# Finitely generated projective modules, over commutative Noetherian ring, of constant rank $1$, if stably isomorphic, then isomorphic?

Let $$M,N$$ be finitely generated projective modules over a commutative Noetherian ring $$R$$ such that $$M_P \cong N_P \cong R_P, \forall P \in Spec(R)$$.

If $$\exists n\ge 1$$ such that $$M \oplus R^n \cong N \oplus R^n$$, then is it true that $$M \cong N$$ ?

• Yes, this is true. One approach uses exterior powers: is this a tool which is accessible for you? – Alex Wertheim May 9 at 23:46
• @AlexWertheim: Sure, you can use exterior powers, I would like to see that ... if there are other approaches, I would like to see those too ! – user102248 May 9 at 23:48

I claim that for any (finitely generated) projective $$R$$-module $$P$$ of constant rank $$1$$, $$P \cong \Lambda^{k+1}(P \oplus R^{k})$$ for any $$k \in \mathbb{N}$$. From this, it will follow that $$M \cong \Lambda^{n+1}(M \oplus R^{n}) \cong \Lambda^{n+1}(N \oplus R^{n}) \cong N$$ We thus focus on proving the claim. Recall that for any two $$R$$-modules $$P, Q$$ and any $$n \in \mathbb{N}$$, there is a natural isomorphism of $$R$$-modules $$\Lambda^{n}(P \oplus Q) \cong \bigoplus_{k=0}^{n} \Lambda^{k}(P) \otimes_{R} \Lambda^{n-k}(Q)$$
From this formula, we deduce $$\Lambda^{k+1}(P \oplus R^{k}) \cong \bigoplus_{i=0}^{k+1} \Lambda^{i}(P) \otimes_{R} \Lambda^{(k+1)-i}(R^{k})$$ The first summand $$\Lambda^{0}(P) \otimes_{R} \Lambda^{k+1}(R^{k})$$ is trivial, since $$\Lambda^{k+1}(R^{k}) = 0$$. The second summand is $$\Lambda^{1}(P) \otimes_{R} \Lambda^{k}(R^{k}) \cong P \otimes_{R} R \cong P$$. We claim that $$\Lambda^{i}(P) = 0$$ for all $$i > 1$$, so that all the higher summands vanish. Indeed, for any maximal ideal $$\mathfrak{m} \subseteq R$$, we have the following sequence of isomorphisms $$R_{\mathfrak{m}}$$-modules: $$(\Lambda^{i}(P))_{\mathfrak{m}} \cong \Lambda^{i}(P_{\mathfrak{m}}) \cong \Lambda^{i}(R_{\mathfrak{m}})$$ since $$P$$ has constant rank $$1$$. But $$\Lambda^{i}(R_{\mathfrak{m}}) = 0$$ for all $$i > 1$$, so every localization of $$\Lambda^{i}(P)$$ at a maximal ideal $$\mathfrak{m} \subset R$$ is $$0$$, whence $$\Lambda^{i}(P) = 0$$, as desired.
• thanks ... it is good to know that the result is in Weibel's K-theory book ... my motivation in fact was also $K$-theoretic, I was trying to prove that $Pic(R)$ embeds in the group of units of the ring $K_0(R)$ ... my question is the injectivity of the natural map – user102248 May 10 at 0:29