Lemma 15.8.9. Let $R$ be a ring. Let $M$ be a finitely presented $R$-module. Let $k \geq 0$. Assume that $\text{Fit}_ k(M) = (f)$ for some nonzerodivisor $f \in R$ and $\text{Fit}_{k - 1}(M) = 0$. Then
$M$ has projective dimension $\leq 1$,
$M' = \mathop{\mathrm{Ker}}(f : M \to M)$ is the $f$-power torsion submodule of $M$,
$M'$ has projective dimension $\leq 1$,
$M/M'$ is finite locally free of rank $k$, and
$M \cong M/M' \oplus M'$.
Proof. Choose a presentation
for some matrix $A$ with coefficients in $R$.
We first prove the lemma when $R$ is local. Set $M' = \{ x \in M \mid fx = 0\} $ as in the statement. By Lemma 15.8.8 we can choose $x_1, \ldots , x_ k \in M$ which generate $M/M'$. Then $x_1, \ldots , x_ k$ generate $M_ f = (M/M')_ f$. Hence, if there is a relation $\sum a_ ix_ i = 0$ in $M$, then we see that $a_1, \ldots , a_ k$ map to zero in $R_ f$ since otherwise $\text{Fit}_{k - 1}(M) R_ f = \text{Fit}_{k - 1}(M_ f)$ would be nonzero. Since $f$ is a nonzerodivisor, we conclude $a_1 = \ldots = a_ k = 0$. Thus $M \cong R^{\oplus k} \oplus M'$. After a change of basis in our presentation above, we may assume the first $n - k$ basis vectors of $R^{\oplus n}$ map into the summand $M'$ of $M$ and the last $k$-basis vectors of $R^{\oplus n}$ map to basis elements of the summand $R^{\oplus k}$ of $M$. Having done so, the last $k$ rows of the matrix $A$ vanish. In this way we see that, replacing $M$ by $M'$, $k$ by $0$, $n$ by $n - k$, and $A$ by the submatrix where we delete the last $k$ rows, we reduce to the case discussed in the next paragraph.
Assume $R$ is local, $k = 0$, and $M$ annihilated by $f$. Now the $0$th Fitting ideal of $M$ is $(f)$ and is generated by the $n \times n$ minors of the matrix $A$ of size $n \times m$. (This in particular implies $m \geq n$.) Since $R$ is local, some $n \times n$ minor of $A$ is $uf$ for a unit $u \in R$. After renumbering we may assume this minor is the first one. Moreover, we know all other $n \times n$ minors of $A$ are divisible by $f$. Write $A = (A_1 A_2)$ in block form where $A_1$ is an $n \times n$ matrix and $A_2$ is an $n \times (m - n)$ matrix. By Algebra, Lemma 10.15.6 applied to the transpose of $A$ (!) we find there exists an $n \times n$ matrix $B$ such that
for some $n \times (m - n)$ matrix $C$ with coefficients in $R$. Then we first conclude $BA_1 = fu 1_{n \times n}$. Thus
Since the determinant of $B$ is a nonzerodivisor we conclude that $A_2 = u^{-1}A_1C$. Therefore the image of $A$ is equal to the image of $A_1$ which is isomorphic to $R^{\oplus n}$ because the determinant of $A_1$ is a nonzerodivisor. Hence $M$ has projective dimension $\leq 1$.
We return to the case of a general ring $R$. By the local case we see that $M/M'$ is a finite locally free module of rank $k$, see Algebra, Lemma 10.78.2. Hence the extension $0 \to M' \to M \to M/M' \to 0$ splits. It follows that $M'$ is a finitely presented module. Choose a short exact sequence $0 \to K \to R^{\oplus a} \to M' \to 0$. Then $K$ is a finite $R$-module, see Algebra, Lemma 10.5.3. By the local case we see that $K_\mathfrak p \cong R_\mathfrak p^{\oplus a}$ for all primes. Hence by Algebra, Lemma 10.78.2 again we see that $K$ is finite locally free of rank $a$. It follows that $M'$ has projective dimension $\leq 1$ and the lemma is proved. $\square$
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