Let $R=\underleftarrow\lim (R/\mathfrak m^i)$ be a complete local ring, with residue field $k=R/\mathfrak m$, and let's assume that $R$ is Noetherian.

If $R$ is a $k$-algebra, then I believe that the following is correct:

The injective hull of $k$ can be described as the set of continuous homomorphisms $$ E(k) = \hom_{\mathrm{cts}}(R,k)=\underrightarrow\lim (\hom(R/\mathfrak m^i,k)), $$ where continuity is with respect to the $\mathfrak m$-adic topology.

Let me now assume that we're in mixed characteristic ($R$ is no longer a $k$-algebra).

Question: Is there a description of $E(k)$ along the same lines as above?
In particular, does the automorphism group of $R$ act on $E(k)$?


1 Answer 1


I believe that the following works in reasonable generality, at least if $R$ is regular, although I am not sure of the precise minimal hypotheses.

Let $n$ be the Krull dimension of $R$. The top local cohomology $H^n_{\mathfrak{m}}(R)$ is canonically isomorphic to $\underset{\to k}{\lim}\text{Ext}^n(R/\mathfrak{m}^k,R)$ and so is natural for isomorphisms of $R$. However, $\text{Aut}(R)$ does not act on the stable Koszul complex that is usually used to compute local cohomology groups, so the action on those groups is quite hard to see (and easy to misunderstand) from that point of view.

There is usually a canonical "coefficient ring" $W\leq R$; at least when $k$ is finite, it is something like the integral closure of the $\mathfrak{m}$-adic completion of the image of $\mathbb{Z}$ in $R$. The module $\Omega^1_{R/W}$ of Kähler differentials is usually free of rank $n-1$ over $R$, so the top exterior power $\Omega^{n-1}_{R/W}$ is an invertible $R$-module. The tensor product $E=\Omega^{n-1}_{R/W}\otimes H^n_{\mathfrak{m}}(R)$ is then an injective $R$-module. There is a canonical residue map $\text{res}\colon E\to k$, which restricts to give an isomorphism $\text{socle}(E)\to k$. By inverting this, we obtain an embedding $\eta\colon k\to E$ of $R$-modules. This exhibits $E$ as an injective hull of $k$. The whole story is $\text{Aut}(R)$-equivariant.


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