The Stacks project

51.18 Structure of certain modules

Some results on the structure of certain types of modules over regular local rings. These types of results and much more can be found in [Huneke-Sharp], [Lyubeznik], [Lyubeznik2].

reference

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. $\square$

reference

Lemma 51.18.2. Let $p$ be a prime number. Let $(A, \mathfrak m, k)$ be a regular local ring with $p = 0$. Denote $F : A \to A$, $a \mapsto a^ p$ be the Frobenius endomorphism. Let $M$ be a $\mathfrak m$-power torsion module such that $M \otimes _{A, F} A \cong M$. Then $M$ is isomorphic to a direct sum of copies of the injective hull $E$ of $k$.

Proof. Choose a set $J$ and an $A$-module homorphism $\varphi : M \to \bigoplus _{j \in J} E$ which maps $M[\mathfrak m]$ isomorphically onto $(\bigoplus _{j \in J} E)[\mathfrak m] = \bigoplus _{j \in J} k$. 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. Recall that $F$ is flat, see Lemma 51.17.6. Let $x_1, \ldots , x_ d$ be a minimal system of generators of $\mathfrak m$. Denote

\[ M_ n = M[x_1^{p^ n}, \ldots , x_ d^{p^ n}] \]

the submodule of $M$ consisting of elements killed by $x_1^{p^ n}, \ldots , x_ d^{p^ n}$. So $M_0 = M[\mathfrak m]$ is a vector space over $k$. Also $M = \bigcup M_ n$ by our assumption that $M$ is $\mathfrak m$-power torsion. Since $F^ n$ is flat and $F^ n(x_ i) = x_ i^{p^ n}$ we have

\[ M_ n \cong (M \otimes _{A, F^ n} A)[x_1^{p^ n}, \ldots , x_ d^{p^ n}] = M[x_1, \ldots , x_ d] \otimes _{A, F} A = M_0 \otimes _ k A/(x_1^{p^ n}, \ldots , x_ d^{p^ n}) \]

Thus $M_ n$ is free over $A/(x_1^{p^ n}, \ldots , x_ d^{p^ n})$. A computation shows that every element of $A/(x_1^{p^ n}, \ldots , x_ d^{p^ n})$ annihilated by $x_1^{p^ n - 1}$ is divisible by $x_1$; for example you can use that $A/(x_1^{p^ n}, \ldots , x_ d^{p^ n}) \cong k[x_1, \ldots , x_ d]/(x_1^{p^ n}, \ldots , x_ d^{p^ n})$ by Algebra, Lemma 10.158.10. Thus the same is true for every element of $M_ n$. Since every element of $M$ is in $M_ n$ for all $n \gg 0$ and since every element of $M$ is killed by some power of $x_1$, we conclude that $M$ is $x_1$-divisible.

Let $x = x_1$. Above we have seen that $M$ is $x$-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 $x e \in M$. There exists an element $z_1 \in M$ with $x z_1 = x e$. Hence $x(z_1 - e) = 0$. Replacing $e$ by $e - z_1$ we may assume $e$ is annihilated by $x$. Thus it suffices to prove that

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

is surjective. If $d = 1$, this is true by construction of $\varphi $. If $d > 1$, then we observe that $E[x]$ is the injective hull of the residue field of the regular ring $A/xA$, see Dualizing Complexes, Lemma 47.7.1. Observe that $M[x]$ as a module over $A/xA$ is $\mathfrak m/(x)$-power torsion and we have

\begin{align*} M[x] \otimes _{A/xA, F} A/xA & = M[x] \otimes _{A, F} A \otimes _ A A/xA \\ & = (M \otimes _{A, F} A)[x^ p] \otimes _ A A/xA \\ & \cong M[x^ p] \otimes _ A A/xA \end{align*}

Argue using flatness of $F$ as before. We claim that $M[x^ p] \otimes _ A A/xA \to M[x]$, $z \otimes 1 \mapsto x^{p - 1}z$ is an isomorphism. This can be seen by proving it for each of the modules $M_ n$, $n > 0$ defined above where it follows by the same result for $A/(x_1^{p^ n}, \ldots , x_ d^{p^ n})$ and $x = x_1$. Thus by induction on $\dim (A)$ we conclude that $\varphi [x]$ is surjective as desired. $\square$


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