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

Question

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 $M \cong N$ ?

Answer 1

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.