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.
 A: The modules $M$ satisfying this property are exactly the projective modules of (constant) rank $1$. In this case, you have automatically $N\simeq M^*$.
See for example this discussion on MO f or furhter information and references : https://mathoverflow.net/questions/104350/equivalent-definitions-of-invertible-modules
A: 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.
If anyone can check my argument I appreciate.
