The Stacks project

47.6 Duality over Artinian local rings

Let $(R, \mathfrak m, \kappa )$ be an artinian local ring. Recall that this implies $R$ is Noetherian and that $R$ has finite length as an $R$-module. Moreover an $R$-module is finite if and only if it has finite length. We will use these facts without further mention in this section. Please see Algebra, Sections 10.52 and 10.53 and Algebra, Proposition 10.60.7 for more details.

Lemma 47.6.1. Let $(R, \mathfrak m, \kappa )$ be an artinian local ring. Let $E$ be an injective hull of $\kappa $. For every finite $R$-module $M$ we have

\[ \text{length}_ R(M) = \text{length}_ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, E)) \]

In particular, the injective hull $E$ of $\kappa $ is a finite $R$-module.

Proof. Because $E$ is an essential extension of $\kappa $ we have $\kappa = E[\mathfrak m]$ where $E[\mathfrak m]$ is the $\mathfrak m$-torsion in $E$ (notation as in More on Algebra, Section 15.88). Hence $\mathop{\mathrm{Hom}}\nolimits _ R(\kappa , E) \cong \kappa $ and the equality of lengths holds for $M = \kappa $. We prove the displayed equality of the lemma by induction on the length of $M$. If $M$ is nonzero there exists a surjection $M \to \kappa $ with kernel $M'$. Since the functor $M \mapsto \mathop{\mathrm{Hom}}\nolimits _ R(M, E)$ is exact we obtain a short exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(\kappa , E) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, E) \to \mathop{\mathrm{Hom}}\nolimits _ R(M', E) \to 0. \]

Additivity of length for this sequence and the sequence $0 \to M' \to M \to \kappa \to 0$ and the equality for $M'$ (induction hypothesis) and $\kappa $ implies the equality for $M$. The final statement of the lemma follows as $E = \mathop{\mathrm{Hom}}\nolimits _ R(R, E)$. $\square$

Lemma 47.6.2. Let $(R, \mathfrak m, \kappa )$ be an artinian local ring. Let $E$ be an injective hull of $\kappa $. For any finite $R$-module $M$ the evaluation map

\[ M \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, E), E) \]

is an isomorphism. In particular $R = \mathop{\mathrm{Hom}}\nolimits _ R(E, E)$.

Proof. Observe that the displayed arrow is injective. Namely, if $x \in M$ is a nonzero element, then there is a nonzero map $Rx \to \kappa $ which we can extend to a map $\varphi : M \to E$ that doesn't vanish on $x$. Since the source and target of the arrow have the same length by Lemma 47.6.1 we conclude it is an isomorphism. The final statement follows on taking $M = R$. $\square$

To state the next lemma, denote $\text{Mod}^{fg}_ R$ the category of finite $R$-modules over a ring $R$.

Lemma 47.6.3. Let $(R, \mathfrak m, \kappa )$ be an artinian local ring. Let $E$ be an injective hull of $\kappa $. The functor $D(-) = \mathop{\mathrm{Hom}}\nolimits _ R(-, E)$ induces an exact anti-equivalence $\text{Mod}^{fg}_ R \to \text{Mod}^{fg}_ R$ and $D \circ D \cong \text{id}$.

Proof. We have seen that $D \circ D = \text{id}$ on $\text{Mod}^{fg}_ R$ in Lemma 47.6.2. It follows immediately that $D$ is an anti-equivalence. $\square$

Lemma 47.6.4. Assumptions and notation as in Lemma 47.6.3. Let $I \subset R$ be an ideal and $M$ a finite $R$-module. Then

\[ D(M[I]) = D(M)/ID(M) \quad \text{and}\quad D(M/IM) = D(M)[I] \]

Proof. Say $I = (f_1, \ldots , f_ t)$. Consider the map

\[ M^{\oplus t} \xrightarrow {f_1, \ldots , f_ t} M \]

with cokernel $M/IM$. Applying the exact functor $D$ we conclude that $D(M/IM)$ is $D(M)[I]$. The other case is proved in the same way. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08YW. Beware of the difference between the letter 'O' and the digit '0'.