Special case of [Theorem 2.4, Lyubeznik]

Lemma 51.18.1. Let $k$ be a field of characteristic $0$. Let $d \geq 1$. Let $A = k[[x_1, \ldots , x_ d]]$ with maximal ideal $\mathfrak m$. Let $M$ be an $\mathfrak m$-power torsion $A$-module endowed with additive operators $D_1, \ldots , D_ d$ satisfying the leibniz rule

$D_ i(fz) = \partial _ i(f) z + f D_ i(z)$

for $f \in A$ and $z \in M$. Here $\partial _ i$ is differentiation with respect to $x_ i$. Then $M$ is isomorphic to a direct sum of copies of the injective hull $E$ of $k$.

Proof. Choose a set $J$ and an isomorphism $M[\mathfrak m] \to \bigoplus _{j \in J} k$. Since $\bigoplus _{j \in J} E$ is injective (Dualizing Complexes, Lemma 47.3.7) we can extend this isomorphism to an $A$-module homomorphism $\varphi : M \to \bigoplus _{j \in J} E$. We claim that $\varphi$ is an isomorphism, i.e., bijective.

Injective. Let $z \in M$ be nonzero. Since $M$ is $\mathfrak m$-power torsion we can choose an element $f \in A$ such that $fz \in M[\mathfrak m]$ and $fz \not= 0$. Then $\varphi (fz) = f\varphi (z)$ is nonzero, hence $\varphi (z)$ is nonzero.

Surjective. Let $z \in M$. Then $x_1^ n z = 0$ for some $n \geq 0$. We will prove that $z \in x_1M$ by induction on $n$. If $n = 0$, then $z = 0$ and the result is true. If $n > 0$, then applying $D_1$ we find $0 = n x_1^{n - 1} z + x_1^ nD_1(z)$. Hence $x_1^{n - 1}(nz + x_1D_1(z)) = 0$. By induction we get $nz + x_1D_1(z) \in x_1M$. Since $n$ is invertible, we conclude $z \in x_1M$. Thus we see that $M$ is $x_1$-divisible. If $\varphi$ is not surjective, then we can choose $e \in \bigoplus _{j \in J} E$ not in $M$. Arguing as above we may assume $\mathfrak m e \subset M$, in particular $x_1 e \in M$. There exists an element $z_1 \in M$ with $x_1 z_1 = x_1 e$. Hence $x_1(z_1 - e) = 0$. Replacing $e$ by $e - z_1$ we may assume $e$ is annihilated by $x_1$. Thus it suffices to prove that

$\varphi [x_1] : M[x_1] \longrightarrow \left(\bigoplus \nolimits _{j \in J} E\right)[x_1] = \bigoplus \nolimits _{j \in J} E[x_1]$

is surjective. If $d = 1$, this is true by construction of $\varphi$. If $d > 1$, then we observe that $E[x_1]$ is the injective hull of the residue field of $k[[x_2, \ldots , x_ d]]$, see Dualizing Complexes, Lemma 47.7.1. Observe that $M[x_1]$ as a module over $k[[x_2, \ldots , x_ d]]$ is $\mathfrak m/(x_1)$-power torsion and comes equipped with operators $D_2, \ldots , D_ d$ satisfying the displayed Leibniz rule. Thus by induction on $d$ we conclude that $\varphi [x_1]$ is surjective as desired. $\square$

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).