The Stacks project

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 $\leq 1$.

Proof. Choose a presentation

\[ R^{\oplus m} \xrightarrow {A} R^{\oplus n} \to M \to 0 \]

We claim the image of $A$ is finite locally free of rank $n - k$. If the claim holds, then the lemma is true by definition of projective dimension. To prove the claim we may replace $R$ by the localization at a prime, see Algebra, Lemma 10.78.2. This reduces us the the case discussed in the next paragraph.

Assume $R$ is local. Set $M' = \{ x \in M \mid fx = 0\} $. 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

\[ BA = B(A_1 A_2) = f \left( \begin{matrix} u 1_{n \times n} & C \end{matrix} \right) \]

for some $n \times (m - n)$ matrix $C$ with coefficients in $R$. Then we first conclude $BA_1 = fu 1_{n \times n}$. Thus

\[ BA_2 = fC = u^{-1}fuC = u^{-1}BA_1C \]

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. This finishes the proof. $\square$


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