# Two finitely generated module over $\mathbb{C}[[x_1, x_2, \dots, x_n]]$

I'm struggling with the following problem.

Let $$A = \mathbb{C}[[x_1, x_2, \dots, x_n]]$$ be the ring of formal power series over $$\mathbb{C}$$. Show that if two finitely generated $$A$$-modules $$M, N$$ satisfy $$M\otimes_AN \cong A$$ as $$A$$-modules, then $$M \cong N \cong A$$.

The part I managed to prove is as follows. Since $$A$$ is the local ring with the maximal ideal $$m=(x_1, x_2, \dots,x_n)$$ and the residue field $$k = A/m \cong \mathbb{C}$$, $$M_k \otimes_k N_k = (M \otimes_A k) \otimes_k (N \otimes_A k) \cong (M \otimes_A N) \otimes_A k \cong A \otimes_A k \cong k$$ as $$k$$-Vector space. Thus the dimensions of $$M_k = M \otimes_A k$$ and $$N_k = N \otimes_A k$$ as $$k$$-Vector space are both $$1$$. So if I could prove that $$M$$ is free $$A$$-Module with the finite rank $$r$$, it follows that $$M \cong A$$ from $$k \cong A^r \otimes_A k \cong (A \otimes_A k)^r \cong k^r$$, so is $$N$$.

How can I prove that $$M$$ is free? I think it may be related to the fact that $$A$$ is Noetherian local ring. (In this case it suffices to show that $$M$$ is flat (Atiyah-Macdonald Exercise 7-15), or $$M$$ is finitely generated projective module over a local ring.)

• Since $M\otimes_A k$ is one dimensional, Nakayama says $M=A/I$ for some ideal $I$. Similarly $N=A/J$. Thus $M\otimes_A N=A/I+J=A$. This says $I+J=0$ and then $I=J=0$. Aug 7, 2020 at 16:17

Given that $$M$$ and $$N$$ are finitely generated, we have that $$M = A \langle x_1, \dots, x_m \rangle$$ for some elements $$x_i$$ in $$M$$ and $$N = A \langle y_1, \dots, y_n \rangle$$ for some elements $$y_j$$ in $$N.$$ Consider the surjections $$\pi : A^m \to M$$ and $$\rho : A^n \to N$$ defined by $$\pi(r_1, \dots, r_m) = r_1 x_1 + \cdots + r_m x_m$$ and $$\rho(r_1, \dots, r_n) = r_1 y_1 + \cdots + r_n y_n.$$ Recall that the functors $$M \otimes_A -$$ and $$- \otimes_A N$$ are right-exact, hence we have surjections $$\pi \otimes_A 1_N : A^m \otimes_A N \to M \otimes_A N$$ and $$1_M \otimes_A \rho : M \otimes_A A^n \to M \otimes_A N.$$ By hypothesis that there exists an isomorphism $$\varphi : M \otimes_A N \to A,$$ it follows that $$\varphi \circ (\pi \otimes_A 1_N) : A^m \otimes_A N \to A$$ and $$\varphi \circ (1_M \otimes_A \rho) : M \otimes_A A^n \to A$$ are surjections. Certainly, $$A$$ is a free $$A$$-module, hence the maps $$\varphi \circ (\pi \otimes_A 1_N)$$ and $$\varphi \circ (1_M \otimes_A \rho)$$ split, i.e., there exist $$A$$-modules $$M'$$ and $$N'$$ such that $$M \otimes_A A^n \cong M' \oplus A$$ and $$A^n \otimes_A M \cong A \oplus N'.$$ Putting this all together gives $$A^n \cong (M \otimes_A N) \otimes_A A^n \cong N \otimes_A (M \otimes_A A^n) \cong N \otimes_A (M' \oplus A) \cong N \oplus (N \otimes_A M'),$$ and analogously, we have that $$A^m \cong M \oplus (M \otimes_A N').$$ Consequently, both $$M$$ and $$N$$ are direct summands of a free $$A$$-module, so they are projective $$A$$-modules. But a finitely generated projective module over a Noetherian local ring is free, so we are done.