The Stacks project

Lemma 15.8.2. Let $R$ be a ring. Let $M$ be a finite $R$-module. Choose a presentation

\[ \bigoplus \nolimits _{j \in J} R \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0. \]

of $M$. Let $A = (a_{ij})_{i = 1, \ldots , n, j \in J}$ be the matrix of the map $\bigoplus _{j \in J} R \to R^{\oplus n}$. The ideal $\text{Fit}_ k(M)$ generated by the $(n - k) \times (n - k)$ minors of $A$ is independent of the choice of the presentation.

Proof. Let $K \subset R^{\oplus n}$ be the kernel of the surjection $R^{\oplus n} \to M$. Pick $z_1, \ldots , z_{n - k} \in K$ and write $z_ j = (z_{1j}, \ldots , z_{nj})$. Another description of the ideal $\text{Fit}_ k(M)$ is that it is the ideal generated by the $(n - k) \times (n - k)$ minors of all the matrices $(z_{ij})$ we obtain in this way.

Suppose we change the surjection into the surjection $R^{\oplus n + n'} \to M$ with kernel $K'$ where we use the original map on the first $n$ standard basis elements of $R^{\oplus n + n'}$ and $0$ on the last $n'$ basis vectors. Then the corresponding ideals are the same. Namely, if $z_1, \ldots , z_{n - k} \in K$ as above, let $z'_ j = (z_{1j}, \ldots , z_{nj}, 0, \ldots , 0) \in K'$ for $j = 1, \ldots , n - k$ and $z'_{n + j'} = (0, \ldots , 0, 1, 0, \ldots , 0) \in K'$. Then we see that the ideal of $(n - k) \times (n - k)$ minors of $(z_{ij})$ agrees with the ideal of $(n + n' - k) \times (n + n' - k)$ minors of $(z'_{ij})$. This gives one of the inclusions. Conversely, given $z'_1, \ldots , z'_{n + n' - k}$ in $K'$ we can project these to $R^{\oplus n}$ to get $z_1, \ldots , z_{n + n' - k}$ in $K$. By Lemma 15.8.1 we see that the ideal generated by the $(n + n' - k) \times (n + n' - k)$ minors of $(z'_{ij})$ is contained in the ideal generated by the $(n - k) \times (n - k)$ minors of $(z_{ij})$. This gives the other inclusion.

Let $R^{\oplus m} \to M$ be another surjection with kernel $L$. By Schanuel's lemma (Algebra, Lemma 10.109.1) and the results of the previous paragraph, we may assume $m = n$ and that there is an isomorphism $R^{\oplus n} \to R^{\oplus m}$ commuting with the surjections to $M$. Let $C = (c_{li})$ be the (invertible) matrix of this map (it is a square matrix as $n = m$). Then given $z'_1, \ldots , z'_{n - k} \in L$ as above we can find $z_1, \ldots , z_{n - k} \in K$ with $z_1' = Cz_1, \ldots , z'_{n - k} = Cz_{n - k}$. By Lemma 15.8.1 we get one of the inclusions. By symmetry we get the other. $\square$

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