Draft lesson.

Weyl polynomial modules and their Koszul calculations

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

The tensor product of polynomial densities and polynomial functions can be zero even though their derived tensor product is not. The missing class occurs in the number of variables, whereas the self-Ext of the function module occurs in degree zero. We compute both using the same finite free resolution. The side of each module and the sign of differentiation determine these degrees and maps.

Use the universe conventions. Let \(k\) be a field and \(n\geq0\) an integer. All modules are small in the fixed universe. Sections 1–4 work over every field; characteristic zero is imposed in Sections 5–6. Write \(A=W_n(k)\) for the unital algebra with generators \(x_1,\ldots,x_n,\partial_1,\ldots,\partial_n\) and relations

For \(n=0\), this means \(A=k\). Put \(E=k^n\), with the ordered basis \(e_1,\ldots,e_n\). Exterior powers are alternating, including in characteristic two. We use cochain differentials of degree \(+1\). A chain resolution \(P_p\) will be viewed as the cochain complex \(P^{-p}=P_p\), with its same boundary.

The complete signs and regularity criterion are in Determinant signs and subset Koszul complexes, Sections 1–5. The normalized Hom differential and the single projective computations of derived Hom and tensor are in Derived Hom and bounded resolution pairs, Sections 1–2 and 6–7. Signed bifunctors and sum and product total complexes, Sections 4 and 6–8, gives their total and balanced tensor conventions. We also prove the finite-resolution exactness and the contractions needed for the present calculations directly.

1. Two ordered bases, without a faithfulness assumption

Proposition 1.1. Both families

are \(k\)-bases of \(A\), over every field.

Proof. To span by the first family, commute an out-of-order adjacent pair \(\partial_i x_j\) using (0.1). The swapped term has one fewer inversion between a derivative and a variable, and the scalar term has shorter word length. Induction first on length and then on this inversion count terminates the procedure. The variables commute among themselves, as do the derivatives, so each resulting word has the indicated multi-index form.

For independence, let \(A\) act on \(k[u_1,\ldots,u_n,z_1,\ldots,z_n]\): \(x_i\) acts by multiplication by \(u_i\), and \(\partial_i\) acts by \(\partial/\partial u_i+z_i\). The latter operators commute; their commutator with multiplication by \(u_j\) is \(\delta_{ij}\), so the defining relations hold. Since differentiation in \(u_i\) kills every polynomial in the \(z\)'s,

A linear relation between the ordered monomials would therefore give a relation between distinct monomials of an ordinary polynomial ring. Every coefficient is zero. This representation is used to prove independence, not assumed faithful in advance; the argument remains valid in positive characteristic.

Define an anti-involution \(t:A\to A\) by reversing products and setting \(t(x_i)=x_i\), \(t(\partial_i)=-\partial_i\). The variable and derivative commutators remain zero. The image of the mixed relation is

Thus the map on the free algebra descends. Its square fixes every generator, hence is the identity. Applying this invertible linear map to the first basis gives \((-1)^{|\beta|}\partial^\beta x^\alpha\), so the second family is also a basis. The signs are units in every field. \(\square\)

In particular \(A\), as a right module over \(k[\partial_1,\ldots,\partial_n]\), is free with basis \(x^\alpha\). This assertion means that every operator has one unique finite expression \(\sum_\alpha x^\alpha q_\alpha(\partial)\). It does not make that polynomial subalgebra central in \(A\).

2. Functions on the left and densities on the right

Let

Here \(J_L\) is a left ideal and \(L\) is a left module; \(J_R\) is a right ideal and \(R\) is a right module. They are not being declared two-sided ideals. In either sum the empty case is zero.

On \(k[x_1,\ldots,x_n]\), multiplication by \(x_i\) and differentiation by \(\partial_i\) satisfy (0.1). The map \(A\to k[x]\), \(a\mapsto a(1)\), is a left-module map from the left regular module to this representation. In the first basis (1.1), its kernel is exactly the span of monomials with \(\beta\ne0\), which is \(J_L\); its image contains every \(x^\alpha\). Consequently

Transport that representation through the anti-involution to obtain a right action \(f\cdot a=t(a)f\). Indeed \((f\cdot a)\cdot b=t(b)t(a)f=t(ab)f=f\cdot(ab)\). We write its underlying polynomials as \(\rho f\), where \(\rho\) is the class of \(1\) and fixes a generator. The second basis (1.1) proves, by applying \(a\mapsto\rho\cdot a\), that its kernel is \(J_R\), so

For example, the right mixed relation acts as

This also checks the minus in (2.3) directly. The term “density” names this specified right module; no integration or analytic completion is used. The polynomial representations in (2.2) and (2.3) need not be faithful representations of the whole algebra. Neither quotient is being used as an algebra quotient.

3. Regular right multiplications and the top quotient

Consider the commuting endomorphisms \(r_i:A\to A\), \(a\mapsto a\partial_i\), of the left regular module. They are left \(A\)-linear: \(r_i(ba)=br_i(a)\). Commutativity of the derivatives gives their commutativity and makes \(r_i\) preserve every earlier image sum.

Theorem 3.1. The list \(r_1,\ldots,r_n\) is an ordered regular sequence of left-module endomorphisms. Its cochain Koszul complex

has zero cohomology except in degree \(n\), where

The given order identifies the last line with \(L\) by \(e_1\wedge\cdots\wedge e_n\mapsto1\); changing the orientation changes that identification by its determinant sign.

Proof. The left ideal \(\sum_{j<i}A\partial_j\) has as a vector-space basis exactly the normal monomials \(x^\alpha\partial^\beta\) with some \(\beta_j>0\), \(j<i\). One inclusion follows by appending \(\partial_j\) to normal words. For the other, factor such a monomial as \((x^\alpha\partial^{\beta-e_j})\partial_j\). Therefore the successive quotient has the basis of normal monomials with \(\beta_j=0\) for every \(j<i\). Appending \(\partial_i\) maps this basis injectively to distinct members of the same quotient basis. This proves the required monicity at every step, including the first. The final quotient is \(A/J_L=L\) with its actual left action, not just a vector space of the same dimension.

Exterior anti-commutation and the commuting \(r_i\) make \(d^2=0\). The complete one-endomorphism criterion in the Koszul prerequisite, Section 4, then proves by induction on these regular prefixes that this cochain complex has only top cohomology. In that top term, each incoming component has image \(A\partial_i\otimes\det E\): its wedge coefficient is invertible on the corresponding determinant line. The incoming image is the sum of those submodules, so its cokernel is exactly (3.2). For \(n=0\), \(K\) is \(k\) in degree zero and the empty determinant is \(k\), giving both assertions. \(\square\)

4. The complete free resolution and why it computes

Put \(P_p=A\otimes_k\bigwedge^pE\), for \(0\le p\le n\), and zero otherwise. Its ordered boundary is

The augmentation \(\epsilon:P_0=A\to L\) is \(a\mapsto a(1)\). All maps are left \(A\)-linear. In the square of the boundary, removing two distinct indices in the two orders gives the same coefficient \(a\partial_i\partial_j\) and opposite exterior signs; removing an index twice contributes zero. Thus \(b^2=0\), and \(\epsilon b_1=0\) by (2.1).

Theorem 4.1. The augmented complex

is exact. Each \(P_p\) is free of rank \(\binom np\) over \(A\). This holds over every field.

Proof. Set \(S=k[t_1,\ldots,t_n]\). By the first ordered basis, the underlying vector complex in (4.2), before augmentation, is

where the latter chain boundary removes \(e_i\) and multiplies by \(t_i\). This identification sends \(x^\alpha\partial^\beta\otimes\omega\) to \(x^\alpha\otimes t^\beta\otimes\omega\). It respects (4.1) because the derivatives are appended on the right. We now verify exactness of this polynomial chain complex without assuming a Koszul theorem for a noncommutative ring.

For one variable let \(U_1=k[t]e\), \(U_0=k[t]\), with boundary \(ge\mapsto tg\). Let \(p\) be evaluation at zero in degree zero and zero in degree one; let \(j:k\to U_0\) be constants. Define \(h:U_0\to U_1\) by

The numerator is divisible by \(t\), so this is a polynomial map over every field. Direct calculation gives \(bh+hb=1-jp\) in both degrees: in degree one, \(h(tg)=ge\). Also \(p\) and \(j\) are chain maps, with \(pj=1\).

Take the tensor product over \(k\) of the \(n\) one-variable complexes. Its boundary on homogeneous tensors is the sum of the factor boundaries with coefficient \((-1)^{p_1+\cdots+p_{i-1}}\) on factor \(i\). Identify a degree-one choice in factors \(i_1<\cdots<i_p\) with the ordered wedge \(e_{i_1}\wedge\cdots\wedge e_{i_p}\). This differential is precisely the boundary in (4.3).

Here is an explicit contraction of that tensor product. Put \(q_i=j_ip_i\) on factor \(i\), and use

A tensor of graded maps acts with the usual sign: the degree-one \(h_i\) contributes \((-1)^{p_1+\cdots+p_{i-1}}\) on a tensor whose preceding factor degrees are \(p_1,\ldots,p_{i-1}\). Each \(q_i\) has degree zero and is a chain map. In \(bH+Hb\), the boundary terms in factors different from \(i\) cancel by this graded sign; the remaining term is

Summing these terms telescopes to \(1-q_1\otimes\cdots\otimes q_n\). Thus the polynomial complex retracts onto its constants in degree zero. Tensor with \(k[x]\) and obtain homology \(k[x]\) only in degree zero. Its projection is exactly the augmentation \(a\mapsto a(1)\), by the ordered basis. These are vector-space contractions, not asserted to be left \(A\)-linear. Exactness as vector spaces proves exactness of the displayed left-module complex, since all its differentials and augmentation are already left-module maps. Its terms have the claimed explicit free bases. For \(n=0\) the tensor product is \(k\), the augmentation is the identity, and the contraction sum is empty. \(\square\)

To connect this resolution with derived functors, let \(P^{-p}=P_p\). Its only cohomology is \(L\) in degree zero. A finite free module is projective: a map to a quotient lifts by choosing a lift of each of its finitely many basis images. It is also flat, since tensoring with it is a finite direct sum of the other module. For an acyclic complex \(Z\), each column of \(\operatorname{Hom}_A(P,Z)\) is therefore acyclic; each column of \(Z\otimes_AP\) is acyclic as well. A finite column filtration proves that each total complex is acyclic: successively adjoining a column gives a short exact sequence of complexes and its cohomology sequence. There are only finitely many columns, even when \(Z\) is unbounded.

Consequently this finite resolution is both K-projective and K-flat in the corresponding inputs. The single-replacement theorems in the prerequisite apply with the actual augmentation (4.2). In particular

The Hom differential is \(d_Lf-(-1)^qf d_P\) in degree \(q\); tensor has the usual signed total differential. These conventions fix every degree and sign in the following calculations. They also show that negative Tor indices and negative Ext degrees vanish here.

5. Self-Ext is the normalized polynomial de Rham complex

Assume now that \(k\) has characteristic zero. For \(q\geq0\), evaluation on the free generators identifies

Write \(D_i=\partial/\partial x_i\), and let \(\alpha_q\) denote this evaluation identification. For an ordered wedge in \(P_{q+1}\), evaluation of \(f b\) gives the alternating sum of \(D_{i_s}\) applied to its omitted-index coefficient: left \(A\)-linearity gives \(f(\partial_i\otimes\omega)=\partial_i f(1\otimes\omega)\). In our standard cochain Hom convention \(L\) has zero differential, so

Define \(\varepsilon_q=(-1)^{q(q+1)/2}\). The actual chain isomorphism from the Hom complex to this de Rham complex is

Indeed \(\varepsilon_{q+1}(-1)^{q+1}=\varepsilon_q\), which proves \(Jd_{\mathrm{Hom}}=d_{\mathrm{dR}}J\). Each multiplier is its own inverse. Degree zero has multiplier one, retaining the ordinary Hom identification. Thus the identification is a chain isomorphism with its signs, not just a correspondence between the terms of two complexes.

Theorem 5.1. For every integer \(q\),

The map \(k\to\operatorname{Ext}^0_A(L,L)\) sends \(\lambda\) to \(\lambda1_L\), and \(R\operatorname{Hom}_A(L,L)\simeq k\) in degree zero.

Proof. Let \(w_i\) denote wedge multiplication by \(e_i^*\) and \(\iota_i\) contraction by its dual basis vector. On an ordered wedge contraction removes the matching label with coefficient \((-1)^{s-1}\). Expanding on basis wedges gives

The first identity includes the case in which a label is already present: insertion then gives zero in one order and removal followed by insertion supplies the surviving term. Summing the second counts each of the \(q\) present labels once.

Put \(h_0=\sum_i x_i\iota_i\), of cochain degree \(-1\). The polynomial identity \(D_i x_j=x_jD_i+\delta_{ij}\) and (5.5) yield

For a polynomial homogeneous of degree \(m\) with exterior degree \(q\), the right side is \((m+q)1\). The differential decreases polynomial degree by one and increases exterior degree by one; \(h_0\) does the reverse. Both therefore preserve the weight \(m+q\). Every element is a finite sum of such homogeneous components.

On each positive-weight component define \(h=h_0/(m+q)\); set \(h=0\) on weight zero. Every positive integer is invertible in \(k\), which is the exact use of characteristic zero. The weight-zero subspace is the constants in exterior degree zero. Its inclusion \(j:k\to\Omega^\bullet\) and projection \(p:\Omega^\bullet\to k\) are chain maps, and (5.6) gives

This proves the stated chain homotopy equivalence, hence every cohomology group. Under (5.1) in degree zero, a constant coefficient is the left-module map \(a(1)\mapsto a(\lambda)=\lambda a(1)\); it is \(\lambda1_L\). The multiplier in (5.3) is one there. Thus the scalar identification is the actual normalized Hom identification. Negative degrees and degrees above \(n\) are zero terms of the computing complex. For \(n=0\) the complex is already \(k\), and the empty contraction gives the same conclusions. \(\square\)

6. Tensor reverses the degree of the surviving class

The balanced relation \(r\otimes a\otimes\omega=(r\cdot a)\otimes1\otimes\omega\) gives inverse isomorphisms

Using (2.3), the boundary of this homological complex is

where \(\iota_i\) now removes \(e_i\) from \(\bigwedge E\), with the same ordered removal signs. The minus is the right density action. It is not introduced by interchanging two tensor inputs.

Theorem 6.1. For every integer \(p\),

The nonzero line is represented by \(\rho\otimes e_1\wedge\cdots\wedge e_n\); the fixed generator and order identify it with \(k\). In cohomological grading,

In particular the ordinary tensor product is zero when \(n>0\), and is \(k\) when \(n=0\).

Proof. Let \(w_i\) be wedge multiplication by \(e_i\), and define the homological degree-one operator \(S_0=-\sum_i x_iw_i\). The minus occurs in both \(B\) and \(S_0\). The same wedge/contraction identity as (5.5) gives

Here \(\sum_i\iota_iw_i=(n-p)1\) on \(\bigwedge^pE\): a missing label contributes one and a present label contributes zero. This is also obtained by subtracting \(\sum_iw_i\iota_i=p1\) from the sum of the first identity in (5.5). Therefore a component with polynomial degree \(m\) and chain degree \(p\) has weight

Both \(B\) and \(S_0\) preserve this weight, since the former decreases both \(m\) and \(p\) by one and the latter increases both by one. Set \(S=S_0/v\) on positive weights and zero on weight zero. Weight zero consists exactly of constant densities in chain degree \(n\). Let \(j\) include that line and let \(p_{\mathrm{top}}\) project to its constant coefficient, killing all other degrees. They are chain maps: the included constants have zero derivative, and there is no boundary entering degree \(n\). Equation (6.5) now gives

All these maps are \(k\)-linear, which is precisely the target category of the tensor computation. This is a homotopy equivalence with the stated line, rather than merely a vanishing test in selected degrees. Chain degree \(n\) is cochain degree \(-n\), hence the shift \([n]\) in (6.4). There are no terms for negative chain degrees or above \(n\), giving every index in (6.3). The augmentation calculation in (4.6) identifies degree-zero homology with the ordinary balanced tensor product, proving the last assertion, including the empty case. \(\square\)

Example 6.2 (one variable). For \(n=1\), tensor is \(\rho k[x]e\xrightarrow{-D}\rho k[x]\). Differentiation is onto in characteristic zero, because \(x^r\) has antiderivative \(x^{r+1}/(r+1)\). Its kernel is the constants. Thus the surviving density is in chain degree one. The normalized Hom computation is \(k[x]\xrightarrow{D}k[x]e^*\), whose kernel is the constants and whose cokernel is zero. The actual degree-zero standard Hom differential before (5.3) is \(-D\). Keeping these three complexes distinct accounts for the same scalar answer in two different degrees.

One can also see the ordinary tensor vanish directly: \(R\otimes_AL\) is generated by \((\rho f)\otimes1\), since all variable factors in \(L\) can be moved to the right input by balancing. Every \(\rho f\) has the form \((\rho g)\cdot\partial_1\) by integration. Then \((\rho f)\otimes1=(\rho g)\otimes\partial_1(1)=0\). This argument needs \(n>0\) and characteristic zero. The derived top class measures what this ordinary quotient forgets.

7. Four graded exercises with complete solutions

Exercise 1 (10 points). In one variable, decide whether \(J_L=A\partial\) is a two-sided ideal. Compute the left action of \(\partial\) on the class of \(x^3\), and the right action of \(\partial\) on \(\rho x^3\). Check the right mixed relation on \(\rho x^2\), in any characteristic.

Solution. If \(J_L\) were a right ideal as well, then \(\partial x\in J_L\), since \(\partial\in J_L\). Also \(x\partial\in J_L\), and \(\partial x-x\partial=1\) would put \(1\) in \(J_L\). The ordered basis proves that \(1\notin J_L\), so it is not two-sided. In the left quotient, \(\partial x^3=x^3\partial+3x^2\) has class \(3x^2\). In the right quotient the action is \(-3\rho x^2\). These equations remain valid when an integer coefficient becomes zero in the field. On \(\rho x^2\), first differentiation and then multiplication by \(x\) gives \(-2\rho x^2\); first multiplication and then differentiation gives \(-3\rho x^2\). Their difference is \(\rho x^2\), as required. Thus the displayed relation holds also in characteristics two and three.

Exercise 2 (14 points). For \(n=2\) over a field of characteristic zero, write \(b_2\) and \(b_1\) of the free resolution and verify their composite. Write the two density boundaries. Find explicit boundaries for the chain-degree-one cycles \(\rho e_1\) and \(\rho(x_1e_1-x_2e_2)\), and identify the top homology generator.

Solution. The free boundaries are

Their composite is \(a\partial_1\partial_2-a\partial_2\partial_1=0\). Tensor gives

The two asked-for preimages are \(\rho x_2e_{12}\) and \(\rho x_1x_2e_{12}\), respectively. Substitution into \(B_2\) gives exactly the specified cycles, including the negative coefficient on \(e_2\); applying \(B_1\) also gives zero. In chain degree two, a cycle has \(D_1f=D_2f=0\). Over characteristic zero this means \(f\) is constant, by its monomial coefficients. There is no incoming term in degree three. Thus \(\rho e_{12}\) generates top homology. Every degree-one cycle is a boundary by (6.7); the exhibited preimages check this explicitly for the chosen examples. Degree-zero homology is zero, again by (6.7).

Exercise 3 (18 points). Let \(k\) have prime characteristic \(\ell>0\), and take \(n=1\). Using the same free resolution, compute every \(\operatorname{Ext}_A^q(L,L)\) and \(\operatorname{Tor}_p^A(R,L)\). Specify polynomial representatives for the nonzero groups. Explain precisely why the characteristic-zero contraction cannot be used.

Solution. The basis and free-resolution proofs did not divide by any integer, so the resolution remains valid. Differentiation gives \(D(x^a)=a x^{a-1}\). Its kernel has basis \(x^{\ell r}\), \(r\geq0\), and is \(k[x^\ell]\). Its image has basis the monomials \(x^a\) with \(a+1\) not divisible by \(\ell\). Therefore its cokernel has unique representatives in \(x^{\ell-1}k[x^\ell]\). The minus on a differential has no effect on these kernel and cokernel statements.

With the normalized Hom identification (5.3),

The first group acts by multiplication by its indicated polynomial; its derivative is zero, so it commutes with the left action. The second group is a cokernel with the displayed representative subspace. Tensor instead gives

All other integer Ext degrees and Tor indices vanish, since both computing complexes have only degrees zero and one in their respective gradings. These identifications are also linear over the central polynomial ring \(k[x^\ell]\): the commutator \([\partial,x^\ell]=\ell x^{\ell-1}\) is zero. In (5.6), for instance, \(x^\ell\) has positive integer weight \(\ell\), but that scalar is zero in \(k\). The same obstruction occurs in (6.5). A division by that weight is unavailable, and the additional classes above prove that no contraction to the characteristic-zero answer could exist.

Exercise 4 (24 points). Work again in characteristic zero. For \(\lambda,\mu\in k^n\), let \(L_\lambda\) have left actions \(x_i f=x_if\), \(\partial_i f=D_if+\lambda_if\), and let \(R_\mu=\rho k[x]\) have right actions \((\rho f)x_i=\rho x_if\), \((\rho f)\partial_i=\rho(-D_if+\mu_if)\). Compute every \(\operatorname{Ext}_A^q(L_\lambda,L_\mu)\) and \(\operatorname{Tor}_p^A(R_\mu,L_\lambda)\), including \(n=0\). Give a contraction when \(\lambda\ne\mu\).

Solution. Adding the specified scalars to the derivative operators does not change any commutator, so these are actual modules. The automorphism fixing the \(x_i\) and translating the \(\partial_i\) by constants transports the ordered-basis and regularity proofs. In particular \(L_\lambda=A/\sum_iA(\partial_i-\lambda_i)\), and replacing every \(\partial_i\) by \(\partial_i-\lambda_i\) in (4.1) gives its exact free resolution. The right quotient is \(R_\mu=A/\sum_i(\partial_i-\mu_i)A\), by the derivative-first basis.

Put \(c_i=\mu_i-\lambda_i\). The normalized Hom differential is \(\delta=\sum_i(D_i+c_i)w_i\) on \(k[x]\otimes\bigwedge E^*\). The density boundary is \(B=\sum_i(-D_i+c_i)\iota_i\) on \(\rho k[x]\otimes\bigwedge E\). If \(c=0\), these are exactly Sections 5–6. Thus the Ext groups are \(k\) in degree zero and zero elsewhere; the Tor groups are \(k\rho\otimes\det E\) in degree \(n\) and zero elsewhere.

If \(c_j\ne0\), the polynomial operator \(T_j=D_j+c_j\) is invertible, with

Each polynomial is killed by a sufficiently high power of \(D_j\), so the sum is finite on each input. Multiplying the geometric sum in either order by \(T_j\) telescopes to the identity. It commutes with all \(D_i+c_i\), by their commutation with \(D_j\) and the same finite sums. The degree-minus-one operator \(h=T_j^{-1}\iota_j\) consequently satisfies \(\delta h+h\delta=1\), by (5.5). This contracts the entire Hom complex.

For tensor use \(U_j=-D_j+c_j\), whose inverse is the same finite geometric formula with \(D_j\) replaced by \(-D_j\). It commutes with every \(-D_i+c_i\). The degree-one operator \(s=U_j^{-1}w_j\) has \(Bs+sB=1\). Thus every Tor group and every Ext group vanishes when \(\lambda\ne\mu\). No completed power series or infinite sum is required. For \(n=0\) there is only the empty parameter vector, the algebra and both modules are \(k\), and both computations are \(k\) in degree zero. This is the equal-parameter case, so the unequal case has no rank-zero instance.

Freely accessible references

• Claude Sabbah, Introduction to algebraic theory of linear systems of differential equations, Chapter I, Section 1.2, especially Proposition 1.2.3, for the one-variable Weyl algebra, its polynomial action and ordered operators. The independence proof in Section 1 above uses extra polynomial variables and covers every characteristic directly.
• Roman Bezrukavnikov, MIT 18.706, Homework 4, Spring 2012, Problem 1(b), for the polynomial-module Ext calculation and its de Rham/Koszul interpretation. The full resolution, both module sides, both contractions and the exercise proofs here are supplied independently.
• Determinant signs and subset Koszul complexes, Sections 1–5; Derived Hom and bounded resolution pairs, Sections 1–2 and 6–7; and Signed bifunctors and sum and product total complexes, Sections 4 and 6–8, for the complete programme proofs of the regularity, differential, derived-value and module-side interfaces used above.

These are known algebraic calculations.