Exterior power on short exact sequence of modules with free middle term

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.

1 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.

• It's because $N\cong R^n/M$. Commented Aug 5, 2020 at 10:25
• Agh I see, since you're replacing $f(M)$ by $M$, so $g(M)=0$, and the result follows ... so one more question: Why is $\wedge^n(M)$ inside the ideal generated by all $\det A$ where the columns of $A$ are in $M$ ?
– uno
Commented Aug 5, 2020 at 10:29
• If you have $a_1\wedge\cdots\wedge a_n$ in $\bigwedge^n M$, then its image in $M$ is $\det A$ times the usual generator of $\bigwedge^n R^n$ where $A$ is the matrix with columns $a_i$. Commented Aug 5, 2020 at 10:32