Resultant forms and elimination

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).

A determinant can remember every point of a finite polynomial scheme, together with its length. A second polynomial, the minimal polynomial, measures how those points and nilpotents can be distinguished by one generic linear function. Comparing the two leads to Noether’s tests for prime and primary ideals. We first build the finite-dimensional construction, then retain all dimensional layers when passing to generic fibres.

Let \(k\) be a field. Rings are commutative with identity. A zero-dimensional ideal is proper, with a nonzero quotient of Krull dimension zero. The proved prerequisites are Noetherian and Artinian rings, Theorem 4.2 for Artinian structure and product decomposition; Krull dimension and Noether normalization, Theorem 4.2 for dimension and transcendence degree; Associated primes and primary decomposition, Theorems 1.2, 3.1, 4.2 and 5.3; and the preceding primary-ideal lesson for contraction and primary localization. We use the Nullstellensatz, Theorems 1.3 and 2.1–2.2, and localization, Theorem 4.2 for Nakayama. Linear algebra and polynomial factorization are elementary background. The finite-field and differential bridges needed here are proved in Section 5. Our free historical edition contains Hentzelt’s work edited by Noether and Noether’s Elimination Theory and General Ideal Theory, linked below.

1. Fixing the sign of the resultant

For \(f=a_m z^m+\cdots+a_0\) and \(g=b_nz^n+\cdots+b_0\), with \(m,n\ge1\), use descending-power bases in the map

\[ S:k[z]_{<n}\oplus k[z]_{<m}\longrightarrow k[z]_{<m+n}, \qquad (u,v)\longmapsto uf+vg. \]

The first block is the \(f\)-block. Define \(\operatorname{Res}(f,g)=\det S\). This specifies a column version of Sylvester’s matrix and its sign once and for all.

Theorem 1.1. If \(a=a_m\ne0\), \(b=b_n\ne0\), and in an algebraic closure \(f=a\prod_{i=1}^m(z-\alpha_i)\), \(g=b\prod_{j=1}^n(z-\beta_j)\), then

\[ \operatorname{Res}(f,g)=a^n b^m\prod_{i,j}(\alpha_i-\beta_j). \tag{1} \]

In particular it vanishes exactly when the polynomials have a common root. Over the universal coefficient ring \(C=\mathbb Z[a_0,\ldots,a_m,b_0,\ldots,b_n]\), there are \(u\in C[z]_{<n}\), \(v\in C[z]_{<m}\) with

\[ uf+vg=\operatorname{Res}(f,g). \tag{2} \]

Proof. First take both polynomials monic with independent roots. A common root makes evaluation annihilate the image of \(S\), so the determinant vanishes when any \(\alpha_i=\beta_j\). In the polynomial ring over \(\mathbb Z\) in all roots, each prime element \(\alpha_i-\beta_j\) therefore divides the determinant; their distinctness implies that their product divides it. Its degree in a fixed \(\alpha_i\) is at most \(n\), since there are \(n\) columns from \(f\), each at most linear in that root. Its degree in a fixed \(\beta_j\) is at most \(m\). Thus the quotient by the product is a constant integer.

To find the constant, set \(f=z^m\), \(g=(z-1)^n\). The \(f\)-columns give the leading \(n\) coordinates as an identity block. The remaining block is multiplication by \(g\) modulo \(z^m\), in descending powers, triangular with diagonal \(g(0)=(-1)^n\). The determinant is \((-1)^{mn}\), exactly the value of the product in (1); the constant is one. Scaling the first and second blocks by \(a,b\) gives their powers \(a^n,b^m\). This polynomial identity specializes to every field, including repeated roots and positive characteristic.

Finally, \(S\operatorname{adj}(S)=(\det S)\mathrm{Id}\) over \(C\). Apply it to the coordinate vector of the constant polynomial one. The two blocks of \(\operatorname{adj}(S)1\) give \(u,v\), proving (2) without dividing by either leading coefficient. \(\square\)

Swapping the two blocks proves \(\operatorname{Res}(g,f)=(-1)^{mn}\operatorname{Res}(f,g)\). If one polynomial is a nonzero constant, the corresponding convention is \(\operatorname{Res}(a,g)=a^n\) and \(\operatorname{Res}(f,b)=b^m\). Formula (2) is a universal identity for positive formal degrees; it does not assert that the empty determinant one for two formal degree-zero polynomials belongs to their universal ideal.

For example,

\[ \operatorname{Res}(z^2+bz+c,z-a)=a^2+ba+c. \]

The discriminant of a degree-\(m\) polynomial with leading coefficient \(a\) is \(a^{2m-2}\prod_{i<j}(\alpha_i-\alpha_j)^2\). Taking \(g=f'\) in (1) gives

\[ \operatorname{disc}(f)=(-1)^{m(m-1)/2}a^{-1}\operatorname{Res}(f,f'), \tag{3} \]

where the derivative is placed in its formal degree-\(m-1\) block. This is a polynomial identity after cancellation of \(a\), and remains valid when the derivative’s actual degree drops in positive characteristic. For \(f=Az^2+Bz+C\), its resultant with \(2Az+B\) is \(-A(B^2-4AC)\), giving the familiar discriminant. When a specialized leading coefficient vanishes, the formal Sylvester determinant can also detect behaviour at infinity; the common-root equivalence in Theorem 1.1 requires the stated actual degrees.

2. Multiplication sees values and lengths

Let \(I\subset k[x_1,\ldots,x_n]\) be zero-dimensional and \(A=k[x]/I\). It is finite-dimensional over \(k\): it is Noetherian Artinian, its finitely many residue fields are finite over \(k\) by Zariski’s lemma, and a composition series has these fields as factors. Multiplication by \(f\) is a \(k\)-linear operator \(M_f\).

Suppose first that \(k\) is algebraically closed. The Artinian product decomposition gives

\[ A=\prod_{\xi\in V(I)}A_\xi, \qquad m_\xi=\dim_k A_\xi. \]

Here \(A_\xi\) is the localization at the point’s maximal ideal. Its maximal ideal is nilpotent and its residue field is \(k\); \(m_\xi\) is its length.

Theorem 2.1 (Stickelberger).

\[ \det(T\mathrm{Id}-M_f)=\prod_{\xi\in V(I)}(T-f(\xi))^{m_\xi}. \tag{4} \]

Proof. On \(A_\xi\), the class of \(f\) is \(f(\xi)+h\) with \(h\) in the nilpotent maximal ideal. Filter this factor by powers of that ideal. On every successive vector-space quotient multiplication by \(h\) is zero, so \(M_f\) acts as the scalar \(f(\xi)\). A basis respecting the filtration gives a block triangular matrix with that scalar throughout its diagonal. The determinant is \((T-f(\xi))^{m_\xi}\). Multiplication respects the product factors, so the full determinant is their product. \(\square\)

The theorem concerns algebraic multiplicity of eigenvalues, not the dimension of their eigenspaces. On \(k[x,y]/(x^2,y)\), multiplication by \(x\) is a nonzero nilpotent operator with characteristic polynomial \(T^2\), while its zero eigenspace has dimension one.

For arbitrary \(k\), base change to \(\bar k\). The matrix entries and determinant commute with base change. In (4), use the geometric points and the dimensions of the local factors of \(A\otimes_k\bar k\). These geometric dimensions incorporate inseparability as well as nilpotent thickness.

3. The resultant form

With independent variables \(u_0,\ldots,u_n\), define

\[ R_I(u)=\det\left(u_0\mathrm{Id}+\sum_{j=1}^nu_jM_{x_j}\right). \tag{5} \]

It is homogeneous of degree \(\dim_k A\), monic in \(u_0\), independent of the chosen basis, and has coefficients in \(k\).

Theorem 3.1. Over \(\bar k\),

\[ R_I(u)=\prod_{\xi\in V_{\bar k}(I)} \left(u_0+\sum_j u_j\xi_j\right)^{m_\xi}. \tag{6} \]

Proof. Apply the maximal-ideal filtration argument from Theorem 2.1 to multiplication by the generic linear function \(\sum u_jx_j\), over \(\bar k(u_1,\ldots,u_n)\), and take \(T=-u_0\), with the determinant signs cancelling. The resulting identity of polynomials is (6). \(\square\)

Unique factorization and the normalization of each factor’s \(u_0\)-coefficient recover the point coordinates and their multiplicities. They do not recover the ideal. For instance \((x^2,y)\) and \((x,y^2)\) both have form \(u_0^2\), despite different nilpotent directions.

Theorem 3.2 (primary criterion). A zero-dimensional ideal is primary exactly when its geometric points form one orbit under \(\operatorname{Aut}(\bar k/k)\). In that case every linear factor in (6) has the same exponent. If \(\mathfrak m=\sqrt I\), \(L=k[x]/\mathfrak m\), and \(\ell\) is the length of \(A\) as a local ring, then

\[ R_I=R_{\mathfrak m}^{\ell} =N_{L/k}\left(u_0+\sum_j u_j\bar x_j\right)^{\ell}. \tag{7} \]

Over an algebraically closed field the criterion says exactly that \(R_I\) is a power of one linear form.

Proof. The quotient’s Artinian factors are indexed by its maximal ideals. A finite Artinian ring is primary as a zero ideal exactly when it is local: in a local factor all nonunits are nilpotent; two nonzero product factors supply zero divisors which are not nilpotent. The embeddings of a finite residue field \(L\) into \(\bar k\) form one orbit, and two different maximal ideals cannot lie in the same orbit since their relation ideals over \(k\) are different. Thus one orbit is equivalent to one maximal ideal.

For a local factor, a composition series has \(\ell\) copies of \(L\). Multiplication preserves it and induces the same multiplication operator on each copy, proving (7) by a triangular determinant. Write \([L:k]=sd\), where \(s\) is the number of embeddings and \(d\) its inseparable degree. After base change, each embedding contributes a local algebra of dimension \(d\); equivalently the norm is the product of its \(s\) linear forms, each to power \(d\). The common exponent in (6) is therefore \(d\ell\). \(\square\)

The squarefree orbit product need not itself have coefficients in \(k\). For \(k=\mathbb F_p(t)\), \(I=(x^p-t,y)\) is maximal and has local length one, but \(R_I=(u_0+u_1t^{1/p})^p=u_0^p+t u_1^p\). Its sole geometric point has multiplicity \(p\). This distinction corrects an unqualified use of “Galois orbit product” over imperfect fields.

4. A finite scheme computed by matrices

Take \(I=(x^2-y,y^2-y)\). Eliminating \(y\) identifies its quotient with \(k[x]/(x^4-x^2)\), of dimension four in every characteristic. With basis \(1,x,x^2,x^3\), columns recording images of basis vectors, we obtain

\[ M_x=\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&1\\0&0&1&0\end{pmatrix}, \qquad M_y=M_x^2=\begin{pmatrix}0&0&0&0\\0&0&0&0\\1&0&1&0\\0&1&0&1\end{pmatrix}. \]

Their characteristic polynomials are \(T^2(T^2-1)\) and \(T^2(T-1)^2\). If \(\operatorname{char}k\ne2\), the geometric points are \((0,0),(1,1),(-1,1)\), with lengths \(2,1,1\). Consequently

\[ R_I=u_0^2(u_0+u_1+u_2)(u_0-u_1+u_2). \tag{8} \]

In characteristic two the last two points coincide: \(x^4-x^2=x^2(x-1)^2\). The two points have length two each, and (8) becomes \(u_0^2(u_0+u_1+u_2)^2\). Both the multiplication calculation and its characteristic qualification matter.

5. Retaining every dimension in a generic fibre

Localizing only at parameters of the largest component can discard embedded components. We instead make the isolated-component filtration from the second lesson explicit. Let \(P=k[x_1,\ldots,x_n]\), \(I\subsetneq P\). Adjoin an invertible generic matrix \(U\), put \(F=k(U)\), and set \(z=Ux\). Write \(I_F\) for the transformed ideal in \(F[z]\).

We first record the finite-field calculations used for generic parameters. They also justify the field and orbit assertions in Section 3.

Finite-field calculation. An irreducible polynomial over a characteristic-zero field is separable: its derivative is nonzero of smaller degree, so it is relatively prime to the polynomial. In characteristic \(p>0\), every monic irreducible polynomial has the form \(g(T^{p^e})\), with \(g\) irreducible and \(g'\ne0\). Repeatedly extract powers of \(T^p\) until the derivative is nonzero. Its distinct roots are the \(p^e\)-th roots of the distinct roots of \(g\), and their multiplicities are \(p^e\).

An embedding of a field extends through a finite simple algebraic extension by choosing a root of the transported minimal polynomial. For a finite tower, the number of embeddings into an algebraic closure is therefore the product of the numbers of distinct roots at the successive steps. Each is the degree of the step divided by a power of \(p\). An extension generated by separable elements has as many embeddings as its degree, because at each step the minimal polynomial divides a separable polynomial. Every element of that extension is separable: restriction to its simple subfield, followed by extension through the remaining tower, bounds the number of embeddings by the number of embeddings of that subfield times the remaining degree. Equality with the full degree forces the simple subfield to have as many distinct embeddings as its degree. In particular, sums, products and inverses of separable elements are separable.

In characteristic \(p>0\), if a finite extension has only one embedding, each simple subfield has only one embedding, since all of its embeddings extend. The preceding polynomial calculation then makes each element purely inseparable, and the extension has degree \(p^g\). For a general finite extension, the ratio of its degree to its number of embeddings is also a power of \(p\), by the tower calculation. In characteristic zero that ratio is one. Any two embeddings of a finite extension of \(k\) into \(\bar k\) are conjugate by an automorphism of \(\bar k/k\): extend the induced isomorphism between their images to an embedding of \(\bar k\) in itself by a maximal partial extension. A missing algebraic element could always be added by choosing a root, so the maximal domain is all of \(\bar k\). Its image is algebraically closed and \(\bar k\) is algebraic over it, so the embedding is onto. This proves the orbit assertion used in Theorem 3.2. The resulting automorphisms of \(A\otimes_k\bar k\) carry its local factors to one another and preserve their dimensions. Applied to a residue field of degree \(sd\), with \(s\) embeddings, this also gives dimension \(d\) to each of its \(s\) geometric factors, as used in (7).

Differential calculation. We use only the construction, polynomial and localization rules, and the first fundamental sequence from Kähler differentials, Theorem 1.1, Theorems 2.2, 3.1 and 5.1. In particular,

\[ E\otimes_C\Omega_{C/K}\longrightarrow\Omega_{E/K} \longrightarrow\Omega_{E/C}\longrightarrow0 \]

is exact for a tower \(K\subset C\subset E\). We now prove the additional facts we need.

For a finite extension \(E/K\) of characteristic \(p\), set \(C=K E^p\), the subfield generated by \(K\) and all \(p\)-th powers in \(E\). Every \(K\)-derivation kills \(C\), so \(\Omega_{E/K}=\Omega_{E/C}\). Choose generators \(b_1,\ldots,b_r\) over \(C\) successively, omitting a generator already in the preceding field. Since \(b_j^p\in C\), each remaining step has degree \(p\). Indeed \(T^p-b_j^p\) has only one distinct root; its irreducible factor is either linear or has degree \(p\), by the preceding polynomial calculation. The monomials \(b_1^{e_1}\cdots b_r^{e_r}\), \(0\le e_j<p\), form a basis over \(C\). Consequently

\[ E=C[T_1,\ldots,T_r]/(T_j^p-b_j^p:1\le j\le r), \qquad \Omega_{E/C}=\bigoplus_{j=1}^r E\,db_j. \]

The presentation is an isomorphism because its spanning monomials map to that basis. Its relations have zero differential, proving the displayed differential formula. Thus \(\Omega_{E/K}=0\) exactly when \(E=K E^p\).

This vanishing implies separability. Iterating \(E=K E^p\) gives \(E=K E^{p^N}\) for every \(N\). If \(\alpha_1,\ldots,\alpha_m\) generate \(E/K\), the irreducible-polynomial calculation makes a sufficiently high \(p\)-power of each \(\alpha_j\) separable over \(K\). For one common \(N\), therefore, \(K E^{p^N}=K(\alpha_1^{p^N},\ldots,\alpha_m^{p^N})\) is separable, by the finite-field calculation. Since it is \(E\), the original extension is separable. Conversely a separable extension has zero differentials by Kähler differentials, Theorem 6.2, whose derivation-extension proof is already present. We have proved the finite criterion

\[ E/K\text{ finite is separable}\quad\Longleftrightarrow\quad \Omega_{E/K}=0. \tag{5a} \]

Finally let \(L/k\) be finitely generated of transcendence degree \(d\), with \(k\) perfect of characteristic \(p\). Choose any transcendence basis \(t\), and put \(D=[L:k(t)]\). Frobenius identifies \(L\) with \(L^p\) and \(k(t)\) with \(k(t)^p\), so \([L^p:k(t)^p]=D\). The monomials in the \(t_j\) with exponents less than \(p\) form a basis of \(k(t)\) over \(k(t)^p=k(t_1^p,\ldots,t_d^p)\). Independence follows by clearing denominators and comparing distinct exponent classes modulo \(p\); their span is a finite-dimensional domain, hence a field, and contains all the \(t_j\). Computing \([L:k(t)^p]\) in two ways now gives

\[ D p^d=[L:L^p]D,\qquad [L:L^p]=p^d. \]

Because \(k\subset L^p\), the same generator and presentation argument over \(L^p\) supplies precisely \(d\) differential basis elements. Thus \(\dim_L\Omega_{L/k}=d\). In characteristic zero this dimension follows from Theorem 6.2 of the differential lesson: any transcendence basis leaves a finite algebraic extension, which is separable by the first paragraph. These arguments prove the dimension assertion without assuming a separating basis exists.

We can now establish linear normalization and its separability qualification.

Lemma 5.1. For each of the finitely many associated primes \(\mathfrak p\) of \(P/I\), of dimension \(d\), the last \(d\) generic coordinates are algebraically independent modulo \(\mathfrak p_F\), and its function field is finite over \(F(z_{n-d+1},\ldots,z_n)\). Any list of more than \(d\) coordinates satisfies a nonzero polynomial relation there. If \(k\) is perfect, the finite extension for the generic list of length \(d\) is separable.

Proof of normalization. Fix one prime, let \(A=P/\mathfrak p\), and let \(L\) be its function field. We prove the generic assertion directly, including finite coefficient fields. If \(n>d\), choose a nonzero relation \(f(x)=0\), with highest homogeneous part \(f_D\). Introduce independent coefficients \(c_2,\ldots,c_n\) and put \(y_j=x_j-c_jx_1\). After substituting \(x_j=y_j+c_jx_1\), the coefficient of \(x_1^D\) is the nonzero polynomial \(f_D(1,c_2,\ldots,c_n)\). It is nonzero because homogeneity makes its monomials have distinct exponent lists in the \(c_j\). Dividing by this coefficient makes \(x_1\) integral over the algebra generated by \(y_2,\ldots,y_n\). All the original generators are then integral, so the inclusion is finite. Its smaller algebra is a domain of dimension \(d\), by equality of dimension under finite integral inclusions, Integral extensions, Theorem 5.1.

Repeat with fresh independent coefficients whenever the number of remaining generators exceeds \(d\). After \(n-d\) steps the remaining \(d\) generators are independent, by the dimension–transcendence-degree theorem, and the full algebra is finite over their polynomial algebra. All these changes are linear and triangular. The final retained coordinates have the shape

\[ w_j=x_{n-d+j}+\sum_{h=1}^{n-d}a_{jh}x_h\quad(1\le j\le d). \]

The \(d(n-d)\) coefficients \(a_{jh}\) are independent variables, together with the coefficients used among the first \(n-d\) eliminated coordinates: at the step eliminating the \(h\)-th coordinate, the new coefficient in each retained coordinate is its fresh coefficient with sign minus, plus an expression in later-step coefficients. Starting at the last step and solving backwards recovers these fresh coefficients from the \(a_{jh}\) and the coefficients among eliminated coordinates. This is an invertible triangular change of the independent coefficient variables. If the latter extra coefficients are denoted \(b\), we have proved finiteness of the function field over \(k(a,b)(w)\). Adjoining the independent \(b\) cannot reduce transcendence degree of \(L(a)/k(a)(w)\); thus this finitely generated extension already has transcendence degree zero, and is finite. When \(n=d\), the original generators are independent and there is no elimination step.

For the last \(d\) rows of a completely generic matrix \(U\), their last \(d\) columns form an invertible matrix \(B\). Multiplication by \(B^{-1}\) writes these rows uniquely as \([a\mid\mathrm{Id}_d]\); their first-block coefficients \(a\), the entries of \(B\), and the remaining rows give independent rational coordinates on the matrix parameter field. The assertion just proved for \(w\) consequently proves it for the last \(d\) coordinates of \(U\). For \(d=0\), the argument says simply that the residue field is finite over the coefficient field. The argument works for each of the finite list of associated primes with the same independent matrix \(U\); no choice of a rational matrix over \(k\) is needed. More than \(d\) coordinates must be dependent by transcendence degree.

Proof of separability. Assume \(k\) perfect, put \(M=L(U)\), \(F=k(U)\), and write \(t\) for the last \(d\) generic coordinates. The differential calculation gives \(\dim_L\Omega_{L/k}=d\). The \(dx_j\) span this space: derivatives of their polynomial and rational expressions follow from Leibniz and the quotient rule. Some \(d\) of them therefore form a basis. Polynomial extension and localization, or directly the universal property of derivations, give

\[ \Omega_{M/k}=(M\otimes_L\Omega_{L/k}) \oplus\bigoplus_{a,b}M\,dU_{ab}. \]

Modulo the second summand, each \(dt_j\) is the corresponding generic linear combination of the \(dx_h\). Their determinant in a differential basis is a nonzero polynomial in the independent last-row coefficients: specialize those rows to select the \(d\) basis generators to obtain determinant one. Hence the \(dt_j\), together with all the \(dU_{ab}\), form a basis of \(\Omega_{M/k}\). Normalization has already proved that \(K=F(t)\) is a rational parameter field and \(M/K\) is finite. The first fundamental sequence identifies \(\Omega_{M/K}\) with the quotient by the span of these basis elements, so it is zero. Criterion (5a) proves that \(M/K\) is separable. In characteristic zero the finite-field calculation gives the same conclusion. The parameter field \(F\) need not be perfect. \(\square\)

Put \(F_i=F(z_{i+1},\ldots,z_n)\), for \(0\le i\le n\), and let

\[ G_i=I_F F_i[z_1,\ldots,z_i]\cap F[z]. \]

These are the saturations by nonzero tail polynomials. A purely transcendental field extension preserves primality and primaryness. For the latter assertion, inject a primary quotient into its Artinian localization at its radical. In the polynomial ring over this local algebra, a polynomial with nonzero residue acts injectively: filter coefficients by powers of the nilpotent maximal ideal; on the first nonzero layer of another polynomial, multiplication is by a nonzero polynomial over a field and cannot kill that vector-valued polynomial. Polynomials with zero residue are nilpotent. Contraction proves primaryness before this localization, and adjoining then localizing the independent coefficient variables preserves it. Faithful flatness preserves irredundancy. Thus the original primary decomposition and associated primes extend faithfully to this generic coefficient field. Lemma 5.1 and the primary localization rule show that \(G_i\) is the intersection of precisely the primary components whose radicals have dimension at least \(n-i\). This is a downward-closed set in prime inclusion, so the contraction is canonical even when embedded primary components are not unique. The ideals form \(G_0\supset G_1\supset\cdots\supset G_n=I_F\). If \(I=0\), then all \(G_i=0\); we treat that prime ideal separately. For nonzero \(I\), \(G_0=F[z]\).

For \(1\le i\le n\), set

\[ H_i=(G_{i-1}/G_i)\otimes_{F[z_{i+1},\ldots,z_n]}F_i. \tag{9} \]

This is a finite-dimensional \(F_i\)-vector space, with commuting multiplication operators from the \(z_j\). Indeed it embeds, by primary decomposition, into the sum of the localized quotients for components of dimension exactly \(n-i\); each is zero-dimensional over \(F_i\), hence finite. Components of larger dimension impose \(G_{i-1}\) and give zero in this injection; those of smaller dimension are killed by localization. Every associated prime of dimension \(n-i\) actually appears in \(H_i\): at that prime, irredundancy supplies an element of the intersection of all other components which is absent from its component; it is in \(G_{i-1}\) and survives in (9). In particular no embedded dimension disappears from this filtration.

Adjoin further independent variables \(v_1,\ldots,v_n\), and on \(H_i\otimes_{F_i}F_i(v)\) consider multiplication by \(L_v=\sum_jv_jz_j\). Denote its monic minimal polynomial by \(E_i(T,v)\) and its characteristic polynomial by \(R_i(T,v)\). For a zero layer both are one. These are the elementary-divisor and norm polynomials in this generic-fibre convention. Primitive numerator forms can be obtained by clearing denominators; irreducibility below always means over \(F_i(v)\), not an accidental factor introduced in a denominator. This convention expresses the same generic minimal-polynomial versus determinant mechanism as the Hentzelt–Noether forms, while avoiding claims of equality between different coordinate normalizations of the historical formulas.

6. Prime and primary tests by the two forms

Theorem 6.1. Let \(I\ne0\) be proper.

  1. The ideal is primary exactly when one layer is nonzero and its norm polynomial is a power of one irreducible polynomial. Equivalently its minimal polynomial is a power of one irreducible polynomial in that single layer.
  2. A prime ideal has one layer and irreducible \(E_i\). Conversely, one layer with irreducible \(R_i\) implies that \(I\) is prime.
  3. Over a perfect coefficient field, one layer with irreducible \(E_i\) is also equivalent to primality; for a prime ideal \(E_i=R_i\).
  4. In characteristic \(p>0\), for a prime ideal over any field, \(R_i=E_i^{p^g}\) for some \(g\ge0\). Over an imperfect field an irreducible \(E_i\) can also occur for a proper primary ideal.

The zero ideal is prime and is the separately excluded empty-filtration case.

Proof of the primary test. Over an algebraic closure of \(F_i\), the support points of a layer have distinct generic values \(\sum v_j\xi_j\): equality as polynomials in independent \(v_j\) forces equality of all coordinates. The determinant proof in Section 2, which applies equally to finite modules supported at maximal ideals, makes its characteristic polynomial the product of these linear factors with positive multiplicities. Each residue field contributes one irreducible factor over \(F_i(v)\), possibly with an inseparable power. Distinct maximal ideals contribute different factors, because a common conjugate generic value would identify their coordinate tuples and hence their relation ideals. The minimal polynomial has the same distinct irreducible factors as the characteristic polynomial: this follows on each local module from scalar-plus-nilpotent multiplication, or from the elementary-divisor decomposition of a linear operator.

By (9), the nonzero layers and their residue-field factors thus record every associated prime of \(P/I\). One factor in one layer is equivalent to one associated prime. A finite module with one associated prime is primary: the zero divisors are the union of its associated primes, and the radical of its annihilator is that single prime. Applied to \(P/I\) this proves the first assertion, using Associated primes and primary decomposition, Theorems 1.2 and 4.2.

Proof of the general prime assertions. In the primary case the only layer is the whole generic quotient \(B=F_i[z_1,\ldots,z_i]/I_FF_i[z_1,\ldots,z_i]\), a finite local algebra. Its residue field is denoted \(L\). If \(I\) is prime, \(B=L\) is a field, so the minimal polynomial of \(L_v\) is irreducible. If the characteristic polynomial is irreducible, its minimal polynomial equals it. Then \(F_i(v)[L_v]\) is a field of degree \(\dim_{F_i}B\), and therefore is all of \(B\otimes F_i(v)\). This algebra is a field, so \(B\) is a field. A primary ideal contracts from its localization at parameters outside its radical; consequently \(I_F\), and by faithful field extension \(I\), is prime.

Proof over a perfect field. Lemma 5.1 makes \(L/F_i\) separable. Its distinct embeddings give distinct values of the generic \(L_v\); hence that value is a primitive element of \(L(v)/F_i(v)\). If \(I\) is prime, this proves equality of minimal and characteristic polynomials.

Now suppose the single-layer minimal polynomial \(h(T,v)\) is irreducible. Reduction modulo the nilpotent maximal ideal of \(B(v)\) gives the same minimal polynomial in its residue field, since that polynomial is irreducible and annihilates the residue value. It is separable by the preceding paragraph. Therefore \(h_T(L_v,v)\ne0\) in the subfield \(F_i(v)[L_v]\subset B(v)\). Differentiate the identity \(h(L_v,v)=0\) with respect to \(v_j\), keeping every element of \(B\) fixed:

\[ h_T(L_v,v)z_j+h_{v_j}(L_v,v)=0. \]

The first coefficient is invertible in that subfield. Hence every \(z_j\) belongs to it. These elements generate \(B(v)\) as an algebra, so \(B(v)\) is the field \(F_i(v)[L_v]\). The preceding contraction argument proves primality.

Proof of the inseparable assertion. For a field \(L/F_i\), the characteristic polynomial of a field element \(a\) is its minimal polynomial to power \([L(v):F_i(v,a)]\): choose a basis over the subfield and repeat its multiplication matrix. With \(a=L_v\), distinct embeddings of \(L\) into an algebraic closure are distinguished by the independent \(v_j\). Indeed all images of the \(z_j\) belong to an algebraic closure of \(F_i\); an equality of their generic linear combinations forces equality of each coordinate. The only embedding of \(L(v)\) fixing \(F_i(v,L_v)\) is therefore the identity. The finite-field calculation of Section 5 proves that this extension is purely inseparable of degree \(p^g\), yielding the formula. \(\square\)

These are the modern forms of Noether’s Theorems XII–XIV in Section 6. “Proper primary” there means primary but not prime. The perfect-field hypothesis refers to the original coefficient field; the parameter field itself may be imperfect. Generic separating parameters, not perfectness of that rational function field, justify the proof.

For a concrete imperfect-field distinction, take \(k=\mathbb F_2(a,b)\), with independent \(a,b\), and \(I=(x^2-a,y^2-b)\). The element \(a\) is not a square, since its formal derivative with respect to \(a\) is one and the derivative of every square is zero. After adjoining \(x\), the field is \(\mathbb F_2(x,b)\), with \(a=x^2\); differentiating with respect to \(b\) shows that \(b\) is still not a square. Thus the quotient is a purely inseparable field of degree four. For \(L_v=v_1x+v_2y\),

\[ E=T^2-v_1^2a-v_2^2b,\qquad R=E^2. \]

The quadratic is irreducible since its constant term is not a square in \(k(v)\): its derivative with respect to \(a\) is \(v_1^2\ne0\), whereas every square has derivative zero. On the other hand, over \(k=\mathbb F_2(a)\), the proper primary quotient \(B=k[x,w]/(x^2-a,w^2)\) has the same pattern \(E=T^2-v_1^2a\), \(R=E^2\). The nonzero nilpotent \(w\) proves that the ideal is not prime, despite irreducible \(E\). These are the two mechanisms distinguished by Noether’s imperfect-field examples.

For a proper primary ideal the norm need not even be a \(p\)-power of the irreducible minimal polynomial. Over \(\mathbb F_3(a)\), the quotient by \((x^3-a,w^2)\) has basis \(x^j,x^jw\), \(0\le j<3\), and therefore dimension six and length two over its degree-three residue field. A generic value \(v_1x+v_2w\) has irreducible minimal polynomial \(T^3-v_1^3a\): its cube is \(v_1^3a\), and differentiation with respect to \(a\) proves that this coefficient is not a cube. The one-root irreducible-polynomial calculation of Section 5 makes the degree three. Its characteristic polynomial is the square of this polynomial, by the composition-series argument. The exponent two is not a power of three. The prime hypothesis in Theorem 6.1(4) is essential.

7. What elimination projects

Let \(k\) be algebraically closed and write variables as \(x,y\). Set \(J=I\cap k[x]\). The image of \(V(I)\) under projection to its \(x\)-coordinates need not be closed, but its Zariski closure is exactly \(V(J)\).

Proof. A polynomial in \(k[x]\) vanishes on that image exactly when, viewed as a polynomial in all variables, it vanishes on \(V(I)\). By the strong Nullstellensatz this ideal is \(\sqrt I\cap k[x]=\sqrt J\); the equality follows directly by taking powers. The closed set defined by the ideal of a set is its closure, proving the assertion. \(\square\)

For \(I=(xy-1)\), the projection omits \(x=0\), yet \(J=0\) and the closure is the whole line. Affine elimination equations determine a closure, not necessarily the image itself.

For projective variables, closedness holds over every base, and homogeneous elimination gives the actual image. Here is a full proof, including the field and local algebra steps.

Theorem 7.1 (projective projection and homogeneous elimination). For any scheme \(Y\), the projection \(\mathbb P^r_Y\to Y\) sends closed subsets to closed subsets, and this remains true after every change of base. On an affine base \(Y=\operatorname{Spec}A\), let \(J\) be a homogeneous ideal of \(S=A[X_0,\ldots,X_r]\), and let \(Z\) be its projective zero set. With \(S_+=(X_0,\ldots,X_r)\), its image is

\[ \pi(Z)=V_A\bigl((J:S_+^\infty)\cap A\bigr), \qquad J:S_+^\infty=\bigcup_{N\ge0}(J:S_+^N). \tag{10} \]

Here \(J:S_+^N\) consists of the elements whose product with every element of \(S_+^N\) is in \(J\). No Noetherian or finite-generation hypothesis on \(A\) or \(J\) is needed.

Proof of the projective field step. Over a field \(K\), a homogeneous ideal \(H\subset K[X_0,\ldots,X_r]\) has empty projective zero set exactly when its quotient has zero degree-\(N\) piece for some \(N\ge0\). Extend to an algebraic closure \(\bar K\). Empty projective zero set means that all common affine zeros are contained in the origin. Conversely a nonzero affine zero gives a projective point by scaling; if the projective scheme has a point, one of its affine charts is a nonzero finite-type \(K\)-algebra. A maximal ideal of that chart has finite residue field by the weak Nullstellensatz, and an embedding of that field in \(\bar K\) supplies a nonzero geometric zero. Thus this geometric criterion detects all projective scheme points, not just \(K\)-rational points.

The strong Nullstellensatz in \(\bar K[X]\) now puts each \(X_j\) in the radical of \(H\bar K[X]\). Choose \(e_j\ge1\) with \(X_j^{e_j}\in H\bar K[X]\). For \(N=1+\sum_{j=0}^r(e_j-1)\), every monomial of degree \(N\) is divisible by at least one of those powers. Hence the degree-\(N\) quotient is zero over \(\bar K\), and is already zero over \(K\): tensoring a nonzero vector space with a field extension cannot make it zero. A zero degree-\(N\) piece makes every higher piece zero by multiplication by the variables, and forces all the \(X_j\) to be nilpotent in the quotient. It therefore leaves no projective prime, since a projective prime must avoid at least one \(X_j\). This proves both directions. The argument also covers \(H=(1)\), where the degree-zero piece is already zero.

Proof near a point of the base. Put \(M_N=(S/J)_N\). This is a finite \(A\)-module, being a quotient of the free module on the finitely many degree-\(N\) monomials. For a prime \(\mathfrak p\subset A\), its fibre is

\[ M_N\otimes_A\kappa(\mathfrak p) =\bigl(\kappa(\mathfrak p)[X]/J(\mathfrak p)\bigr)_N. \]

Indeed taking this quotient and its fixed graded piece commutes with tensor product by right exactness. Suppose the projective fibre is empty. The field step gives an \(N\) for which this tensor product is zero. Nakayama, Localization, local properties and support, Theorem 4.2, gives \((M_N)_{\mathfrak p}=0\). Choose a finite generating list for \(M_N\). Each generator is killed by some element outside \(\mathfrak p\), by the localization equality criterion; their product \(f\notin\mathfrak p\) kills them all. Consequently \((M_N)_f=0\). Multiplication by the variables gives \((M_m)_f=0\) for every \(m\ge N\), so every projective fibre over the open neighbourhood \(D(f)\) is empty. The complement of the image is therefore open.

For the exact formula, a finite module \(M\) is zero after localization at \(\mathfrak p\) exactly when \(\operatorname{Ann}_A(M)\) contains an element outside \(\mathfrak p\): use the same finite generating list to prove the nontrivial direction. The field step and Nakayama therefore give

\[ \pi(Z)=\bigcap_{N\ge0}V_A(\operatorname{Ann}_A M_N) =V_A\left(\bigcup_{N\ge0}\operatorname{Ann}_A M_N\right). \]

The annihilators are increasing, because degree-\((N+1)\) monomials are products of variables with degree-\(N\) monomials. Finally, \(\operatorname{Ann}_A M_N=(J:S_+^N)\cap A\): both sides say that multiplying every degree-\(N\) monomial by the coefficient gives an element of \(J\), and those monomials generate \(S_+^N\) as an ideal. This proves (10).

Proof over general bases and under base change. Every closed subset \(W\) of \(\mathbb P^r_A\) is cut out by homogeneous equations. To verify this topological assertion, take a point outside \(W\) in the chart \(X_i\ne0\). An equation of \(W\) on that affine chart is nonzero at the point. Homogenize it by clearing powers of \(X_i\), then multiply by one further \(X_i\). The resulting homogeneous equation still does not vanish at that point, vanishes on \(W\) within the chart, and vanishes on all points outside the chart. The collection of these equations cuts out exactly \(W\). The affine-base argument applies to its homogeneous ideal. For arbitrary \(Y\), apply it on every affine open: the intersection of the image with that open is the image of the restricted closed subset, so it is closed there. Closedness is local on this open cover. Finally, after any map \(Y'\to Y\), the pulled-back projective space is \(\mathbb P^r_{Y'}\), to which the same proof applies. This proves universal closedness. \(\square\)

In the classical case of an algebraically closed coefficient field, formula (10) is the homogeneous elimination ideal on each affine chart of the base. Saturation removes the irrelevant affine origin of the projective coordinates. Thus projective elimination describes an image that is already closed, while affine elimination may describe only its closure.

Consider over \(\mathbb C\) the conics \(y^2-x=0\), \(y^2+x^2-2=0\). Their resultant in \(y\) is

\[ (x^2+x-2)^2=(x-1)^2(x+2)^2. \]

The points are \((1,\pm1)\) and \((-2,\pm i\sqrt2)\). The Jacobian determinant is \(-2y(1+2x)\), nonzero at all four points, so each is simple. The squares in the elimination resultant reflect two points above each projected coordinate, rather than a double point at every intersection. Here both polynomials are monic in the eliminated variable, so no leading-coefficient exception interferes.

Higher-dimensional Chow forms organize incidence with complementary linear spaces. They extend the theme of a form recording geometric intersections; we make no additional Chow-form theorem a premise of the proofs here.

8. Exercises

  1. Basic. Compute \(\operatorname{Res}(x^2+bx+c,x-a)\) and explain its vanishing.
  2. Intermediate. Compute the multiplication matrices for \((x^2-y,y^2-y)\), and check the characteristic polynomials in characteristic two as well as characteristic zero.
  3. Intermediate. Compute the forms for \((x^2,y)\) and \((x^2-1,y-x)\). Identify what changes in characteristic two.
  4. Intermediate. Eliminate \(y\) from \(x^2+y^2=1\), \(y=x^2-1\) over \(\mathbb C\), finding all points and local lengths.
  5. Advanced. Prove the universal ideal-membership identity for the resultant using the adjugate, including degree bounds for its two coefficients. Explain why this proof survives specialization of leading coefficients.

9. Solutions

1. Formula (1) gives \(f(a)=a^2+ba+c\); the two sign changes from \(\prod(\alpha_i-a)\) cancel. It vanishes exactly when \(a\) is a root of the quadratic.

2. The basis and matrices in Section 4 follow from \(y=x^2\), \(x^4=x^2\). Direct matrix multiplication gives \(M_y=M_x^2\). Determinants give \(T^2(T^2-1)\) and \(T^2(T-1)^2\) over \(\mathbb Z\), so reduction modulo any characteristic is valid. In characteristic two both become \(T^2(T-1)^2\). The two length-two local factors yield these multiplicities, as (4) requires.

3. The first quotient has basis \(1,x\), with \(M_x^2=0\), \(M_y=0\), hence form \(u_0^2\). The second is \(k[x]/(x^2-1)\), with \(y=x\), giving \(R=u_0^2-(u_1+u_2)^2\). In characteristic different from two this factors as \((u_0+u_1+u_2)(u_0-u_1-u_2)\), representing the simple points \((1,1),(-1,-1)\). In characteristic two it is the square of the first linear form, representing one point of length two.

4. The monic linear relation substitutes directly into the circle polynomial. The resultant is \(x^2+(x^2-1)^2-1=x^2(x^2-1)\). The quotient is \(\mathbb C[x]/(x^2(x-1)(x+1))\), so the points are \((0,-1),(1,0),(-1,0)\), with lengths \(2,1,1\). The repeated root at zero is tangency, now a genuine local multiplicity rather than two separate points with the same projection.

5. Form the universal matrix \(S\) over \(C\). The vector \(w=\operatorname{adj}(S)e\), where \(e\) represents one, has entries in \(C\). Its first \(n\) entries specify a polynomial of degree less than \(n\), the remaining \(m\) a polynomial of degree less than \(m\). The matrix identity \(Sw=(\det S)e\) is exactly \(uf+vg=\operatorname{Res}(f,g)\). Every specialization is a ring homomorphism preserving this identity. It uses no inverted coefficient; only the interpretation of determinant vanishing as an actual common root can fail after degrees drop.

The historical papers

Read works 22 and 24 in the freely accessible complete English corpus edition, with its editable TeX. Work 22, On the Theory of Polynomial Ideals and Resultants, Sections 2–6, treats modules, norm and elementary-divisor forms. Work 24, Elimination Theory and General Ideal Theory, Sections 5–6, places the dimensional pieces in primary decomposition and states the prime and proper-primary criteria as Theorems XII–XIV. The edition is a machine-assisted working translation; its original and third-party materials retain their own rights. The independent formulation above explains our generic-fibre convention and proves its imperfect-field qualifications; it does not identify every modern determinant with every historical normalization.

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.