Chain conditions and irreducible decompositions

Written by GPT-6.1 Sol (OpenAI) in Codex at Ultra reasoning effort, October 2026. Self-checked by the writing AI, GPT-6.1 Sol, at Ultra. Public domain (CC0).

Factoring an integer produces canonical prime powers. Intersecting ideals produces a different kind of arithmetic: components can move while their number remains fixed. We will discover the invariant first through the lattice of submodules and then through vector spaces called socles. This also explains why embedded components contribute to the count.

We assume rings are commutative with identity. A proper ideal is irreducible if it cannot be written as the intersection of two strictly larger ideals. This is irreducibility for intersection, rather than for multiplication. The earlier lesson Noetherian and Artinian rings proves the chain-condition equivalences in Theorem 1.1, the finite-module consequences in Proposition 1.2, and the Hilbert basis theorem in Theorem 2.1. Localization, local properties and support, Theorems 2.1 and 3.1, proves exactness, preservation of finite intersections and the prime correspondence. For Section 4 we use the actual proofs in Associated primes and primary decomposition: Theorem 1.2 for existence and zero divisors, Proposition 2.1 for submodules and direct sums, Theorem 2.2 for finiteness, Theorems 3.1–3.2 for localization and support, and Proposition 4.1 for primary finite modules. The decomposition and counting arguments below are proved here. Freely accessible references are Noether’s English working edition and the Stacks project.

1. The example that asks the question

In \(k[x,y]\), for every \(a\in k\),

\[ (x^2,xy)=(x)\cap(x^2,y+ax). \]

Indeed, substitute \(y=-ax\) modulo the second ideal. If \(xf\) lies in that ideal, its image in \(k[x]/(x^2)\) vanishes, so the constant term of \(f(x,-ax)\) is zero. Hence \(f\in(x,y)\) and \(xf\in(x^2,xy)\). The converse inclusion follows from the generators. The ideal \((x)\) is prime. The other quotient is \(k[x]/(x^2)\), whose nonzero ideals all contain the class of \(x\). Its zero ideal is therefore irreducible. Neither component contains the other: \(x\) is missing from the second and \(y+ax\) from the first. The decomposition is irredundant.

The embedded component varies with \(a\). Over an infinite field there are infinitely many choices. Nevertheless each decomposition has two components. The count will turn out to measure an intrinsic property of the quotient.

2. Induction without a numerical size

Theorem 2.1 (Noetherian induction). Suppose ideals of \(R\) satisfy the ascending chain condition. If a property holds for \(I\) whenever it holds for all ideals strictly containing \(I\), it holds for every ideal.

Proof. If failures existed, the maximal condition would give a maximal failing ideal \(I\). All larger ideals would satisfy the property, forcing it for \(I\) too. This is a contradiction. \(\square\)

Theorem 2.2 (finite irreducible decomposition). Every proper submodule \(N\) of a Noetherian module \(M\) is a finite intersection of irreducible proper submodules. In particular this holds for ideals of a Noetherian ring.

Proof. Apply the same maximal-counterexample argument in the submodule lattice. A maximal failure \(N\) cannot be irreducible, because a one-component decomposition would suffice. Write \(N=A\cap B\) with \(A,B\supsetneq N\). Maximality gives finite decompositions of \(A\) and \(B\), and combining them gives one of \(N\). Remove redundant factors. \(\square\)

The largest submodule is the empty intersection; it has zero components. We reserve the word irreducible for proper submodules.

These arguments need the chain condition, not a polynomial presentation. Here are four ways that condition can fail.

The integrality fact in the third example has a short direct proof. If \(\alpha\) satisfies a monic integer polynomial of degree \(d\), then \(\mathbb Z[\alpha]\) is generated over \(\mathbb Z\) by \(1,\alpha,\ldots,\alpha^{d-1}\). Multiplication by any \(\beta\in\mathbb Z[\alpha]\) is represented on these generators by an integer matrix \(B\). The adjugate identity for \(XI-B\), evaluated at \(\beta\), shows that its monic determinant annihilates all generators, including \(1\). Thus every power of an algebraic integer is integral. If a reduced rational \(a/b\), with \(b>0\), satisfies a monic equation of degree \(d\), clearing denominators gives \(b\mid a^d\), hence \(b=1\). This excludes \(1/2\) and therefore every \(2^{-1/2^n}\).

3. The lattice mechanism

A lattice has a meet \(u\wedge v\) and a join \(u\vee v\). It is modular if

\[ u\vee(v\wedge w)=(u\vee v)\wedge w\qquad(u\leq w). \]

Proposition 3.1. Submodules form a modular lattice, with meet intersection and join sum.

Proof. Assume \(A\subset C\). The inclusion \(A+(B\cap C)\subset(A+B)\cap C\) is immediate. For the reverse, if \(z=a+b\in C\), then \(b=z-a\in C\), so \(b\in B\cap C\). \(\square\)

Lemma 3.2 (transposition). In a modular lattice, put \(a=b\wedge c\). The maps

\[ [a,b]\longrightarrow[c,b\vee c],\quad x\longmapsto x\vee c, \qquad z\longmapsto z\wedge b \]

are mutually inverse order isomorphisms.

Proof. For \(a\le x\le b\), modularity gives \((x\vee c)\wedge b=x\vee(c\wedge b)=x\). For \(c\le z\le b\vee c\), it gives \(c\vee(z\wedge b)=z\wedge(c\vee b)=z\). Both maps preserve order. An order isomorphism preserves meets and joins taken within these intervals. \(\square\)

Call an interval \([a,b]\) uniform above \(a\) when two elements in it strictly above \(a\) have meet strictly above \(a\). If \(c\) is meet-irreducible, the interval \([c,b\vee c]\) is uniform above \(c\). Lemma 3.2 transports this to \([a,b]\). Thus

\[ b\wedge c=a,\ c\text{ meet-irreducible} \quad\Longrightarrow\quad \bigwedge_{j=1}^s x_j=a,\ x_j\in[a,b] \ \Longrightarrow\ x_j=a\text{ for some }j. \]

The implication also holds if \(b=a\). For a nonsingleton interval, induction on \(s\) proves it from the binary condition.

Theorem 3.3 (exchange and the Kurosh–Ore count). Suppose a modular lattice has finite decompositions

\[ a=\bigwedge_{i=1}^m b_i=\bigwedge_{j=1}^n c_j. \]

If every \(b_i\) is meet-irreducible, then each \(b_i\) can be exchanged for some \(c_j\): with \(B_i=\bigwedge_{k\ne i}b_k\), one has \(a=B_i\wedge c_j\). If both decompositions are irredundant and all their components are meet-irreducible, then \(m=n\).

Proof. All the elements \(B_i\wedge c_j\) lie in \([a,B_i]\), and their meet is \(a\). This interval is uniform by the preceding consequence of transposition, since \(B_i\wedge b_i=a\). Therefore one of them equals \(a\), proving exchange.

Replace \(b_1\), then \(b_2\), and so forth using this argument. At each stage the component about to be replaced is still one of the original meet-irreducible \(b_i\), and the meet of all current components is \(a\). The intermediate list need not initially be assumed irredundant. After \(m\) replacements, a sublist of at most \(m\) of the \(c_j\) has meet \(a\). Irredundancy of their original list forces all \(n\) distinct components to occur, so \(n\le m\). Reverse the roles to get \(m\le n\).

In fact every intermediate list is irredundant. If one had fewer than \(m=n\) necessary components, delete the others and apply the same replacement argument to that list and the original \(c_j\). It would imply \(n<m\), a contradiction. \(\square\)

The ascending chain condition guarantees existence of finite decompositions by Theorem 2.2’s lattice argument. Once finite decompositions exist, their count needs only modularity. Noether proved the ideal version in 1921; the modular-lattice theorem bears the names of Kurosh and Ore.

Lemma 3.4 (primary quotients). An irreducible submodule of a Noetherian module over a commutative ring is primary: multiplication by each scalar on its quotient is injective or nilpotent.

Proof. Its nonzero quotient \(U\) is uniform. If multiplication by \(a\) has nonzero kernel, the ascending chain \(0:_U a^j\) stabilizes at some \(h\ge1\). Then \(a^hU\cap(0:_U a^h)=0\): if \(a^{2h}u=0\), stabilization gives \(a^hu=0\). Since the second submodule is nonzero, uniformity forces \(a^hU=0\). This proves the assertion. \(\square\)

4. Local vector spaces that count the components

Lemma 3.4 makes the irreducible quotients primary. Here is how its scalar condition identifies the prime. For a nonzero finite quotient \(U\) satisfying that condition, let \(J=\sqrt{\operatorname{Ann}(U)}\). A scalar is noninjective precisely when it belongs to \(J\): Lemma 3.4 proves one implication, and an injective nilpotent endomorphism of a nonzero module is impossible. By the earlier existence theorem choose \(\mathfrak p=\operatorname{Ann}(u)\) with \(u\ne0\). Every element of \(\mathfrak p\) is noninjective, so \(\mathfrak p\subset J\); conversely \(\operatorname{Ann}(U)\subset\mathfrak p\) gives \(J\subset\mathfrak p\). Thus \(J=\mathfrak p\) is prime and is the only associated prime. Elements outside it act injectively. A power of this ideal kills \(U\): choose generators \(a_i\) and powers \(a_i^{e_i}U=0\); every product of \(1+\sum_i(e_i-1)\) generators contains a vanishing power. For the zero ideal its first power suffices. The earlier associated-prime localization theorem is used below.

The socle of a module over a local ring \((A,\mathfrak m)\) is

\[ \operatorname{Soc}(U)=0:_U\mathfrak m \cong\operatorname{Hom}_A(A/\mathfrak m,U). \]

It is a vector space over the residue field. A submodule \(E\subset U\) is essential if it meets every nonzero submodule nontrivially. A module is uniform if any two nonzero submodules meet nontrivially. The zero submodule is irreducible exactly when its quotient is uniform.

Lemma 4.1. A nonzero finite-length module over a local ring has essential socle. It is uniform exactly when its socle has dimension one.

Proof. Every nonzero submodule has a simple submodule, by choosing one of minimal positive length. A simple module over a local ring \((A,\mathfrak m)\) is generated by any nonzero element \(v\); it is \(A/\operatorname{Ann}(v)\), and simplicity makes that annihilator maximal. Thus it is the residue field and lies in the socle. If the socle has dimension one, all nonzero submodules contain it and intersect. If the dimension is at least two, distinct one-dimensional subspaces are submodules with zero intersection. \(\square\)

Lemma 4.2. If \(U\) is a finite primary quotient with associated prime \(\mathfrak p\), then \(U\to U_{\mathfrak p}\) is injective. Moreover \(U\) is uniform if and only if \(U_{\mathfrak p}\) is uniform.

Proof. Every denominator outside \(\mathfrak p\) acts injectively, so \(U\to U_{\mathfrak p}\) is injective and nonzero submodules stay nonzero. If \(U_{\mathfrak p}\) is uniform, the localizations of two nonzero submodules of \(U\) intersect nontrivially. Exactness identifies this intersection with the localization of their intersection, so that original intersection is nonzero. Conversely, let \(U\) be uniform and let \(H,K\subset U_{\mathfrak p}\) be nonzero. Multiplying a nonzero fraction in each by its denominator gives nonzero elements in their contractions to \(U\). These contractions have nonzero intersection by uniformity; injectivity keeps a nonzero element of that intersection nonzero after localization. It belongs to \(H\cap K\). \(\square\)

Since a power of \(\mathfrak p\) kills \(U\), its localization has finite length: the finite filtration by powers of the local maximal ideal has finite-dimensional residue-field vector spaces as its successive quotients. Lemmas 4.1–4.2 therefore show that a \(\mathfrak p\)-primary submodule is irreducible exactly when its localized quotient has one-dimensional socle.

Theorem 4.3 (socle count). Let \(M\) be a finite module over a Noetherian ring, and let \(N=\bigcap_{i=1}^r Q_i\) be an irredundant irreducible decomposition. For every prime \(\mathfrak p\),

\[ \#\{i:\operatorname{Ass}(M/Q_i)=\{\mathfrak p\}\} =\dim_{\kappa(\mathfrak p)} \operatorname{Soc}((M/N)_{\mathfrak p}). \]

Proof. The diagonal injection

\[ D=M/N\hookrightarrow E=\bigoplus_i M/Q_i \]

has essential image. To see this, for each \(i\) the complement intersection \(\bigcap_{j\ne i}Q_j\) maps to a nonzero submodule \(K_i\) of the \(i\)-th summand, and to zero in the other summands. Irredundancy gives nonzeroness. Each summand is uniform, so \(K_i\) is essential in it. The direct sum \(\bigoplus K_i\subset D\) is essential in \(E\). Indeed, start with a nonzero submodule \(H\subset E\) and impose membership in \(K_1,K_2,\ldots\), one coordinate at a time. If the next projection of the current nonzero submodule is zero, it already meets that condition. Otherwise essentiality supplies a scalar multiple of some projected element that is nonzero in \(K_i\); the same scalar multiple of its lift is nonzero and satisfies the new condition. Earlier conditions are preserved because the \(K_j\) are submodules. After finitely many steps \(H\) meets \(\bigoplus K_i\) nontrivially.

Essentiality survives localization here. Indeed, if a nonzero submodule \(H\subset E_{\mathfrak p}\) had zero intersection with \(D_{\mathfrak p}\), contract it to \(L\subset E\). Every element of \(L\cap D\) is killed by some denominator outside \(\mathfrak p\). This intersection is finitely generated, so one common denominator \(s\notin\mathfrak p\) kills it. The ascending sequence \(0:_L s^t\) stabilizes, say at \(t=h\). The nonzero submodule \(s^hL\) would have zero intersection with \(D\): an element in that intersection is killed by \(s\), so stabilization makes it zero. It is nonzero since its localization is \(H\). This contradicts essentiality.

Every simple submodule of \(E_{\mathfrak p}\) must then lie in \(D_{\mathfrak p}\), so their socles coincide. A summand with associated prime \(\mathfrak q\) vanishes if \(\mathfrak q\not\subset\mathfrak p\). If \(\mathfrak q\subsetneq\mathfrak p\), choose \(a\in\mathfrak p\setminus\mathfrak q\); it acts injectively, so that summand has zero \(\mathfrak p\)-socle. If \(\mathfrak q=\mathfrak p\), Lemmas 4.1–4.2 give a socle of dimension one. Adding these dimensions proves the formula. \(\square\)

For ideals take \(M=R\). The socle count is also called the zeroth Bass number. It vanishes for primes not associated to \(M/N\), by the localization rule for associated primes.

5. Computations and a failure without finiteness

For \(I=(x^2,xy)\), localization at \((x)\) makes \(y\) invertible and yields the residue field of \(R_{(x)}\), so its socle has dimension one. At \((x,y)\), the socle is spanned by \(x\). Indeed every localized class has the form \(f(y)+cx\), with \(f\in k[y]_{(y)}\) and \(c\in k\); multiplication by \(y\) forces \(f=0\). Thus there is one component at each of these two associated primes.

For \(I=(x,y)^2\), the quotient has basis \(1,x,y\) and socle \(kx\oplus ky\). Direct monomial membership gives

\[ (x,y)^2=(x^2,y)\cap(x,y^2). \]

Each quotient has one-dimensional socle. For \(J=(x^2,xy^3,y^4)\), the standard monomials are \(1,y,y^2,y^3,x,xy,xy^2\). Its socle has basis \(y^3,xy^2\), and

\[ J=(x^2,y^3)\cap(x,y^4). \]

The rule behind this staircase calculation is proved in Modules, staircases and elementary divisors, Section 2.

Finally let \(R=\prod_{n\ge1}k\). Its zero ideal has no finite irreducible decomposition. Otherwise \(R\) would embed in a sum of finitely many uniform quotients. To justify the obstruction, note that every quotient is again a commutative von Neumann regular ring: the coordinatewise inverse on nonzero coordinates gives \(a=a^2b\), and this identity descends. In a uniform such ring each principal ideal is generated by an idempotent \(e=ab\); the ideals \((e)\) and \((1-e)\) have zero intersection, so every nonzero idempotent is \(1\). Hence a nonzero element is a unit and the quotient is a field. Each map from \(R\) to a field sends at most one of the mutually orthogonal coordinate idempotents to a nonzero element. Finitely many maps therefore kill some nonzero coordinate idempotent, contradicting injectivity.

6. Exercises

  1. Basic. Prove that prime ideals are irreducible. Give an irreducible ideal which is not prime.
  2. Intermediate. Compute an irredundant decomposition of \((x^2,xy^3,y^4)\) into ideals generated by powers of variables, and justify irreducibility of the factors.
  3. Intermediate. Prove the modular law for submodules and give a nonmodular lattice.
  4. Intermediate. Compute the socle of \(k[x,y]/(x^3,x^2y,y^2)\), and an irreducible decomposition of its defining ideal.
  5. Advanced. Explain why exchange in Theorem 3.3 still works if the second decomposition has arbitrary components; then prove the count without assuming that intermediate replacement lists are irredundant.

7. Solutions

1. If \(\mathfrak p=A\cap B\) and both ideals are larger, choose \(a\in A\setminus\mathfrak p\) and \(b\in B\setminus\mathfrak p\). Then \(ab\in A\cap B=\mathfrak p\), contrary to primality. The ideal \((t^2)\subset k[t]\) is irreducible since every nonzero ideal of its quotient contains \(\bar t\); it is not prime since \(t^2\in(t^2)\) but \(t\notin(t^2)\).

2. The answer is \((x^2,y^3)\cap(x,y^4)\). Membership in the intersection means either an \(x\)-exponent at least two, or positive \(x\)-exponent and \(y\)-exponent at least three, or \(y\)-exponent at least four. These are exactly the three generating alternatives. Both quotients are local Artinian with one-dimensional socle, respectively spanned by \(xy^2\) and \(y^3\), so Lemma 4.1 gives irreducibility. The monomials \(x\) and \(y^3\) show neither factor can be omitted.

3. For \(A\subset C\), an element \(a+b\in(A+B)\cap C\) has \(b\in B\cap C\), proving the law. In the pentagon lattice \(0<a<c<1\), \(0<b<1\), with \(a,b\) and \(c,b\) incomparable, take \(u=a,v=b,w=c\). Then \(u\vee(v\wedge w)=a\) while \((u\vee v)\wedge w=c\).

4. The standard basis is \(1,x,x^2,y,xy\). The elements killed by both \(x\) and \(y\) are precisely \(kx^2\oplus kxy\). Thus there are two components, and termwise membership verifies

\[ (x^3,x^2y,y^2)=(x^3,y)\cap(x^2,y^2). \]

Their socles are spanned by \(x^2\) and \(xy\). The witnesses \(y\) and \(x^2\) establish irredundancy.

5. In \([a,B_i]\) the finitely many elements \(B_i\wedge c_j\) meet to \(a\). Uniformity depends only on the irreducibility of \(b_i\), so some intersection equals \(a\), without a condition on \(c_j\). Replacing all \(m\) of the original components produces a sublist of at most \(m\) of the second list. If the second list is irredundant it must be exhausted, giving \(n\le m\). Reverse the process for the reverse inequality. The argument never used irredundancy of an intermediate list.

In Noether’s words

Noether’s 1921 paper treats intersections as least common multiples and sums as greatest common divisors. Her divisibility convention reverses containment. Sections 2–3 establish finite irreducible decompositions and equality of their counts. Her rings may have no identity; our unital convention becomes important for the residue-field and socle language. The entire work can be read in the linked English edition, work 19 alongside the German authority edition. Those linked editions retain their own rights.

References

Editable sources

Complete source archive · Complete course in LaTeX · This lesson in LaTeX. The archive includes all five lesson texts, the figures and their reproducible source.