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Postmodern Algebra

Sam Macdonald

Section 20.3 Canonical Modules

Definition 20.24.

A canonical module over a Cohen-Macaulay local ring \((R, \mathfrak{m}, k)\) is a finitely generated module \(\omega_{R}\) such that \(\operatorname{Hom}_{R}\left(\omega_{R}, E_{R}(k)\right) \cong \mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)\text{.}\)

Example 20.25.

Example 4.24. For a Cohen-Macaulay local ring \((R, \mathfrak{m}, k)\text{,}\) the rank one free module \(R\) is a canonical module if and only if \(R\) is Gorenstein. There are missing footnotes here

Example 20.26.

Example 4.25. If \(R\) is complete local (and CM), then, by Matlis duality, \(\omega_{R}=\mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)^{\vee}\) is noetherian and \(\omega_{R}^{\vee}=\mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)^{\vee \vee}=\mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)\text{,}\) so \(\omega_{R}\) is a canonical module. Moreover, if \(M\) is noetherian and \(M^{\vee} \cong \mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)\text{,}\) then \(M \cong M^{\vee \vee} \cong \mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)^{\vee}=\omega_{R}\text{,}\) so all canonical modules of \(R\) are isomorphic.

Example 20.27.

Example 4.26. If \(R\) is artinian local (and hence complete), then all of \(R\) is \(\mathfrak{m}\)-torsion, so \(R=\) \(\mathrm{H}_{\mathfrak{m}}^{0}(R)\text{.}\) Since \(E_{R}(k)^{\vee} \cong R, E_{R}(k)\) is a canonical module in this case.
In our original proof of local duality in the regular case, we noted that the proof worked in greater generality. Our definition of canonical module is designed to exactly fulfill that situation.

Proof.

This follows from the lemma on the behavior of the injective hull of the residue field under such maps and Hom-tensor adjunction:
\begin{equation*} \operatorname{Hom}_{A}\left(M, E_{A}(l)\right)=\operatorname{Hom}_{A}\left(M \otimes_{R} R, E_{A}(l)\right)=\operatorname{Hom}_{R}\left(M, \operatorname{Hom}_{A}\left(R, E_{A}(l)\right)=\operatorname{Hom}_{R}\left(M, E_{R}(k)\right) .\right. \end{equation*}
These isomorphisms are all valid as \(A\)-modules or as \(R\)-modules.
The following proposition is the key source of canonical modules.

Proof.

Since \(R\) and \(\omega_{A}\) are finitely generated \(A\)-modules, the \(A\)-modules \(\operatorname{Ext}_{A}^{\bullet}\left(R, \omega_{A}\right)\) are noetherian \(A\)-modules. They are \(R\)-modules as well, and noetherian \(R\)-modules as such.
We then apply local duality over \(A\text{,}\) and the previous lemma:
\begin{equation*} \mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(R)=\mathrm{H}_{\mathfrak{n}}^{\operatorname{dim}(R)}(R)=\operatorname{Ext}_{A}^{\operatorname{dim}(A)-\operatorname{dim}(R)}\left(R, \omega_{A}\right)^{\vee_{A}}=\operatorname{Ext}_{A}^{\operatorname{dim}(A)-\operatorname{dim}(R)}\left(R, \omega_{A}\right)^{\vee_{R}} \end{equation*}
Since local duality is functorial, the map \(R \stackrel{\cdot r}{\longrightarrow} R\) for any \(r \in R\) induces the same map on both sides, so this is an isomorphism of \(R\)-modules.

Example 20.32.

Example 4.31. In the first homework, you showed that for \(A=K\left[X_{2 \times 3}\right]\) and \(R=A / I_{2}(X), R\) is Cohen-Macaulay, and \(\operatorname{Ext}_{A}^{2}(R, A)\) is the cokernel of the map \(A^{3} \rightarrow A^{2}\) given by the matrix \(X^{T}\text{.}\) Evidently, this two-generated module is a canonical module for \(R\text{.}\)

Example 20.33.

Example 4.32. Let \(R=K\left[x^{3}, x^{2} y, x y^{2}, y^{3}\right]\text{,}\) and \(A=K\left[x^{3}, y^{3}\right]\text{.}\) We can write \(R=A \oplus x^{2} y A \oplus\) \(x y^{2} A\text{.}\) We have that \(\operatorname{Hom}_{A}(R, A)\) is a canonical module for \(R\text{.}\) To understand this module more concretely, note that \(\operatorname{Hom}_{A}(R, A)\) is generated by the maps \(\phi_{1}, \phi_{x^{2} y}, \phi_{x y^{2}}\) dual to the basis of \(R\) over \(A\) specified above. Evidently, \(\phi_{1}=x^{2} y \cdot \phi_{x^{2} y}=x y^{2} \cdot \phi_{x y^{2}}\text{,}\) so we can ignore \(\phi_{1}\) as a generator. The maps \(\phi_{x^{2} y}, \phi_{x y^{2}}\) both decrease degrees by 3 , so neither is a multiple of the other. One relation between the maps is given above. It is easy to find a few more: \(x^{3} \cdot \phi_{x^{2} y}=x^{2} y \cdot \phi_{x y^{2}}\) and \(x y^{2} \cdot \phi_{x^{2} y}=y^{3} \cdot \phi_{x y^{2}}\text{.}\)
Our next goal is to show that canonical modules are unique up to isomorphism when they exist. We want to do this by reducing to the complete case, where we have already observed this. To this end, we will collect a few facts on completion and Matlis duality.

Proof.

Let \(\phi: \widehat{M} \rightarrow \widehat{N}\) be an isomorphism, in particular, surjective. First, we note that
\begin{equation*} \operatorname{Hom}_{\widehat{R}}(\widehat{M}, \widehat{N}) \cong \widehat{R} \otimes_{R} \operatorname{Hom}_{R}(M, N) \cong \operatorname{Hom}_{R}(M, N) \end{equation*}
Thus, we can pick \(\alpha \in \operatorname{Hom}_{R}(M, N)\) such that \(\phi-\widehat{\alpha} \in \mathfrak{m H o m}_{R}(M, N)\text{.}\) If \(n \in N\) satisfies \(n=\phi\left(m^{\prime}\right)\text{,}\) with \(m^{\prime} \in \widehat{M}\text{,}\) take \(m \in M\) with \(m-m^{\prime} \in \mathfrak{m} \widehat{M}\text{,}\) so \(\alpha(m) \in n+\mathfrak{m} N\text{.}\) Thus, \(N \subseteq \operatorname{im}(\alpha)+\mathfrak{m} N\text{,}\) and hence \(\operatorname{im}(\alpha)=N\) by NAK.
Similarly, we can find a surjection \(\psi: N \rightarrow M\text{,}\) and hence a surjection \(\psi \circ \alpha: M \rightarrow M\text{.}\) Since \(M\) is finitely generated, \(\psi \circ \alpha: M \rightarrow M\) must be an isomorphism, hence injective. Then, \(\alpha: M \rightarrow N\) must be injective too, hence an isomorphism.
  1. We have already seen that \(R^{\vee \vee} \cong \widehat{R}\text{.}\) Then, given a presentation
\begin{equation*} R^{a} \stackrel{A}{\longrightarrow} R^{b} \rightarrow M \rightarrow 0 \end{equation*}
applying double duality gives
and thus \(M^{\vee \vee} \cong \widehat{M}\text{.}\)
  1. Recall that \(E_{R}(k) \cong \widehat{R} \otimes_{R} E_{R}(k) \cong E_{\widehat{R}}(k)\text{.}\) Since \(M^{\vee}\) is artinian, and applying flat base change
\(\operatorname{Hom}_{R}\left(M, E_{R}(k)\right) \cong \widehat{R} \otimes_{R} \operatorname{Hom}_{R}\left(M, E_{R}(k)\right) \cong \operatorname{Hom}_{\widehat{R}}\left(\widehat{R} \otimes_{R} M, \widehat{R} \otimes_{R} E_{R}(k)\right) \cong \operatorname{Hom}_{\widehat{R}}\left(\widehat{M}, E_{\widehat{R}}(k)\right)\text{.}\)

Proof.

The first statement follows immediately from part (3) above.
If \(M\) and \(N\) are canonical modules for \(R\text{,}\) then \(\widehat{M}\) and \(\widehat{N}\) are canonical modules for \(\widehat{R}\text{.}\) As noted earlier, this implies that \(\widehat{M} \cong \widehat{N}\text{,}\) so by part (1) of the previous lemma, \(M \cong N\text{.}\)
Here is one application of the uniqueness of canonical modules.

Proof.

Let \(c=\operatorname{dim}(R)-\operatorname{dim}(R / I)\text{.}\) More naturally, the claim is that \(P_{\bullet} \cong \operatorname{Hom}_{R}\left(P_{\bullet}, R\right)\) after reindexing the latter (changing the indices from cohomological \(0,1, \ldots, c\) to homological \(c, c-\) \(1, \ldots, 0)\text{.}\) The complex \(\operatorname{Hom}_{R}\left(P_{\bullet}, R\right)\) has homology \(\operatorname{Ext}_{R}^{\bullet}(R / I, R)\text{.}\) The complex vanishes past position \(c\) by Auslander-Buchsbaum. The homology below degree \(c\) vanishes by Rees (or local duality). The homology at position \(c\) is a canonical module for \(R / I\text{,}\) so must be \(R / I\) itself, since \(R / I\) is Gorenstein. Thus, \(\operatorname{Hom}_{R}\left(P_{\bullet}, R\right)\) after the reindexing is a minimal free resolution for \(R / I\text{.}\) Any two minimal free resolutions are isomorphic (exercise!).

Proof.

Proof. For the first statement, compute \(\operatorname{Ext}_{R}^{t}(R / \underline{f} R, \omega) \cong H^{t}(\underline{f} ; \omega) \cong \omega_{R} / \underline{f} \omega_{R}\text{.}\)
We prove (2) by induction on \(t\text{.}\) Suppose both statements are true for \(f^{\prime}=f_{1}, \ldots, f_{t-1}\text{,}\) and set \(\bar{M}\) to be \(M / \underline{f}^{\prime} M\text{.}\) Then, \(\operatorname{ann}_{\bar{\omega}}\left(f_{t}\right) \cong \operatorname{Hom}_{\bar{R}}\left(\bar{R} / f_{t} \bar{R}, \bar{\omega}\right)\text{.}\) Applying (1), \(\bar{\omega} \cong \omega_{\bar{R}}\text{.}\) We apply local duality to see that \(\operatorname{Hom}_{\bar{R}}\left(\bar{R} / f_{t} \bar{R}, \bar{\omega}\right)^{\vee} \cong \mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(\bar{R})}\left(\bar{R} / f_{t} \bar{R}\right)=0\text{.}\) Thus, \(f_{t}\) is a nonzerodivisor on \(\bar{\omega}\text{.}\)

Proof.

Proof. Since \(\widehat{R}\) is faithfully flat over \(R\text{,}\) it suffices to check to isomorphism after tensoring with \(\widehat{R}\text{.}\) Since both sides are finitely generated, this is the same as completing, which turns \(R\) into \(\widehat{R}\text{,}\) and \(\operatorname{Hom}_{R}\left(\omega_{R}, \omega_{R}\right)\) into \(\operatorname{Hom}_{\widehat{R}}\left(\omega_{\widehat{R}}, \omega_{\widehat{R}}\right)\text{.}\) Thus, we can assume that \(R\) is complete. Then, if \(f_{1}, \ldots, f_{d}\) is a SOP for \(R\text{,}\) write \(R_{t}=R /\left(f_{1}^{t}, \ldots, f_{d}^{t}\right)\text{.}\) Then \(\omega_{R}=\lim _{\longleftarrow} \omega_{R} \otimes_{R} R_{t}=\lim \omega_{R_{t}}\text{.}\)
We claim that \(\operatorname{Hom}_{R}\left(\omega_{R}, \omega_{R}\right) \cong \lim _{\longleftarrow} \operatorname{Hom}_{R_{t}}\left(\omega_{R_{t}}, \omega_{R_{t}}\right)\text{.}\) Indeed, an \(R\)-linear map \(f: \omega_{R} \rightarrow \omega_{R}\) induces a sequence of maps \(f_{t}=f \otimes_{R} R_{t}: \omega_{R_{t}} \rightarrow \omega_{R_{t}}\) such that \(f_{t+1} \otimes_{R_{t+1}} R_{t}=f_{t}\text{.}\) This gives an element in the inverse limit. Conversely, given such a sequence of maps \(\left\{f_{n}\right\}\) with the same compatibility property, one can define an \(R\)-linear map \(f\) from \(\omega_{R}\) to \(\omega_{R}\) by the rule
\begin{equation*} f(w) \bmod \left(x_{1}^{t}, \ldots, x_{d}^{t}\right) \omega_{R}:=f_{n}(w) \bmod \left(x_{1}^{t}, \ldots, x_{d}^{t}\right) \omega_{R} \end{equation*}
Since \(R_{t}\) is artinian, \(\omega_{R_{t}} \cong E_{R_{t}}\text{.}\) Then, \(\operatorname{Hom}_{R_{t}}\left(E_{R_{t}}, E_{R_{t}}\right) \cong R_{t}\) by the natural map, and passing to the inverse limit gives the isomorphism.
A CM local ring has a canonical module if and only if it is a quotient of a Gorenstein ring. Given a ring \(R\) and a module \(M\text{,}\) the trivial extension or idealization of \(R\) by \(M\) is the \(\operatorname{ring} R \rtimes M\) with \(R\)-module structure \(R \rtimes M \cong R \oplus M\) and multiplication \((r, m)\left(r^{\prime}, m^{\prime}\right)=\left(r r^{\prime}, r m^{\prime}+r^{\prime} m\right)\text{.}\) It is easy to check that this structure makes \(R \rtimes M\) into a ring in which \(0 \oplus M\) is an ideal with square zero, and \(R \cong(R \rtimes M) /(0 \oplus M)\text{.}\) If \((R, \mathfrak{m})\) is local, then \((R \rtimes M, \mathfrak{m} \oplus M)\) is local as well.

Proof.

Proof. Note that \(R \rtimes \omega_{R}\) is a finitely generated \(R\)-module. Since a regular sequence on \(R\) is also regular on \(\omega_{R}\text{,}\) this ring is also Cohen-Macaulay.
We compute a canonical module for \(R \rtimes \omega_{R}\) as \(\operatorname{Hom}_{R}\left(R \rtimes \omega_{R}, \omega_{R}\right)\text{.}\) As an \(R\)-module, this Hom is isomorphic to \(\operatorname{Hom}_{R}\left(R, \omega_{R}\right) \oplus \operatorname{Hom}_{R}\left(\omega_{R}, \omega_{R}\right) \cong \omega \oplus R\text{.}\) I’ll leave to you as an exercise the routine check that the module structure on this \(\omega \oplus R\) induced by the premultiplication action on \(\operatorname{Hom}_{R}\left(R \rtimes \omega_{R}, \omega_{R}\right)\) agrees with the structure of \(R \rtimes \omega_{R}\text{.}\)
We now show that canonical modules localize. We could work with the construction above to do this, but we will give a slightly more flexible proof. We prepare for this with a lemma.

Proof.

Proof. Since height \((I)=\min \{\operatorname{height}(\mathfrak{p}) \mid \mathfrak{p} \in \operatorname{Min}(I)\}\) and \(\operatorname{dim}(R / I)=\max \{\operatorname{dim}(R / \mathfrak{p}) \mid \mathfrak{p} \in\) \(\operatorname{Min}(I)\}\text{,}\) it suffices to show the equality for prime ideals, so let \(\mathfrak{p}\) be prime. By the worksheet on CM rings, we know that height \((\mathfrak{p})=\operatorname{depth}_{\mathfrak{p}}(R)\text{,}\) and \(\operatorname{dim}(R)=\operatorname{dim}(R / \mathfrak{q})\) for all \(\mathfrak{q} \in \operatorname{Ass}(R)\text{.}\) Setting \(h=\operatorname{height}(\mathfrak{p})\text{,}\) take a regular sequence \(f_{1}, \ldots, f_{h} \subseteq \mathfrak{p}\text{.}\) Since \(f\) is a regular sequence, height \((f)=h\text{,}\) so \(\mathfrak{p}\) must be a minimal prime of \((f)\text{.}\) Then \(R /(f)\) is CM of \(\operatorname{dimension} \operatorname{dim}(R)-h\text{.}\) But \(\mathfrak{p} \in \operatorname{Ass}(R /(f))\text{,}\) so \(\operatorname{dim}(R / \mathfrak{p})=\operatorname{dim}(R)-h\text{,}\) as required.

Proof.

Proof. Let \((S, \mathfrak{n}, l)\) be a Gorenstein ring mapping onto \(R\text{,}\) so that \(\omega_{R} \cong \operatorname{Ext}_{S}^{\operatorname{dim}(S)-\operatorname{dim}(R)}(R, S)\text{.}\) Let \(\mathfrak{p} \in \operatorname{Spec}(R)\text{,}\) and \(\mathfrak{q}\) be the contraction of \(\mathfrak{p}\) in \(S\text{.}\) We note that if \(R=S / I\text{,}\) since \(S\) is Cohen-Macaulay, by the previous lemma,
\begin{equation*} \operatorname{dim}\left(S_{\mathfrak{q}}\right)-\operatorname{dim}\left(R_{\mathfrak{p}}\right)=\operatorname{height}_{S_{\mathfrak{q}}}\left(I S_{\mathfrak{q}}\right)=\operatorname{height}_{S}(I)=\operatorname{dim}(S)-\operatorname{dim}(R) \end{equation*}
Then,
\begin{equation*} \left(\omega_{R}\right)_{\mathfrak{p}} \cong\left(\operatorname{Ext}_{S}^{\operatorname{dim}(S)-\operatorname{dim}(R)}(R, S)\right)_{\mathfrak{p}} \cong \operatorname{Ext}_{S_{\mathfrak{q}}}^{\operatorname{dim}(S)-\operatorname{dim}(R)}\left(R_{\mathfrak{p}}, S_{\mathfrak{q}}\right) \cong \operatorname{Ext}_{S_{\mathfrak{q}}}^{\operatorname{dim}\left(S_{\mathfrak{q}}\right)-\operatorname{dim}\left(R_{\mathfrak{p}}\right)}\left(R_{\mathfrak{p}}, S_{\mathfrak{q}}\right) \end{equation*}
is a canonical module for \(R_{\mathfrak{p}}\text{,}\) since \(S_{\mathfrak{q}}\) is Gorenstein.

Exercise 20.42.

Exercise 4.41. Show that the minimal injective resolution of a canonical module has the same form as the minimal injective resolution of a RLR we found in the last homework.

Exercise 20.43.

Exercise 4.42. Show that if \((R, \mathfrak{m}, k)\) is a Cohen-Macaulay local ring that is a quotient of a Gorenstein ring, a module \(M\) is a canonical module for \(R\) if and only if \(M\) is finitely generated and
\begin{equation*} \operatorname{Ext}_{R}^{i}(k, M) \cong \begin{cases}k & \text { if } i=\operatorname{dim}(R) \\ 0 & \text { if } i \neq \operatorname{dim}(R)\end{cases} \end{equation*}

Proof.

We note that if \(R\) is reduced, and \(\mathfrak{p} \in \operatorname{Min}(R)\text{,}\) then \(R_{\mathfrak{p}}\) is a field, so "e.g." is valid.
First, we show that \(\omega_{R}\) is isomorphic to an ideal. The point is that \(\left(\omega_{R}\right)_{\mathfrak{p}} \cong \omega_{R_{\mathfrak{p}}} \cong R_{\mathfrak{p}}\) for all minimal primes \(\mathfrak{p}\) of \(R\text{,}\) using the exercise above and the hypothesis. If \(W\) is the set of nonzerodivisors of \(R\text{,}\) which is the same as the set of nonzerodivisors on \(\omega_{R}\text{,}\) then \(W^{-1} \omega_{R} \cong\) \(\prod \omega_{R_{\mathfrak{p}}} \cong \prod R_{\mathfrak{p}} \cong W^{-1} R\text{.}\) Then, an isomorphism \(W^{-1} \omega_{R} \cong W^{-1} R\) restricts to an injection from \(\alpha: \omega_{R} \hookrightarrow W^{-1} R\text{.}\) If \(w_{1}, \ldots, w_{t}\) is a generating set for \(\omega_{R}\text{,}\) and \(\alpha\left(w_{i}\right)=r_{i} / s_{i}\text{,}\) then \(s_{1} \cdots s_{t} \alpha\) is an injective map from \(\omega_{R}\) to \(R\text{.}\)
Now, we show that such an ideal has all associated primes of height one. To obtain a contradiction, suppose we have some \(\mathfrak{q} \in \operatorname{Ass}(R / I)\) of height at least two. We have that \(R_{\mathfrak{q}}\) is \(\mathrm{CM}\) of dimension at least two, with canonical module \(I R_{\mathfrak{q}}\text{,}\) so \(\operatorname{depth}\left(I R_{\mathfrak{q}}\right) \geq 2\text{.}\) On the other hand, \(\operatorname{depth}\left(R_{\mathfrak{q}} / I R_{\mathfrak{q}}\right)=0\text{,}\) and \(\operatorname{depth}\left(R_{\mathfrak{q}}\right) \geq 2\text{,}\) so \(\operatorname{depth}\left(I R_{\mathfrak{q}}\right)=1\text{.}\) This is the desired contradiction.
We recall the following fact about factorization:

Exercise 20.45.

Exercise 4.44. If \(R\) is a UFD, and \(I \subseteq R\) is such that every associated prime of \(R / I\) has height one, then \(I\) is principal.
The following is now evident.

Proof.

\(R\) has a canonical module isomorphic to an unmixed height one ideal, which is necessarily principal. Thus, the canonical module is isomorphic to \(R\) itself.
It turns out that one cannot drop the Cohen-Macaulay hypothesis in the previous theorem.

Example 20.47.

Example 4.46. Let \(S=\mathbb{F}_{2}\left[x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}\right]\text{,}\) and \(\mathfrak{n}=\left(x_{1}, \ldots, y_{3}\right)\text{.}\) Let \(G=\{e, \sigma\}\) be a group of order two, where \(e\) is the identity. Let \(G\) act on \(S\) by the rule \(\sigma\left(x_{i}\right)=y_{i}\) and \(\sigma\left(y_{i}\right)=x_{i}\) for all \(i\text{.}\) Consider the ring of invariants \(S^{G}\) with maximal ideal \(\mathfrak{m}\text{,}\) the contraction of \(\mathfrak{n}\text{.}\) We claim that \(\left(S_{\mathfrak{m}}^{G}, \mathfrak{m}, \mathbb{F}_{2}\right)\) is a UFD that is not Cohen-Macaulay.
\(S_{\mathfrak{m}}^{G}\) is a UFD: We show that \(S^{G}\) is a UFD. Consider an element \(f \in S^{G}\text{.}\) Since \(S\) is a UFD, \(f\) admits a factorization into \(S\)-irreducibles:
\begin{equation*} f=a_{1} \cdots a_{t} \end{equation*}
Then,
\begin{equation*} a_{1} \cdots a_{t}=f=\sigma(f)=\sigma\left(a_{1}\right) \cdots \sigma\left(a_{t}\right) . \end{equation*}
We say that such an \(R\) is generically Gorenstein
Each \(\sigma\left(a_{i}\right)\) must be irreducible, since one could apply \(\sigma\) to a nontrivial factorization to get a nontrivial factorization of \(a_{i}\text{.}\) Thus, there is some \(\tau \in \mathcal{S}_{t}\) such that \(\sigma\left(a_{i}\right)=u_{i} a_{\tau(i)}\) for some units \(u_{i}\text{.}\) But, the only unit in \(S\) is 1 . Thus, we can regroup the \(a_{i}\) ’s in such a way that
\begin{equation*} f=b_{1} \cdots b_{r} \cdot c_{1} \cdots c_{s} \cdot \sigma\left(c_{1}\right) \cdots \sigma\left(c_{s}\right) \end{equation*}
with all \(b_{i}\) ’s and \(c_{i}\) ’s irreducible in \(S, \sigma\left(b_{i}\right)=b_{i}\text{,}\) and \(\sigma\left(c_{i}\right) \neq c_{i}\text{.}\) In particular, any irreducible element of \(S^{G}\) must be of the form \(b_{1}\) or \(\left(c_{1} \sigma\left(c_{1}\right)\right)\text{.}\) It is then clear that any factorization of \(f\) into \(S^{G}\)-irreducibles must be of the (unique!) form
\begin{equation*} f=b_{1} \cdots b_{r} \cdot\left(c_{1} \sigma\left(c_{1}\right)\right) \cdots\left(c_{s} \sigma\left(c_{s}\right)\right) . \end{equation*}
\(S_{\mathfrak{m}}^{G}\) is Cohen-Macaulay: By the Theorem on symmetric polynomials, \(\mathbb{F}_{2}\left[x_{1}, y_{1}\right]\) is rank two free module over the polynomial subring \(\mathbb{F}_{2}\left[x_{1}+y_{1}, x_{1} y_{1}\right]\text{.}\) After self-tensoring, we see that \(S\) is a rank eight free module over the polynomial subring \(A=\mathbb{F}_{2}\left[\left\{x_{i}+y_{i}, x_{i} y_{i} \mid i=1,2,3\right\}\right]\text{.}\) Since \(A\) consists of invariants, \(A \subseteq S^{G} \subseteq S\text{,}\) so \(A\) is a Noether normalization of \(S^{G}\text{.}\) Hence, \(\left\{x_{i}+y_{i}, x_{i} y_{i} \mid i=1,2,3\right\}\) is a SOP for \(S_{\mathfrak{m}}^{G}\text{.}\)
We claim that \(x_{1}+y_{1}, x_{2}+y_{2}, x_{3}+y_{3}\) is not a regular sequence in \(S_{\mathfrak{m}}^{G}\text{.}\) Indeed, we have the relation
\begin{equation*} \left(x_{1}+y_{1}\right)\left(x_{2} y_{3}+x_{3} y_{2}\right)+\left(x_{2}+y_{2}\right)\left(x_{1} y_{3}+x_{3} y_{1}\right)+\left(x_{3}+y_{3}\right)\left(x_{1} y_{2}+x_{2} y_{1}\right)=0 \end{equation*}
since every monomial in the expansion appears twice and \(S^{G}\) has characteristic two. Each of the (-) elements above invariant, so this is a relation in \(S^{G}\text{,}\) and gives a relation in \(S_{\mathfrak{m}}^{G}\text{.}\)
Now, we observe that the set of linear forms fixed by the group action is generated over \(\mathbb{F}_{2}\) by \(x_{1}+y_{1}, x_{2}+y_{2}, x_{3}+y_{3}\text{.}\) The 2 -form \(x_{1} y_{2}+x_{2} y_{1}\) cannot be generated by these linear forms, so \(x_{1} y_{2}+x_{2} y_{1} \notin\left(x_{1}+y_{1}, x_{2}+y_{2}\right) S^{G}\text{.}\) We claim that \(x_{1} y_{2}+x_{2} y_{1} \notin\left(x_{1}+y_{1}, x_{2}+y_{2}\right) S_{\mathfrak{m}}^{G}\) as well; we leave this as an easy exercise for you. Then, we conclude that \(x_{3}+y_{3}\) is a zerodivisor in \(S_{\mathfrak{m}}^{G} /\left(x_{1}+y_{1}, x_{2}+y_{2}\right) S_{\mathfrak{m}}^{G}\text{.}\) Therefore, \(S_{\mathfrak{m}}^{G}\) is not Cohen-Macaulay.

Example 20.48.

Example 4.47. We now give a brief indication of a \(\mathrm{CM}\) local ring with no canonical module. The hard work for this example is due to Ferrand and Raynaud, Fibres formelles d’un anneau locale nothérian, who give an example of a 1-dimensional local domain \((R, \mathfrak{m}, k)\) such that \(\widehat{R}\) is not generically Gorenstein: for some minimal prime \(\mathfrak{p}\) of \(\widehat{R}, \widehat{R}_{\mathfrak{p}}\) is not Gorenstein.
Note that \(R\) is Cohen-Macaulay, simply because it contains a nonzerodivisor. Suppose, that \(R\) has a canonical module \(\omega\text{.}\) Then, since \(R\) is a domain, and hence generically Gorenstein, there is an ideal \(I\) of pure height one, which hence is \(\mathfrak{m}\)-primary, such that \(\omega \cong I\text{.}\) Then, \(\widehat{R}\) is CM, with canonical module \(\widehat{\omega} \cong \widehat{I}\text{,}\) which is also \(\mathfrak{m}\)-primary. Thus, \(\widehat{I}\) is not contained in any \(\mathfrak{p} \in \operatorname{Min}(\widehat{R})\text{.}\) Thus, for \(\mathfrak{p} \in \operatorname{Min}(\widehat{R})\text{,}\) we have
\begin{equation*} \widehat{R}_{\mathfrak{p}}=\widehat{I}_{\mathfrak{p}} \cong \widehat{\omega}_{\mathfrak{p}} \end{equation*}
which contradicts the fact that \(\widehat{R}\) is not generically Gorenstein.
Later examples of T. Ogoma exhibit Cohen-Macaulay UFDs that are not Gorenstein, hence, that have no canonical module.
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