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 In particular, the injective hull $E$ of $\kappa $ is a finite $R$-module.
\[ \text{length}_ R(M) = \text{length}_ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, E)) \]
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
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 is an isomorphism. In particular $R = \mathop{\mathrm{Hom}}\nolimits _ R(E, E)$.
\[ M \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, E), E) \]
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
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. $\square$
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