, then, since \(\bar{\mathfrak{R}}_i\) is assigned to the same group as \(\mathfrak{R}_i\), Theorem XI gives that \(\mathfrak{L}_i\) is relatively prime to \(\bar{\mathfrak{R}}_i\), and \(\bar{\mathfrak{L}}_i\) is relatively prime to \(\mathfrak{R}_i\). From \[\mathfrak{R}_i\mathfrak{L}_i\equiv0(\bar{\mathfrak{R}}_i), \qquad \bar{\mathfrak{R}}_i\bar{\mathfrak{L}}_i\equiv0(\mathfrak{R}_i)\] therefore follows \[\mathfrak{R}_i\equiv0(\bar{\mathfrak{R}}_i), \qquad \bar{\mathfrak{R}}_i\equiv0(\mathfrak{R}_i), \qquad \mathfrak{R}_i=\bar{\mathfrak{R}}_i.\] Thus:

Theorem XII. Every ideal can be represented uniquely as the least common multiple of finitely many mutually relatively prime and relatively-prime irreducible ideals.

Definition VI. If the shortest representation \(\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}]\) is reduced with respect to \(\mathfrak{L}\), then \(\mathfrak{R}\) is called an isolated ideal if no associated prime ideal of \(\mathfrak{R}\) is contained in an associated prime ideal of \(\mathfrak{L}\); equivalently, if \(\mathfrak{L}\) is relatively prime to \(\mathfrak{R}\).

Then the representation \(\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}]\) satisfies the conditions of the representation \(\mathfrak{M}=[\mathfrak{R}_i, \mathfrak{L}_i]\) in Lemma V, and Lemma V proves:

Lemma VI. If \(\mathfrak{R}\) is an isolated ideal of the shortest representation \(\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}]\), reduced with respect to \(\mathfrak{L}\), then the representation is also reduced with respect to \(\mathfrak{R}\).

Thus isolated ideals occur only in reduced representations. The associated prime ideals arising in the decompositions of \(\mathfrak{R}\) and \(\mathfrak{L}\) into irreducible ideals complement one another to the uniquely determined associated prime ideals arising in the corresponding decomposition of \(\mathfrak{M}\). Hence no associated prime ideal belonging to the decomposition of \(\mathfrak{R}\) into irreducible ideals is contained in the remaining associated prime ideals of \(\mathfrak{M}\). Conversely, if this condition is satisfied and \(\mathfrak{R}\) occurs in at least one representation \(\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}]\) reduced with respect to \(\mathfrak{R}\), and hence also in a reduced representation \(\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}^*]\), then \(\mathfrak{R}\) is isolated by Definition VI. This gives the following definition, independent of the particular complement \(\mathfrak{L}\).

Definition VIa. \(\mathfrak{R}\) is called an isolated ideal if the prime ideals belonging to its decomposition into irreducible ideals are not contained in the other associated prime ideals belonging to the corresponding decomposition of \(\mathfrak{M}\), and if \(\mathfrak{R}\) occurs in at least one representation \(\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}]\) reduced with respect to \(\mathfrak{R}\).³⁰

Suppose \[\mathfrak{M}=[\mathfrak{R}, \mathfrak{L}]=[\bar{\mathfrak{R}}, \bar{\mathfrak{L}}]\] are two representations of \(\mathfrak{M}\) by isolated ideals \(\mathfrak{R}\) and \(\bar{\mathfrak{R}}\), with complements \(\mathfrak{L}\) and \(\bar{\mathfrak{L}}\), such that the associated prime ideals of \(\mathfrak{R}\) and \(\bar{\mathfrak{R}}\) agree. Replace \(\mathfrak{L}\) and \(\bar{\mathfrak{L}}\) by divisors \(\mathfrak{L}^*\) and \(\bar{\mathfrak{L}}^{*}\) so that the representations become reduced. Then the associated prime ideals of \(\mathfrak{L}^*\) and \(\bar{\mathfrak{L}}^{*}\) also agree; by Theorem XI, \(\mathfrak{L}^*\) is relatively prime to \(\bar{\mathfrak{R}}\) and \(\bar{\mathfrak{L}}^{*}\) is relatively prime to \(\mathfrak{R}\). Since \[\mathfrak{R}\mathfrak{L}^*\equiv0(\bar{\mathfrak{R}}), \qquad \bar{\mathfrak{R}}\bar{\mathfrak{L}}^{*}\equiv0(\mathfrak{R}),\] we obtain \[\mathfrak{R}\equiv0(\bar{\mathfrak{R}}), \qquad \bar{\mathfrak{R}}\equiv0(\mathfrak{R}), \qquad \mathfrak{R}=\bar{\mathfrak{R}}.\] Thus isolated ideals are uniquely determined by their associated prime ideals. This sharpens Theorems VII and IX on decompositions into irreducible, respectively greatest primary, ideals; by the addendum to Theorem IX it is enough to assume a shortest representation. In summary:

Theorem XIII. In every shortest representation of an ideal as the least common multiple of irreducible, respectively greatest primary, ideals, the isolated irreducible, respectively greatest primary, ideals are uniquely determined; the ambiguity concerns only the non-isolated irreducible, respectively greatest primary, ideals.³¹ In general, isolated ideals are uniquely determined by their associated prime ideals.

If in such a shortest representation by irreducible, respectively greatest primary, ideals the ideals \(\mathfrak{B}_i\), respectively \(\mathfrak{Q}_j\), are non-isolated, then by definition their complements \(\mathfrak{U}_i\), respectively \(\mathfrak{L}_j\), are divisible by \(\mathfrak{P}_i\), respectively \(\mathfrak{P}_j\). Hence \[\mathfrak{U}_i^{\varrho_i}\equiv0(\mathfrak{B}_i) \qquad \text{respectively} \qquad \mathfrak{L}_j^{\sigma_j}\equiv0(\mathfrak{Q}_j).\] Conversely, if these relations hold, then \(\mathfrak{P}_i\), respectively \(\mathfrak{P}_j\), is contained in at least one associated prime ideal of the complement, and hence, by the addendum to Theorem IX, in an associated prime ideal of the divisor \(\mathfrak{L}_j^*\) of \(\mathfrak{L}_j\) that supplies a representation reduced with respect to \(\mathfrak{L}_j^*\). Thus \(\mathfrak{B}_i\), respectively \(\mathfrak{Q}_j\), is non-isolated. In particular, irreducible ideals \(\mathfrak{B}_i\) whose associated prime ideal occurs more than once in the decomposition of \(\mathfrak{M}\) are always non-isolated. Non-isolated primary ideals are therefore also characterized by the fact that a power of every complement is divisible by them; isolated primary ideals by the fact that this cannot occur.

If the underlying ring domain \(\Sigma\) has a unit, that is, an element \(\varepsilon\) such that \(\varepsilon a=a\) for every element of \(\Sigma\),³² then coprime ideals can be defined as follows.

Definition VIII. Two ideals \(\mathfrak{R}\) and \(\mathfrak{S}\) are called coprime if their greatest common divisor is the unit ideal \(\mathfrak{D}=(\varepsilon)\) consisting of all elements of \(\Sigma\). An ideal is called coprime-irreducible if it cannot be represented as the least common multiple of pairwise coprime ideals.

Note that two coprime ideals are always mutually relatively prime. By definition there are elements \(r\equiv0(\mathfrak{R})\) and \(s\equiv0(\mathfrak{S})\) with \(\varepsilon=r+s\). From \(\mathfrak{T}\mathfrak{R}\equiv0(\mathfrak{S})\) it follows that \(\mathfrak{T} r\equiv0(\mathfrak{S})\), hence \(\mathfrak{T}\varepsilon=\mathfrak{T}\equiv0(\mathfrak{S})\); correspondingly, from \(\bar{\mathfrak{T}}\mathfrak{S}\equiv0(\mathfrak{R})\) it follows that \(\bar{\mathfrak{T}}\equiv0(\mathfrak{R})\).³³ Thus Lemma V implies that every representation by pairwise coprime ideals is also reduced.

The uniqueness proof rests on the analogue of Theorem X:

Theorem XIV. If \(\mathfrak{R}\) is coprime to each of the ideals \(\mathfrak{S}_1, \ldots,\mathfrak{S}_\lambda\), then \(\mathfrak{R}\) is also coprime to \(\mathfrak{S}=[\mathfrak{S}_1\cdots\mathfrak{S}_\lambda]\). Conversely, if \(\mathfrak{R}\) and \(\mathfrak{S}\) are coprime, then \(\mathfrak{R}\) is coprime to each \(\mathfrak{S}_j\). If \(\mathfrak{R}=[\mathfrak{R}_1\cdots\mathfrak{R}_\mu]\) and every \(\mathfrak{R}_i\) is coprime to every \(\mathfrak{S}_j\), then \(\mathfrak{R}\) and \(\mathfrak{S}\) are coprime; here too the converse holds.

Indeed, if \(\mathfrak{R}\) is coprime to every \(\mathfrak{S}_j\), then there are elements \(s_j\) such that \[s_j\equiv0(\mathfrak{S}_j), \qquad s_j\equiv\varepsilon(\mathfrak{R}).\] Consequently \[s_1s_2\cdots s_\lambda\equiv0(\mathfrak{S}), \qquad s_1s_2\cdots s_\lambda\equiv\varepsilon(\mathfrak{R}), \qquad (\mathfrak{R}, \mathfrak{S})=(\varepsilon).\] Since \((\mathfrak{R}, \mathfrak{S})\) is divisible by each \((\mathfrak{R}, \mathfrak{S}_j)\), the converse follows. Repeated application of the same argument gives the second part. For if \(\mathfrak{R}_i\) is coprime to \(\mathfrak{S}_1, \ldots,\mathfrak{S}_\lambda\) for a fixed \(i\), then \(\mathfrak{R}_i\) is coprime to \(\mathfrak{S}\); if this holds for every \(i\), then, since coprimeness is mutual, \(\mathfrak{S}\) is coprime to \(\mathfrak{R}\). Conversely, coprimeness of \(\mathfrak{R}\) and \(\mathfrak{S}\) gives coprimeness of \(\mathfrak{S}\) with \(\mathfrak{R}_i\), and hence of \(\mathfrak{R}_i\) with \(\mathfrak{S}\).

The proof of existence and uniqueness³⁴ of the decomposition into coprime-irreducible ideals is, as in the corresponding proof for relatively-prime ideals, a matter of a unique grouping. Since, however, the relatively-prime irreducible ideals \(\mathfrak{R}_1, \ldots,\mathfrak{R}_\sigma\) of \(\mathfrak{M}\) are uniquely defined by Theorem XII, it is unnecessary here to go back to the associated prime ideals.

We collect the uniquely defined relatively-prime irreducible ideals \(\mathfrak{R}_1, \ldots,\mathfrak{R}_\sigma\) of \(\mathfrak{M}\) into groups so that every ideal of one group is coprime to every ideal of any different group, while a single group cannot be split into two subgroups such that every ideal of one subgroup is coprime to every ideal of the other subgroup. Such a grouping is obtained as follows. By definition a group \(G\) must contain, together with every ideal \(\mathfrak{R}\), all ideals not coprime to \(\mathfrak{R}\). Let these be \(\mathfrak{R}_{i_1}\); let \(\mathfrak{R}_{i_1i_2}\) be the ideals not coprime to these; in general let \(\mathfrak{R}_{i_1\ldots i_\lambda}\) be not coprime to \(\mathfrak{R}_{i_1\ldots i_{\lambda-1}}\). Since only finitely many ideals occur, this process terminates after finitely many steps. The set of ideals so obtained forms a group \(G\) with the desired properties: all ideals outside \(G\) are, by construction, coprime to all ideals in \(G\). If \(G\) were split into two subgroups \(G^{(1)}\) and \(G^{(2)}\), and if \(\mathfrak{R}_{i_1\ldots i_\lambda}\) belonged to \(G^{(1)}\), then, because coprimeness is mutual, \(G^{(1)}\) would also have to contain \(\mathfrak{R}_{i_1\ldots i_{\lambda-1}}, \ldots,\mathfrak{R}_{i_1}\), hence also \(\mathfrak{R}\) and therefore the whole group \(G\). This proves irreducibility. Repeating the construction with the remaining ideals gives a grouping \(G_1, \ldots,G_\tau\) of all \(\mathfrak{R}\).

The grouping is unique: if \(G'_1, \ldots,G'_\tau\) is a second grouping and \(\mathfrak{R}_{i_1\ldots i_\lambda}\) is an element of \(G'_i\), then \(G'_i\) contains \(\mathfrak{R}\) and hence \(G_1\), and, by the irreducibility condition, cannot contain elements different from \(G_1\); therefore \(G'_i=G_1\).

Let \(\mathfrak{T}_i\) be the least common multiple of the ideals \(\mathfrak{R}\) united in a group \(G_i\). We show that \[\mathfrak{M}=[\mathfrak{T}_1\cdots\mathfrak{T}_\tau]\] is a representation of \(\mathfrak{M}\) by coprime-irreducible ideals. Decomposing the \(\mathfrak{T}\) into the \(\mathfrak{R}\) first shows that \([\mathfrak{T}_1\cdots\mathfrak{T}_\tau]\) really represents \(\mathfrak{M}\). By Theorem XIV the \(\mathfrak{T}\) are pairwise coprime, and each \(\mathfrak{T}\) is coprime-irreducible. This proves the existence of such a representation.

For uniqueness, let \(\mathfrak{M}=[\bar{\mathfrak{T}}_1\cdots\bar{\mathfrak{T}}_\tau]\) be a second representation of this type. Decomposing the \(\bar{\mathfrak{T}}\) into relatively-prime irreducible ideals \(\mathfrak{R}\), Theorem XIV shows that ideals \(\mathfrak{R}\) occurring in different \(\bar{\mathfrak{T}}\) are coprime to one another, hence also mutually relatively prime; therefore they agree with the uniquely defined relatively-prime irreducible ideals \(\mathfrak{R}\) of \(\mathfrak{M}\). The ideals \(\bar{\mathfrak{T}}_i\) also induce, by Theorem XIV, a grouping \(G'_i\) of the \(\mathfrak{R}\) with the stated properties. Since this grouping is unique and each \(\bar{\mathfrak{T}}_i\) is uniquely determined by the group \(G'_i=G_i\), one has \(\bar{\mathfrak{T}}_i=\mathfrak{T}_i\); uniqueness is proved.

For pairwise coprime ideals, the least common multiple is moreover equal to the product. For by Theorem XIV, if \(\mathfrak{M}=[\mathfrak{T}_1\cdots\mathfrak{T}_\tau]\), the complement \(\mathfrak{L}_i\) is also coprime to \(\mathfrak{T}_i\); hence there are elements \[t_i\equiv0(\mathfrak{T}_i), \qquad l_i\equiv0(\mathfrak{L}_i), \qquad \varepsilon=t_i+l_i.\] From \(f\equiv0(\mathfrak{T}_i)\) and \(f\equiv0(\mathfrak{L}_i)\) it follows that \[f=f\varepsilon=ft_i+fl_i, \qquad f\equiv0(\mathfrak{L}_i\mathfrak{T}_i).\] Conversely \(\mathfrak{L}_i\mathfrak{T}_i\) is divisible by \([\mathfrak{L}_i, \mathfrak{T}_i]\), whence \([\mathfrak{L}_i, \mathfrak{T}_i]=\mathfrak{L}_i\mathfrak{T}_i\); continuing the procedure on \(\mathfrak{L}_i\) gives \[\mathfrak{M}=[\mathfrak{T}_1\cdots\mathfrak{T}_\tau] =\mathfrak{T}_1\mathfrak{T}_2\cdots\mathfrak{T}_\tau.\] Therefore:

Theorem XV. Every ideal can be represented uniquely as a product of finitely many pairwise coprime and coprime-irreducible ideals.

We now show that the content of the first three paragraphs, which concerns irreducible ideals and not primary and prime ideals, remains valid under weaker hypotheses. Those paragraphs do not use the commutative law of multiplication and refer only to the property of ideals of being modules. They therefore remain valid for modules with respect to non-commutative domains, which are now to be defined. The definition of modules is to be based on a double domain \((\Sigma,T)\) with the following properties.

\(\Sigma\) is an abstractly defined non-commutative ring; that is, \(\Sigma\) is a system of elements \(a,b,c,\ldots\) for which two modes of combination are defined, ring addition \((\#)\) and ring multiplication \((\times)\), satisfying the laws stated in § 1, with the exception of the commutative law 4 for ring multiplication.

\(T\) is a system of elements \(\alpha,\beta,\gamma,\ldots\), for which, in connection with \(\Sigma\), two combinations are likewise defined: addition, which from any two elements \(\alpha,\beta\) produces a unique third element \(\alpha+\beta\); and multiplication of an element \(\alpha\) of \(T\) by an element \(c\) of \(\Sigma\), which uniquely produces an element \(c\alpha\) of \(T\).³⁵

For these combinations the following laws hold:

1. The associative law of addition: \((\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)\).

2. The commutative law of addition: \(\alpha+\beta=\beta+\alpha\).

3. The law of unrestricted and unique subtraction: there is in \(T\) one and only one element \(\xi\) satisfying the equation \(\alpha+\xi=\beta\) (one writes \(\xi=\beta-\alpha\)).

4. The associative law of multiplication: \(a\cdot(b\cdot\gamma)=(a\times b)\cdot\gamma\).

5. The distributive law: \[(a\# b)\cdot\gamma=a\cdot\gamma+b\cdot\gamma,\qquad c\cdot(\alpha+\beta)=c\cdot\alpha+c\cdot\beta.\]

From these conditions there follow, as is well known, the existence of the zero element and the validity of the distributive law also for the combination of subtraction and multiplication: \[(a-b)\cdot\gamma=a\cdot\gamma-b\cdot\gamma,\qquad c\cdot(\alpha-\beta)=c\cdot\alpha-c\cdot\beta,\] where \((-)\) denotes subtraction in \(\Sigma\). If \(\Sigma\) contains a unit \(\varepsilon\), then \(\varepsilon\cdot\alpha=\alpha\) is to hold for every element \(\alpha\) of \(T\).

By a module \(M\) in \((\Sigma,T)\) we mean a system of elements of \(T\) satisfying the two conditions:

1. Along with \(\alpha\), \(M\) contains \(c\cdot\alpha\), where \(c\) is any element of \(\Sigma\).

2. Along with \(\alpha\) and \(\beta\), \(M\) contains the difference \(\alpha-\beta\); hence also \(n\alpha\) for every integer \(n\).³⁶

By this definition, \(T\) itself forms a module in \((\Sigma,T)\). If in particular the domain \(T\) and the combinations established there coincide with the domain \(\Sigma\) and the combinations valid there, then the module \(M\) becomes a right ideal \(M\) in \(\Sigma\). If \(\Sigma\) is also assumed commutative, the ordinary concept of ideal results, and is thus a special case of the concept of module.³⁷

For modules all definitions of § 1 remain valid. Thus \(\alpha\equiv 0\,(M)\), respectively \(N\equiv 0\,(M)\), means that \(\alpha\), respectively every element of \(N\), is an element of \(M\); in other words, \(\alpha\), respectively \(N\), is divisible by \(M\). \(M\) is a proper divisor of \(N\) if \(M\) contains elements different from those of \(N\); from \(N\equiv 0\,(M)\) and \(M\equiv 0\,(N)\) follows \(M=N\). The definitions of greatest common divisor and least common multiple likewise remain literally the same. If, in particular, the module \(M\) contains a finite number of elements \(\alpha_1,\ldots,\alpha_e\) such that \(M=(\alpha_1\ldots\alpha_e)\), that is, \[\alpha=c_1\alpha_1+\cdots+c_e\alpha_e+n_1\alpha_1+\cdots+n_e\alpha_e\] for every \(\alpha\equiv 0\,(M)\), where the \(c_i\) are elements of \(\Sigma\) and the \(n_i\) are integers, then \(M\) is called a finite module, and \(\alpha_1,\ldots,\alpha_e\) a module basis.

In what follows we assume, just as in § 1, only those domains \((\Sigma,T)\) that satisfy the finiteness condition: every module in \((\Sigma,T)\) is finite, and hence has a module basis.

For such a domain \((\Sigma,T)\), Theorem I on finite chains holds also for modules, as the proof given there shows, and therefore all hypotheses for §§ 2 and 3 are satisfied. Definition I and Lemma I on shortest and reduced representations carry over directly; likewise Theorem II on the representability of every module as the least common multiple of finitely many irreducible modules, where Lemma II shows that every shortest representation of this kind is at the same time reduced. Theorem III also remains valid, expressing reducibility of a module through properties of its complement; from it follows Lemma III, and finally Theorem IV, which asserts equality of the number of components in two different shortest representations of a module as the least common multiple of irreducible modules. Theorem IV also gives Lemma IV as the converse of Lemma I on reduced representations.

Addendum. The same argument shows that all these theorems and definitions remain valid if, for two-sided domains \(T\), that is, domains which are both right- and left-sided, one everywhere understands by a module a two-sided module: one which, along with \(\alpha\), also contains \(c\alpha\) and \(\alpha c\), and along with \(\alpha\) and \(\beta\) also contains \(\alpha-\beta\).

While all these theorems rest only on the concept of divisibility and of least common multiple, the further uniqueness theorems rest essentially on the product concept and therefore do not admit a direct transfer. Thus the definition of a primary ideal and of a prime ideal cannot be transferred to modules, since the product of two elements of \(T\) is not defined. The formally possible transfer to non-commutative rings loses its meaning, since here the existence of the associated prime ideal cannot be proved³⁸ and the proof that an irreducible ideal is primary also fails. These two facts form the basis of the following uniqueness theorems. By contrast, if the non-commutative ring has a unit, coprime and coprime-irreducible ideals can be defined, and the arguments used in the proof of Theorem II show that every ideal can be represented as the least common multiple of finitely many pairwise coprime and coprime-irreducible ideals.³⁹

Finally, we mention a sufficient criterion for the finiteness condition to hold in \((\Sigma,T)\): if \(\Sigma\) contains a unit and satisfies the finiteness condition, and if \(T\) itself is a finite module in \((\Sigma,T)\), then every module in \((\Sigma,T)\) is finite.

Indeed, from the existence of the unit in \(\Sigma\), for \[\mathfrak{M}=(f_1, \ldots,f_e), \qquad f=\bar b_1f_1+\cdots+\bar b_ef_e+n_1f_1+\cdots+n_ef_e,\] where the \(\bar b_i\) are elements of \(\Sigma\) and the \(n_i\) are integers, there follows also a representation \[f=b_1f_1+\cdots+b_ef_e\] with the \(b_i\) in \(\Sigma\). For since \(f_i=\varepsilon f_i\), one has \(n_if_i=n_i\varepsilon f_i\), and since \(n_i\varepsilon=(\varepsilon+\cdots+\varepsilon)\) belongs to \(\Sigma\), \(b_i=\bar b_i+n_i\varepsilon\) is an element of \(\Sigma\). The hypothesis on \(T\) says that every element \(\alpha\) of \(T\) admits a representation \[\alpha=\bar a_1\tau_1+\cdots+\bar a_k\tau_k+n_1\tau_1+\cdots+n_k\tau_k,\] and hence \[\alpha=a_1\tau_1+\cdots+a_k\tau_k,\] the second representation following from the first just as above because \(\varepsilon\tau_i=\tau_i\).

As \(\alpha\) runs through the elements of a module \(M\) in \((\Sigma,T)\), the coefficients \(a_k\) of \(\tau_k\) run through an ideal \(\mathfrak{M}_k\) of \(\Sigma\); by the preceding, for every \(a_k\equiv 0\,(\mathfrak{M}_k)\) one has \[a_k=b_1a_k^{(1)}+\cdots+b_ea_k^{(e)}.\] Let \(\alpha^{(i)}\) be an element of \(M\) for which the coefficient of \(\tau_k\) is \(a_k^{(i)}\). Then \(\alpha-b_1\alpha^{(1)}-\cdots-b_e\alpha^{(e)}\) is an element belonging to \(M\) which depends only on \(\tau_1, \ldots,\tau_{k-1}\). The procedure can be repeated for the totality of these elements, which no longer contain \(\tau_k\) and which form a module \(M'\), and the assertion is proved by finitely many repetitions.

1. Let the underlying ring domain \(\Sigma\) consist of all polynomials in \(x_1, \ldots,x_n\) with arbitrary complex coefficients. By Hilbert’s basis theorem (Ann. 36) the finiteness condition holds for it. The issue is the relation of our theorems to the known theorems of elimination theory and module theory.

This relation is established by the following special case of a known theorem of Hilbert:⁴⁰

If \(f\) vanishes for every finite system of values of \(x_1, \ldots,x_n\) that is a zero of all polynomials of a prime ideal \(\mathfrak{P}\) (a zero of \(\mathfrak{P}\)), then \(f\) is divisible by \(\mathfrak{P}\). In other words, a prime ideal \(\mathfrak{P}\) consists of the totality of polynomials which vanish at its zeros.⁴¹

Thus, if a product \(fg\) vanishes for all zeros of \(\mathfrak{P}\), at least one factor vanishes; the zeros form an irreducible algebraic variety. Conversely, if this definition of irreducible variety is adopted, then the totality of polynomials vanishing on an irreducible variety forms a prime ideal. Prime ideal and irreducible variety therefore correspond one-to-one. Since \(\mathfrak{Q}\equiv 0\,(\mathfrak{P})\) and \(\mathfrak{P}^e\equiv 0\,(\mathfrak{Q})\), the zeros of a primary ideal coincide with those of its associated prime ideal.⁴² Therefore the representation of an ideal as the least common multiple of greatest primary ideals gives a resolution of all zeros of the ideal into irreducible varieties; and, as Lasker showed, the converse also holds. The proof of uniqueness of the associated prime ideals thus corresponds here to the fundamental theorem of elimination theory on the unique decomposability of an algebraic variety into irreducible varieties, and for special polynomial domains, where no unique product representation of the polynomials by irreducible polynomials of the domain and consequently no elimination theory exists, it can serve as an equivalent of that theorem of elimination theory.

The irreducible varieties corresponding to the isolated primary ideals are exactly those which occur in the “minimal resolvent”;⁴³ for the zeros of every non-isolated greatest primary ideal are at the same time zeros of at least one isolated one, namely one whose associated prime ideal is divisible by that of the non-isolated ideal. The uniqueness of isolated primary ideals therefore gives, in the exponents, new invariant multiplicity numbers. The uniqueness theorems for the resolution of primary ideals into irreducible ones can likewise be regarded as supplements to elimination theory in the sense of multiplicity.

After these remarks the different decompositions can be interpreted in terms of their behavior with respect to algebraic varieties. Pairwise coprime ideals correspond to varieties having no common zero; for mutually relatively prime ideals, no irreducible variety of one ideal is at the same time a common zero; the greatest primary ideals vanish only on irreducible varieties which are all distinct from one another; in the decomposition into irreducible ideals the same irreducible varieties may occur with multiplicity.

It should also be noted that, instead of the general polynomial domain, the domain of all homogeneous forms may also be taken as the basis; for one readily verifies that the general theorems remain valid for the combinations there in force, where addition is defined only for forms of the same dimension.⁴⁴

A simplest example of the four different decompositions - for which the formulas follow below - is given, according to the preceding, by a line and two intersecting lines skew to it, one of the latter containing, with higher multiplicity, a point different from the point of intersection. The decomposition into coprime-irreducible ideals corresponds to the decomposition into the line and the variety skew to it; in the decomposition into relatively-prime irreducible ideals this variety splits into the two lines; the decomposition into greatest primary ideals corresponds to separating off the point of higher multiplicity, while the decomposition into irreducible ideals forces a resolution of that point.

Taking this point as the origin, the running line as the \(y\)-axis, the line intersecting it parallel to the \(x\)-axis, and the line skew to it parallel to the \(z\)-axis, such a configuration is represented, for example, by the following irreducible ideals⁴⁵:

The associated prime ideals are:

The greatest primary ideals are:

The relatively-prime irreducible ideals are:

The coprime-irreducible ideals are:

This gives the total ideal:

\(\displaystyle \mathfrak{M}=[\mathfrak{S}_1,\mathfrak{S}_2]=\mathfrak{S}_1\cdot\mathfrak{S}_2\)

\(\displaystyle =((x-1)(y-1)x^3,(y-1)x^2y,(y-1)xy^2,(x-1)z,y(y-1)x^3,yz),\)

Here the ideals \(\mathfrak{B}_1\), \(\mathfrak{B}_2\), and \(\mathfrak{B}_3\) are isolated, and hence uniquely determined also in the decompositions into irreducible and greatest primary ideals. The ideals \(\mathfrak{B}_4\) and \(\mathfrak{B}_5\), respectively \(\mathfrak{Q}_4\), are non-isolated; they are not uniquely determined, but can, for example, be replaced by

Likewise, \(\mathfrak{Q}_4\) can be replaced by

2. Just as the general, and the integral, polynomial domains satisfy the finiteness condition, so does every finite integral domain of polynomials, by Hilbert’s theorem on module bases;⁴⁶ the coefficients may be assigned to an arbitrary field. We give one more example, for the domain of all even polynomials, the simplest domain in which, because \(x^2y^2=(xy)^2\), no unique product representation of the polynomials by irreducible polynomials of the domain exists. It is the same configuration as in the preceding example, now given by the irreducible ideals \[\begin{gathered} \mathfrak{B}_1=(x^2-1,xy,y^2,yz),\quad \mathfrak{B}_2=(y^2-1,xz,yz,z^2),\\ \mathfrak{B}_3=(x^2,xy,xz,yz,z^2),\quad \mathfrak{B}_4=(x^4,x^3y,y^2,xz,yz,z^2),\\ \mathfrak{B}_5=(x^2,y^2,xz,yz,z^2). \end{gathered}\] The associated prime ideals are \[\mathfrak{P}_1=\mathfrak{B}_1, \quad \mathfrak{P}_2=\mathfrak{B}_2, \quad \mathfrak{P}_3=\mathfrak{B}_3, \quad \mathfrak{P}_4=\mathfrak{P}_5=(x^2,xy,y^2,xz,yz,z^2).\] That \(\mathfrak{P}_1\) is a prime ideal follows from the fact that every polynomial of the domain has the form \[f=\varphi(z^2)+xz\,\psi(z^2)\pmod{\mathfrak{P}_1}.\] Thus if \[f_1=\varphi_1(z^2)+xz\,\psi_1(z^2), \qquad f_2=\varphi_2(z^2)+xz\,\psi_2(z^2),\] then from \(f_1f_2\equiv 0\,(\mathfrak{P}_1)\) one obtains the equations \[\varphi_1\varphi_2+z^2\psi_1\psi_2=0, \qquad \varphi_2\psi_1+\varphi_1\psi_2=0,\] and hence \(f_1\equiv 0\,(\mathfrak{P}_1)\) or \(f_2\equiv 0\,(\mathfrak{P}_1)\). In exactly the same way one sees that \(\mathfrak{P}_2\) is a prime ideal; \(\mathfrak{P}_3\) is a prime ideal because every polynomial of the domain modulo \(\mathfrak{P}_3\) is congruent to a polynomial in \(y^2\); \(\mathfrak{P}_4\) consists of all polynomials of the domain. Thus \(\mathfrak{B}_1\), \(\mathfrak{B}_2\), and \(\mathfrak{B}_3\) are irreducible as prime ideals, while \(\mathfrak{B}_4\) and \(\mathfrak{B}_5\) each have only the single proper divisor \(\mathfrak{P}_4\), and hence are necessarily irreducible.

From the irreducible ideals one obtains the greatest primary ideals \[\begin{gathered} \mathfrak{Q}_1=\mathfrak{B}_1, \quad \mathfrak{Q}_2=\mathfrak{B}_2, \quad \mathfrak{Q}_3=\mathfrak{B}_3, \quad \mathfrak{Q}_4=[\mathfrak{B}_4,\mathfrak{B}_5]=(x^4,x^3y,y^2,xz,yz,z^2), \end{gathered}\] the relatively-prime irreducible ideals \[\mathfrak{R}_1=\mathfrak{Q}_1, \quad \mathfrak{R}_2=\mathfrak{Q}_2, \quad \mathfrak{R}_3=[\mathfrak{Q}_3,\mathfrak{Q}_4]=(x^4,x^3y,x^2y^2,xy^3,xz,yz,z^2),\] and the coprime-irreducible ideals \[\mathfrak{S}_1=\mathfrak{R}_1, \quad \mathfrak{S}_2=[\mathfrak{R}_2,\mathfrak{R}_3] =((y^2-1)x^4,(y^2-1)x^3y,(y^2-1)x^2y^2, (y^2-1)xy^3,xz,yz,z^2).\] As in the first example, \[[\mathfrak{B}_4,\mathfrak{B}_5]=[\mathfrak{D}_4,\mathfrak{D}_5], \qquad [\mathfrak{Q}_3,\mathfrak{Q}_4]=[\mathfrak{Q}_3, \overline{\mathfrak{Q}}_4],\] where \[\mathfrak{D}_4=(x^4,xy+\lambda x^2,\ldots), \qquad \mathfrak{D}_5=(x^2+\mu xy,\ldots), \qquad \lambda\mu\ne1,\] and \[\overline{\mathfrak{Q}}_4=(x^4,x^3y,y^2+\lambda xy,\ldots).\] The remaining irreducible, respectively greatest primary, ideals \(\mathfrak{B}_1\), \(\mathfrak{B}_2\), \(\mathfrak{B}_3\) are uniquely determined as isolated ideals.

1. Let the underlying domain \(\Sigma\) consist of all even rational integers. Thus \(\Sigma\) can be put in one-to-one correspondence with the domain of all rational integers, by assigning to each number \(2a\) in \(\Sigma\) the number \(a\). It follows at once that every ideal in \(\Sigma\) is a principal ideal \((2a)\), where in the basis representation \(2c=n\cdot2a\) for every element \(2c\) of the ideal, the odd integers \(n\) are only abbreviations for finite sums.

The prime ideals of the domain are \(\mathfrak{P}_0=\mathfrak{D}=(2)\) and \(\mathfrak{P}=(2p)\), where \(p\) denotes an odd prime number; every prime ideal is therefore divisible by \(\mathfrak{P}_0\), but by no further prime ideal. The primary ideals are \[\mathfrak{Q}_{e_0}=(2\cdot2^{e_0}) \qquad\text{and}\qquad \mathfrak{Q}_e=(2p^e).\] They are at the same time irreducible ideals, and by what was said about the prime ideals, two ideals belonging to different odd prime numbers are mutually relatively prime, but no \(\mathfrak{Q}\) is relatively prime to any \(\mathfrak{Q}_{e_0}\).⁴⁷ Thus the \(\mathfrak{Q}_{e_0}\) are the only non-isolated primary ideals. The unique decomposition of \(a\) into prime powers corresponds to the unique representation of the ideal \((2a)\) by greatest primary, at the same time irreducible, ideals: \[(2a)=[(2\cdot2^{e_0}),(2p_1)^{e_1},\ldots,(2p_\alpha)^{e_\alpha}].\] Thus here, in contrast to the examples from the polynomial domain, even the non-isolated greatest primary ideals are uniquely determined. If \(e_0=0\) this is at the same time a representation by mutually relatively prime ideals, while if \(e_0>0\) the ideal is relatively-prime-irreducible.

Thus, while in the domain of all integers the four different decompositions coincide, here this happens only for the two decompositions into greatest primary and into irreducible ideals on the one hand, and, for \(e_0>0\), for coprime-irreducible (every ideal is coprime-irreducible, since the domain has no unit) and relatively-prime-irreducible ideals on the other; for \(e_0=0\), however, the coprime-irreducible and relatively-prime-irreducible decompositions become different. At the same time one already obtains here an example in which one prime ideal can be divisible by another without being identical with it; more generally, one sees that product representation does not follow from divisibility. The latter fact - a consequence of the absence of a unit in the domain - is also why no unique product representation of the numbers of the domain by irreducible numbers of the domain exists, even though every ideal is a principal ideal; the introduction of least common multiples therefore proves necessary. It should also be noted that the situation remains exactly the same if, instead of all even integers, one takes all integers divisible by a fixed prime number or power of a prime.

Irreducible and primary ideals also become different, however, if \(\Sigma\) consists of all numbers divisible by a composite number \(g=p_1^{\sigma_1}\cdots p_\nu^{\sigma_\nu}\). Again every ideal is a principal ideal \((g\cdot a)\), and the prime ideals are again \(\mathfrak{P}_0=\mathfrak{D}=(g)\) and \(\mathfrak{P}=(g\cdot p)\), where \(p\) is a prime different from the primes occurring in \(g\). But the irreducible ideals are \(\mathfrak{B}_{\lambda_i}=(g\cdot p_i^{\lambda_i})\) and \(\mathfrak{B}_e=(g\cdot p^e)\); the primary ideals are \(\mathfrak{Q}_e=\mathfrak{B}_e\) and the ideals, different from the irreducible ones, \[\mathfrak{Q}_{\lambda_1\ldots\lambda_\nu}=(g\cdot p_1^{\lambda_1}\cdots p_\nu^{\lambda_\nu}),\] where the \(\mathfrak{B}_{\lambda_i}\) and \(\mathfrak{Q}_{\lambda_1\ldots\lambda_\nu}\) all have the same associated prime ideal \(\mathfrak{P}_0=(g)\). Here too the decomposition into irreducible ideals is unique, and consequently so is the decomposition into greatest primary ideals; thus here again the non-isolated ideals are uniquely determined.

2. An example of a non-commutative ring domain is supplied by the ideal theory in non-commutative polynomial domains treated in the paper of Noether–Schmeidler. In particular one has “completely reducible” ideals, that is, ideals for which the components of the decomposition are pairwise coprime and have no proper divisors; the components are therefore a fortiori irreducible. Thus, in addition to the isomorphism proved there according to § 9, one obtains equality of the number of components in two different decompositions. For the decomposition of systems of partial or ordinary linear differential expressions, which arises as a special case of that paper, this gives a result which does not seem to have been noticed even in the known case of a single ordinary linear differential expression.

At the same time the system \(T\) of all residue classes of a fixed ideal \(\mathfrak{M}\), together with the non-commutative polynomial domain \(\Sigma\), supplies a double domain \((\Sigma,T)\), where \(T\) has the module property with respect to \(\Sigma\). For the difference of two residue classes is again a residue class, and likewise the product of a residue class by an arbitrary polynomial; whereas the product of two residue classes does not exist (loc. cit., § 3). Thus the systems of residue classes called “subgroups” there furnish examples of modules in double domains \((\Sigma,T)\), where the underlying ring domain \(\Sigma\) is non-commutative.

This is a conception of elementary divisor theory dictated by the general developments, but the theory itself is assumed as known.

Let \(\Sigma\) be the domain of all integral matrices with \(n^2\) elements, with addition and multiplication defined in the usual sense for matrices. Then \(\Sigma\) is a non-commutative ring domain; the ideals are therefore in general one-sided, with two-sided ideals only as a special case.⁴⁸

We first show that every ideal is principal. For this purpose, in the case of right ideals, to each matrix \(A=(a_{ik})\) we assign the module \[A=(a_{11}\xi_1+\cdots+a_{1n}\xi_n, \ldots, a_{n1}\xi_1+\cdots+a_{nn}\xi_n)\] of integral linear forms. Conversely, to this module there corresponds every matrix that supplies a basis of \(A\), hence, beside \(A\), also \(UA\), where \(U\) is unimodular. More generally, to the product \(PA\) there corresponds a module \(B\) which is a multiple of \(A\). A single linear form from \(A\) is given by such matrices \(P\) as contain only one non-zero row.

Let now \(A_1,A_2, \ldots,A_\nu, \ldots\) be all elements of an ideal \(\mathfrak{M}\); let \(A_1,A_2, \ldots,A_\nu, \ldots\) be the modules assigned to them; let \(A\) be their greatest common divisor and \(UA\) the most general matrix assigned to this module \(A\). To each single linear form in \(A\) there then corresponds, by the definition of greatest common divisor, a matrix \(P_1A_{i_1}+\cdots+P_\sigma A_{i_\sigma}\), where, as above, the \(P\) have only one non-zero row. It follows that the matrix \(A\) corresponding to a basis of \(A\) also admits such a representation, now with general \(P\), and hence becomes an element of \(\mathfrak{M}\). Since, furthermore, every module \(A_i\) is divisible by \(A\), every matrix \(A_i\) is divisible by \(A\); \(A\), and in general \(UA\), forms a basis of \(\mathfrak{M}\). If one is dealing with left ideals, then correspondingly the columns of each matrix are to be regarded as the basis of a module; every ideal is a principal ideal, for which, beside \(A\), also \(AV\) is a basis, with \(V\) an arbitrary unimodular matrix.

For the connection with elementary divisor theory, we now base the following on two-sided ideals; hence in particular \(PAQ\) belongs to the ideal whenever \(A\) does. By the preceding, the most general basis of such an ideal is \(UAV\), where \(U\) and \(V\) are unimodular. The basis elements therefore exhaust a class of equivalent matrices,⁴⁹ and there is a one-to-one relation between ideal and class, hence also between ideal and the elementary divisor system \((a_1\mid a_2\mid\cdots\mid a_n)\) of the class, where the \(a_i\) are, as is well known, non-negative integers, each dividing the following one. Thus the matrix of the class that occurs in the normal form determined by the elementary divisors may be regarded as a special basis of the ideal; divisibility of ideals, respectively of classes, implies divisibility of the elementary divisors, and conversely.

But Du Pasquier has shown, loc. cit. § 11, that for every two-sided ideal the rank is \(n\) and all elementary divisors agree. To be able to consider the case of general elementary divisors, we must therefore start not from the ideals but directly from the two-sided classes (which are likewise to be denoted by capital German letters). A class \(\mathfrak A=UAV\) is divisible by another class \(\mathfrak B\) if \(A=PBQ\).

In general: the least common multiple (the greatest common divisor) of two classes is obtained by forming the least common multiple (the greatest common divisor) of the corresponding elementary divisor systems.⁵⁰ For let \[(a_1\mid a_2\mid\cdots\mid a_n),\quad (b_1\mid b_2\mid\cdots\mid b_n),\quad (c_1\mid c_2\mid\cdots\mid c_n),\] where \(c_i=[a_i,b_i]\), be respectively the elementary divisor systems of \(\mathfrak A\), \(\mathfrak B\), and \(\mathfrak C^*\), and let \(\mathfrak C=[\mathfrak A,\mathfrak B]\). Then \(\mathfrak C^*\) is divisible by \(\mathfrak A\) and \(\mathfrak B\), hence by \(\mathfrak C\); conversely, the elementary divisors of \(\mathfrak C\) are divisible by those of \(\mathfrak C^*\), so \(\mathfrak C\) is divisible by \(\mathfrak C^*\) and therefore \(\mathfrak C=\mathfrak C^*\). The proof for the greatest common divisor is obtained correspondingly.

The unique representation of the elementary divisors \(a_i\) as least common multiples of prime powers therefore corresponds to a representation of \(\mathfrak A\) as the least common multiple of classes \(\mathfrak Q\) whose elementary divisors are given by powers of one prime number, in symbols \[\mathfrak Q\sim(p^{r_1}\mid p^{r_2}\mid\cdots\mid p^{r_e}\mid0\mid\cdots\mid0), \qquad r_1\leqq r_2\leqq\cdots\leqq r_e.\] If in particular the rank \(\varrho\) is equal to \(n\), then one has here a decomposition into coprime and coprime-irreducible classes which is unique despite the non-commutative domain.

The classes \(\mathfrak Q\) can be further decomposed into classes corresponding to the elementary divisor systems \[\begin{gathered} \mathfrak B_1\sim(p^{r_1}\mid\cdots\mid p^{r_1}),\quad \mathfrak B_2\sim(1\mid p^{r_2}\mid\cdots\mid p^{r_2}),\quad \ldots,\\ \mathfrak B_e\sim(1\mid\cdots\mid1\mid p^{r_e}\mid\cdots\mid p^{r_e}),\quad \mathfrak B_{e+1}\sim(1\mid\cdots\mid1\mid0\mid\cdots\mid0), \end{gathered}\] where in \(\mathfrak B_\nu\) the number \(1\) occurs \((\nu-1)\) times. If, for example, \(r_1=r_2=\cdots=r_\mu=0\), then \(\mathfrak B_1, \ldots,\mathfrak B_\mu\) are equal to the unit class and are to be omitted from the decomposition; the same holds for \(\mathfrak B_{e+1}\) if \(\varrho=n\). If, furthermore, say \(r_\nu=r_{\nu+1}=\cdots=r_{\nu+\lambda}\), then \(\mathfrak B_{\nu+1}, \ldots,\mathfrak B_{\nu+\lambda}\) are proper divisors of \(\mathfrak B_\nu\), and are likewise to be discarded. Denote by \(\mathfrak B_{i_1}, \ldots,\mathfrak B_{i_k}\) those that remain and now give a shortest representation. Then \[\mathfrak Q=[\mathfrak B_{i_1}, \ldots,\mathfrak B_{i_k}]\] is the unique decomposition of \(\mathfrak Q\) into irreducible classes. Indeed, suppose \[\mathfrak B_\nu\sim(1\mid\cdots\mid1\mid p^{r_\nu}\mid\cdots\mid p^{r_\nu})\] can be represented as the least common multiple of \[\mathfrak C\sim(1\mid\cdots\mid1\mid p^{s_1}\mid\cdots\mid p^{s_\lambda}) \quad\text{and}\quad \mathfrak D\sim(1\mid\cdots\mid1\mid p^{t_1}\mid\cdots\mid p^{t_\mu}).\] Then in the elementary divisor systems of \(\mathfrak C\) and \(\mathfrak D\) the number \(1\) must stand in the first \((\nu-1)\) places; in the \(\nu\)-th place one of the exponents \(s_\nu\) or \(t_\nu\), say \(s_\nu\), must become equal to \(r_\nu\). But since \[r_\nu=s_1\leqq s_2\leqq\cdots\leqq s_\lambda\leqq r_\nu,\] one obtains \(\mathfrak C=\mathfrak B_\nu\), and \(\mathfrak B_\nu\) is therefore irreducible.⁵¹ The same holds for \(\mathfrak B_{e+1}\), where \(p\) is to be replaced by \(0\). But, as the formation of the least common multiple shows, each of these irreducible classes gives a definite exponent and the place where this exponent first occurs in the elementary divisor system of \(\mathfrak Q\), respectively \(\mathfrak A\), while \(\mathfrak B_{e+1}\) gives the rank. Since these numbers are uniquely determined by the elementary divisor system of \(\mathfrak Q\), respectively \(\mathfrak A\), and since the relation between elementary divisors and class is one-to-one, the decomposition of \(\mathfrak Q\), and likewise of any class, into irreducible classes is unique.

In summary: Every two-sided class \(\mathfrak A\) of integral matrices with bounded number of entries can be represented uniquely as the least common multiple of finitely many irreducible two-sided classes. Each irreducible class represents a fixed prime divisor of the elementary divisor system of \(\mathfrak A\), an associated exponent, and the place where that exponent first occurs. The irreducible class corresponding to the divisor \(0\) gives the rank of \(\mathfrak A\).