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 , , .

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. $\square$

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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