# Invertible modules over a noetherian ring

Let $$A$$ be a commutative noetherian ring and let $$M, N$$ be two finitely generated $$A$$-modules such that $$M\otimes_A N\cong A$$

I would like to know which are all such $$A$$-modules.

It seems to me that $$M$$ (and so $$N$$) has to be free and then since $$M\otimes_A N\cong A$$, it follows that they are both of rank $$1$$, i.e. isomorphic to $$A$$ itself.

My understanding of the problem

Using the noetherian hypothesis on $$A$$ we know that $$M, N$$ are also finitely presented and so we have two exact sequences $$\begin{equation*}A^{\oplus^{m_1}}\to A^{\oplus^{m_2}}\to M\to 0 \\ A^{\oplus^{n_1}}\to A^{\oplus^{n_2}}\to N\to 0 \end{equation*}$$ We could try to tensor the first sequence by $$N$$ and the second one by $$M$$ to obtain $$\begin{equation*}N^{\oplus^{m_1}}\to N^{\oplus^{m_2}}\to A\to 0 \\ M^{\oplus^{n_1}}\to M^{\oplus^{n_2}}\to A\to 0 \end{equation*}$$but maybe this is a dead end. I don't have any further idea. Any hint/solution is appreciated.

The modules $$M$$ satisfying this property are exactly the projective modules of (constant) rank $$1$$. In this case, you have automatically $$N\simeq M^*$$.
Perhaps I found another way to prove it. First of all we can assume that $$A$$ is a local ring and let $$m$$ be its unique maximal ideal.
The isomorphism $$M\otimes_A N\cong A$$ induced an isomorphism $$M/mM\otimes N/mN\cong \mathcal{K}$$ where $$\mathcal{K}:=A/m$$ is the residue field and both factor in the latter tensor product are $$\mathcal{K}$$-vector spaces. In particular they are of dimension $$1$$ and by Nakayama's lemma $$M$$ (and $$N$$) is cyclic, i.e. is generated by one element $$x$$. So we can write $$M=(x):=\{ax \ | a\in A\}$$. In particular is easy to prove that $$M\cong A/Ann_A(x)$$ where $$Ann_A(x):=\{a\in A \ | ax=0\}$$ is the annihilator of $$x$$ in $$A$$. Suppose now that there exists an element $$a\in A$$ such that $$ax=0$$ in $$A$$ ( so $$a\in Ann_A(x)$$). Consider the map $$f:A\to A$$ given by $$f(b):=ab$$, then it follows that $$\begin{equation*}1_M\otimes f:A\otimes_A M\cong M\to A\otimes_A M\cong M \end{equation*}$$is the zero map since $$M\cong (x)$$ and $$ax=0$$. From this follows that $$f$$ is the zero map and since $$1\in A$$ then $$a=0$$. We conclude that $$M\cong A$$ as desired.