The Stacks project

Proof. By Lemma 47.15.8 the complex $\omega _ B^\bullet $ is dualizing for $B$. We have

by Lemma 47.13.1. Since $\kappa '$ is isomorphic to a finite direct sum of copies of $\kappa $ as an $A$-module and since $\omega _ A^\bullet $ is normalized, we see that this complex only has cohomology placed in degree $0$. Thus $\omega _ B^\bullet $ is a normalized dualizing complex as well. $\square$

Proof. Special case of Lemma 47.16.1. $\square$

Proof. Since $\kappa $ is the unique simple object of the category we have $F(\kappa ) \cong \kappa $. Since our category is abelian, we find that $F$ is exact. Hence $F(E)$ has the same length as $E$ for all finite length modules $E$. Since $\mathop{\mathrm{Hom}}\nolimits (E, \kappa ) = \mathop{\mathrm{Hom}}\nolimits (F(E), F(\kappa )) \cong \mathop{\mathrm{Hom}}\nolimits (F(E), \kappa )$ we conclude from Nakayama's lemma that $E$ and $F(E)$ have the same number of generators. Hence $F(A/\mathfrak m^ n)$ is a cyclic $A$-module. Pick a generator $e \in F(A/\mathfrak m^ n)$. Since $F$ is $A$-linear we conclude that $\mathfrak m^ n e = 0$. The map $A/\mathfrak m^ n \to F(A/\mathfrak m^ n)$ has to be an isomorphism as the lengths are equal. Pick an element

which maps to a generator for all $n$ (small argument omitted). Then we obtain a system of isomorphisms $A/\mathfrak m^ n \to F(A/\mathfrak m^ n)$ compatible with all $A$-module maps $A/\mathfrak m^ n \to A/\mathfrak m^{n'}$ (by $A$-linearity of $F$ again). Since any finite length module is a cokernel of a map between direct sums of cyclic modules, we obtain the isomorphism of the lemma. $\square$

Proof. By induction on the length of $N$ we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(N, \omega _ A^\bullet )$ is a module of finite length sitting in degree $0$. Thus $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ induces an anti-equivalence on the category of finite length modules. Since the same is true for $\mathop{\mathrm{Hom}}\nolimits _ A(-, E)$ by Proposition 47.7.8 we see that

is an equivalence as in Lemma 47.16.3. Hence it is isomorphic to the identity functor. Since $\mathop{\mathrm{Hom}}\nolimits _ A(-, E)$ applied twice is the identity (Proposition 47.7.8) we obtain the statement of the lemma. $\square$

Proof. We prove this by induction on $d$. If $d = 0$, this follows from Lemma 47.16.4 and Matlis duality (Proposition 47.7.8) which guarantees that $\mathop{\mathrm{Hom}}\nolimits _ A(M, E)$ is nonzero if $M$ is nonzero.

Assume the result holds for modules with support of dimension $< d$ and that $M$ has depth $> 0$. Choose an $f \in \mathfrak m$ which is a nonzerodivisor on $M$ and consider the short exact sequence

Since $\dim (\text{Supp}(M/fM)) = d - 1$ (Algebra, Lemma 10.63.10) we may apply the induction hypothesis. Writing $E^ i = \mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, \omega _ A^\bullet )$ and $F^ i = \mathop{\mathrm{Ext}}\nolimits ^ i_ A(M/fM, \omega _ A^\bullet )$ we obtain a long exact sequence

By induction $E^ i/fE^ i = 0$ for $i + 1 \not\in \{ -\dim (\text{Supp}(M/fM)), \ldots , -\text{depth}(M/fM)\} $. By Nakayama's lemma (Algebra, Lemma 10.20.1) and Algebra, Lemma 10.72.7 we conclude $E^ i = 0$ for $i \not\in \{ -\dim (\text{Supp}(M)), \ldots , -\text{depth}(M)\} $. Moreover, in the boundary case $i = - \text{depth}(M)$ we deduce that $E^ i$ is nonzero as $F^{i + 1}$ is nonzero by induction. Since $E^ i/fE^ i \subset F^{i + 1}$ we get

(see lemma used above) we also obtain the dimension estimate (2).

If $M$ has depth $0$ and $d > 0$ we let $N = M[\mathfrak m^\infty ]$ and set $M' = M/N$ (compare with Lemma 47.11.6). Then $M'$ has depth $> 0$ and $\dim (\text{Supp}(M')) = d$. Thus we know the result for $M'$ and since $R\mathop{\mathrm{Hom}}\nolimits _ A(N, \omega _ A^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _ A(N, E)$ (Lemma 47.16.4) the long exact cohomology sequence of $\mathop{\mathrm{Ext}}\nolimits $'s implies the result for $M$. $\square$

Proof. We will use the results of Lemma 47.16.5 without further mention. Fix a finite module $M$. If $M$ is Cohen-Macaulay, then only $\mathop{\mathrm{Ext}}\nolimits ^{-d}_ A(M, \omega _ A^\bullet )$ can be nonzero, hence (1) $\Rightarrow $ (3). The implication (3) $\Rightarrow $ (2) is immediate. Assume (2) and let $N = \mathop{\mathrm{Ext}}\nolimits ^{-\delta }_ A(M, \omega _ A^\bullet )$ be the nonzero $\mathop{\mathrm{Ext}}\nolimits $ where $\delta = \text{depth}(M)$. Then, since

(Lemma 47.15.3) we conclude that $M = \mathop{\mathrm{Ext}}\nolimits _ A^{-\delta }(N, \omega _ A^\bullet )$. Thus $\delta \geq \dim (\text{Supp}(M))$. However, since we also know that $\delta \leq \dim (\text{Supp}(M))$ (Algebra, Lemma 10.72.3) we conclude that $M$ is Cohen-Macaulay.

To prove the final statement, it suffices to show that $N = \mathop{\mathrm{Ext}}\nolimits ^{-d}_ A(M, \omega _ A^\bullet )$ is in $CM_ d$ for $M$ in $CM_ d$. Above we have seen that $M[0] = R\mathop{\mathrm{Hom}}\nolimits _ A(N[d], \omega _ A^\bullet )$ and this proves the desired result by the equivalence of (1) and (3). $\square$

Proof. Immediate from Lemma 47.16.4. $\square$

Proof. Use the short exact sequence $0 \to I \to A \to B \to 0$ and Lemmas 47.16.4 and 47.16.2. $\square$

Proof. Use the short exact sequence $0 \to A \to A \to B \to 0$ and Lemma 47.16.2. $\square$

Proof. Since $A_\mathfrak p$ has dimension zero, there exists an integer $n > 0$ such that $\mathfrak p^ nA_\mathfrak p$ is zero. Set $B = A/\mathfrak p^ n$ and $\omega _ B^\bullet = R\mathop{\mathrm{Hom}}\nolimits _ A(B, \omega _ A^\bullet )$. Since $B_\mathfrak p = A_\mathfrak p$ we see that

The second equality holds by More on Algebra, Lemma 15.99.2. By Lemma 47.16.2 we may replace $A$ by $B$. After doing so, we see that $\dim (A) = e$. Then we see that $H^ i(\omega _ A^\bullet )_\mathfrak p$ can only be nonzero if $i = -e$ by Lemma 47.16.5 parts (1) and (2). On the other hand, since $(\omega _ A^\bullet )_\mathfrak p$ is a dualizing complex for the nonzero ring $A_\mathfrak p$ (Lemma 47.15.6) we see that the remaining module has to be nonzero. $\square$