## A Fitting remark

In this post I want to talk about an innocent commutative algebra lemma. Let $R$ be a DVR with uniformizer $\pi$, and let $M$ be a finite torsion $R$-module, so $M\simeq \oplus_{i=1}^{n}R/\pi^{k_i}$ for some uniquely determined sequence $k_1 \geq \dots \geq k_n > 0$. I’ll somewhat abusively refer to the $k_i$‘s as the “elementary divisors” of $M$.

Lemma.  If $N \subseteq M$ is an $R$-submodule generated by $j$ elements, then $\ell(N) \leq \sum_{1\leq i \leq j} k_i$. Furthermore, if equality holds, then $N$ is a direct summand of $M$.

(Here and throughout, $\ell$ denotes $R$-module length.)

Proof. We first prove the inequality by induction on $n$. Fix a surjection $f:M\to R/\pi^{k_{1}}$ such that $\ker f\simeq\oplus_{2\leq i\leq n}R/\pi^{k_{i}}$.  Choose a minimal basis $n_{1},\dots,n_{j}$ of $N$ such that $f(n_{1})$ generates $f(N)\subseteq R/\pi^{k_{1}}$. Choose some $a_{i}\in R$ with $f(n_{i})=a_{i}f(n_{1})$ for all $2\leq i\leq j$, and make the substitution $n_{i}\to n_{i}-a_{i}n_{1}$ for $i\geq2$. Having done this, we get a basis $n_{1},\dots,n_{j}\in N$ with $f(n_{1})$ generating $f(N)$ and $n_{i}\in\ker f$ for all $2\leq i\leq j$. Let $N'\subseteq\ker f\cap N$ denote the $R$-submodule generated by $n_{2},\dots,n_{j}$; applying our induction hypothesis to the modules $N' \subseteq \ker f$, we get the inequality $\ell(N')\leq k_{2}+\cdots k_{j}$. Since $N'$ and $Rn_{1}\subset N$ generate $N$, we get an inequality $\ell(N)=\ell(N'+Rn_{1})\leq\ell(N')+\ell(Rn_{1})$. But $\ell(Rn_{1})\leq k_{1}$, since $\pi^{k_{1}}$ annihilates $M$, so
$\ell(N)\leq\ell(N')+\ell(Rn_{1})\leq k_{2}+\cdots+k_{j}+k_{1},$
as desired.

For the second claim, we argue by induction on $j$; the case $j=1$ is easy (argument: $N$ must project isomorphically onto a direct summand of $M$ of the form $R/\pi^{k_1}$). Maintain the previous notation, and assume we have an equality $\ell(N)=k_{1}+\cdots+k_{j}$. Since $\ell(N')\leq k_{2}+\cdots+k_{j}$ and $\ell(Rx_{1})\leq k_{1}$, the chain of inequalities
$\ell(N)=\ell(N'+Rn_{1})\leq\ell(N')+\ell(Rn_{1})\leq k_{1}+\cdots+k_{j}=\ell(N)$
then forces $\ell(N')=k_{2}+\cdots+k_{j}$ and $\ell(Rn_{1})=k_{1}$. Since $N'$ and $Rx_{1}$ are generated by $j-1$ and $1$ element, respectively, they are both direct summands of $M$ by the induction hypothesis. Finally, the equality $\ell(N'+Rn_{1})=\ell(N')+\ell(Rn_{1})$ implies that $N'\cap Rn_{1}=0$, so $N\cong N'\oplus Rn_{1}\subseteq M$ is a direct summand of $M$. $\square$

This lemma really really really looks like it should be well-known, but I couldn’t find it stated in the literature.  Presumably I was just typing the wrong things into google.  Can some reader provide a reference?  If you can find a reference in a textbook (not a research paper), this will settle a bet between me and AJdJ.  Also, it would be really nice if there were a “coordinate-free” proof which didn’t involve choosing a basis for $N$.  Before finding the argument given above, I spent a while trying to make a proof based on the theory of Fitting ideals; the latter seem quite natural in this context, since one has the equality $\mathrm{Fitt}(N)=(\pi^{\ell(N)})$ for any finite $R$-module.  Can the reader make such a proof work?

OK literally while writing the previous two sentences I hit upon the following argument for the first part of the lemma. It clearly suffices to show the complementary inequality $\ell(M/N)\geq\sum_{j For this we use Fitting ideals as follows. Recall that for any finite torsion module $Q$ over $R$ with elementary divisors $k_{i}$, we have an equality $\mathrm{Fitt}_{j}(Q)=(\pi^{\sum_{j; in particular, $\mathrm{Fitt}(Q)=\mathrm{Fitt}_{0}(Q)=(\pi^{\ell(Q)})$, and $\mathrm{Fitt}_{m}(Q)=R$ if $Q$ is generated by $\leq m$ elements. Returning to the situation at hand, we have an inclusion $\mathrm{Fitt}_{j}(N)\mathrm{Fitt}(M/N)\subseteq\mathrm{Fitt}_{j}(M)=(\pi^{\sum_{j (this is a special case of Proposition XIII.10.7 in Lang’s Algebra). But $\mathrm{Fitt}_{j}(N)=R$ since by assumption $N$ is generated by $j$ elements, so we get
$(\pi^{\ell(M/N)})=\mathrm{Fitt}(M/N)\subseteq\mathrm{Fitt}_{j}(M)=(\pi^{\sum_{j
and this immediately implies the desired inequality.

Great! Is there a Fitting ideal proof of the second part?