Let $(R,\mathfrak m,k)$ be a Noetherian local ring. For a finitely generated $R$-module $M$, let $\wedge^j(M)$ denote its $j$-th exterior power. Recall that $\wedge^j(R^{\oplus j})\cong R,\forall j\ge 1$.
Now suppose we have an exact sequence of finitely generated $R$-modules
$0\to M \xrightarrow{f} R^{\oplus n}\xrightarrow{g} N\to 0 $ . So we have an induced map $\wedge^n(f): \wedge^n(M)\to \wedge^n(R^{\oplus n})\cong R $ . Let $a\in \operatorname{Im}(\wedge^n(f))\subseteq R$ . Then how to prove that $aN=0$ ?
(If needed , I'm willing to assume that $f(M)\subseteq \mathfrak m R^{\oplus n}$ . )
My thoughts: Since $g$ is surjective, we have $\wedge^n(g)$ is surjective. Since $g\circ f=0$, we also have by functoriality that $\wedge^n(g)\circ\wedge^n(f)=0$ . We also have an exact sequence
$R^{\oplus n}\otimes M\cong \wedge^{n-1}(R^{\oplus n})\otimes \ker g \to \wedge^n(R^n)\xrightarrow{\wedge^n(g)} \wedge^n(N)\to 0$ .
Apart from this, I can't think of anything else. Please help.
Best Answer
We can think of $M$ as a submodule of $R^n$. Then the image of $\bigwedge^n M$ in $R$ is the ideal generated by all $\det A$ where $A$ runs through the $n$ by $n$ matrices whose columns are in $M$. So the result boils down to the assertion that $(\det A) R^n\subseteq M$ whenever $A$ is such a matrix. But for $u\in R^n$, $Au\in M$, and for $u\in R^n$ $$(\det A)v=A(\text{adj}\,A)v=Au\in M$$ where $u=(\text{adj}\,A)v$.
This argument works over all commutative rings.