If over a commutative ring $R$ we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ are finitely generated projective?

We have finite generation, because if $M\otimes N$ is generated by $\sum_i a_{ij} x_i^j\otimes y_i^j$, then we have $x_{i_1}^{j_1}\otimes y_{i_2}^{j_2}\otimes x^{j_3}_{i_3}$ generating $M^n$. Projecting onto $M$, we see that $x_i^j$ generate $M$. My question is can we show that this map splits?

I am able to verify if in the case that $n=1$. I can't personally think of an example where it is not true, but I am having problems constructing a splitting. Any other solution is appreciated too.

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    $\begingroup$ Interesting. I think one can show that $M, N$ are flat (the point being that tensoring with $M \otimes N$ is exact, and this functor is the composition of tensoring with $N$ and tensoring with $M$). So if the statement is false I think a counterexample has to involve f.g. flat modules which are not projective, which are somewhat exotic (see e.g. math.stackexchange.com/questions/400776/…); in particular $R$ cannot be Noetherian. $\endgroup$ Mar 31, 2014 at 6:30
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    $\begingroup$ @QiaochuYuan, you can do the same argument with the functor $\hom_R(M,\hom_R(N,\mathord-))$, which is isomorphic to $\hom_R(M\otimes_RN,\mathord-)$, which in turn is $\hom_R(R^n,\mathord-)$. $\endgroup$ Mar 31, 2014 at 7:00
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    $\begingroup$ @Qiaochu: I don't understand your argument for flatness. $\endgroup$ Apr 7, 2014 at 11:18
  • $\begingroup$ I don't see how $M$ is finitely generated. It seems to me that every element of $M^n \cong M \otimes (M \otimes N)$ is of the form $x \otimes (\sum_{i = 1}^n r_i(x_i \otimes y_i)).$ $\endgroup$ Aug 14, 2020 at 22:47

2 Answers 2


$\def\id{\operatorname{id}}$Suppose $M\otimes N$ is isomorphic to $R^n$. Pick a basis $\{x_1,\dots,x_n\}$ of $M\otimes N$, with $x_i=\sum_{j=1}^{r_i}m_{i,j}\otimes n_{i,j}$ for each $i\in\{1,\dots,n\}$. Let $r=r_1+\cdots+r_n$, let $\{e_{i,j}:1\leq i\leq n, 1\leq j\leq r_i\}$ be a basis of $R^r$, and consider the map $f:R^r\to M$ which maps $e_{i,j}$ to $m_{i,j}$. Then $f\otimes\id_N:R^r\otimes N\to M\otimes N$ is obviously surjective. Now $M\otimes N$ is free, so $f\otimes\id_N$ is a split surjection, and therefore so is $f\otimes\id_N\otimes\id_M:R^r\otimes N\otimes M\to M\otimes N\otimes M$. As $R^r\otimes N\otimes M\cong R^r\otimes R^n\cong R^{rn}$ is free, this means that $M\otimes N\otimes M\cong R^n\otimes M\cong M^n$ is projective, as is every one of its obvious $n$ direct summands, each isomorphic to $M$.

Later. We can do this more simply as follows; this is what i wanted to do before out of my comment above, but somehow messed up (in a way, it is a transposed variant of the first paragraph)

Suppose $\phi:M\otimes N\to R^n$ is an isomorphism, let $p:R^n\to R$ be the projection onto the first summand, and suppose $\xi=\sum_{i=1}^rm_i\otimes n_i\in M\otimes N$ is such that $\phi(\xi)=(1,0,\dots,0)$. Then the map $\psi:R^r\otimes N\to R$ such that $\psi(e_i\otimes n)=p(\phi(m_i\otimes n))$ is surjective. Since $R$ is free, $\psi$ is split, so that so is $\psi\otimes\id_M:R^r\otimes N\otimes M\to R\otimes M$. The domain of this last map is free, isomorphic to $R^{rn}$, and the codomain is isomorphic to $M$, so we see that $M$ is projective.

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    $\begingroup$ This is, mutatis mutandi, what one does to show that invertible modules are projective, iirc. $\endgroup$ Mar 31, 2014 at 7:37
  • $\begingroup$ This is a truly beautiful way to prove this. Thanks! $\endgroup$
    – pax
    Mar 31, 2014 at 7:40
  • $\begingroup$ Notice that it is enough to suppose that $M\otimes N$ is free, possibly of infinite rank. $\endgroup$ Mar 31, 2014 at 8:00

Let $M,N$ be two $R$-modules such that $M \otimes N$ is free of rank $>0$. Let $p : F \to M$ be any epi with $F$ free. Then $N \otimes p : N \otimes F \to N \otimes M$ is epi, hence it splits. Then also $M \otimes N \otimes p$ splits, which implies that $M \otimes N \otimes M$ is a direct summand of $M \otimes N \otimes F$ and is therefore projective. But since $R$ is a direct summand of $M \otimes N$, it follows that $R \otimes M = M$ is a direct summand of $M \otimes N \otimes M$, hence also projective.

Some additions: If $M \otimes N$ is locally free of finite rank $>0$, then $M$ is locally free of finite rank (i.e. finitely generated projective). One may ask if we can drop the "finite rank" here or, perhaps more interesting, what happens when we assume only that $M \otimes N$ is projective and faithfully flat. In the above proof, then since $M \otimes N \otimes M$ is flat we conclude that $M$ is flat, in fact faithfully flat since $M \otimes N$ is. Now the question is if tensoring with a faithfully flat module (not a ring extension) descends the property of being projective. I only know this in the finitely generated case.

  • $\begingroup$ «Now it is well-known that $N$ is finitely generated projective since $M \otimes N$ is» assumes precisely the content of the question, no? :-) (or rather something a bit stronger, in fact, as the OP has that the tensor product is free, not just projective) $\endgroup$ Apr 7, 2014 at 16:01
  • $\begingroup$ I've improved the answer. The first part is essentially a simplification of your argument (and as usual: without elements). $\endgroup$ Apr 8, 2014 at 7:50

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