Suppose that i have a commutative ring $R$ and $R$-modules $M$ and $N$ such that $M$ and $M\otimes_R N$ are finitely generated projective modules. Suppose also that $ {\rm rk}_R(M)={\rm rk}_R(M\otimes_R N).$

Is it true that $N$ is a projective finitely generated module?


Let $R=A\times B$ be a product of two rings, $M=A\times \{ 0 \}$ and $N$ any $B$-module viewed as $R$-module via the projection $R\to B$. Then $M$ is projective over $R$, $M\otimes_R N=0$. But you can't conclude that $N$ is projective.

You should suppose that $M_p\ne 0$ for any prime ideal $p$ of $R$.

Edit Positive answer under the above condition. Note that the condition on the ranks is not sufficient as shows any example with $M=0$.

What is important is that $M$ is faithfully flat. This is true because it is flat by projectivity and faithfully flat by the condition $M_p\ne 0$ for all prime ideal $p$ of $R$.

Now suppose $M\otimes N$ is projective. Let us prove first that $N$ is flat: let $N_1\to N_2$ be an injective $R$-linear map. Let $L$ be the kernel of $N_1\otimes N\to N_2\otimes N$. Then $L\otimes M=0$ because $M$ and $M\otimes N$ are flat. By faithfull flatness of $M$, this implies that $L=0$. Hence $N$ is flat.

Fact. Let $K$ be any $R$-module such that $M\otimes_R K$ is finitely generated. Then $K$ is finitely generated.

Proof: $M\otimes K$ has a finite generating family of the form $m_i\otimes k_i$ with $k_i\in K$ and $m_i\in M$. If $K'$ is the submodule of $K$ generated by the $k_i$'s, then $M\otimes K'=M\otimes K$, hence $M\otimes (K/K')=0$. Again by faithfull flatness of $M$, this implies that $K/K'=0$ and $K=K'$ is finitely generated.

Applying this fact to $N$, we see that $N$ is finitely generated. Let $R^n\to N$ be a surjective $R$-linear map with kernel $P$. We have an exact sequence $$ 0\to M\otimes P\to M^n \to M\otimes N\to 0.$$ As $M\otimes N$ is projective, this exact sequence splits and $M\otimes P$ is a direct summand of $M^n$, hence finitely generated. By the above fact, $P$ is finitely generated.

So $N$ is flat and finitely presented. It is known that this condition is equivalent to be projective (and finitely generated).

  • $\begingroup$ In my case i also have the condition ${\rm rk}_R(M)={\rm rk}_R(M\otimes_R N)\ne 0,$ in particular $M_p\ne 0$ for any prime ideal $p$ of $R.$ $\endgroup$ – Hector Pinedo Oct 31 '12 at 23:18
  • $\begingroup$ @navigetor23: thanks, I will correct. $\endgroup$ – user18119 Nov 1 '12 at 23:32
  • $\begingroup$ thanks a lot for your help. The only step that is not clear to me is why the condition $M_p\ne 0$ for all $p\in {\rm Spec}(R)$ implies that $M$ is faithfully flat. $\endgroup$ – Hector Pinedo Nov 2 '12 at 16:42
  • $\begingroup$ @HectorPinedo: because $M_p\ne 0$ implies, by Nakayama, that $M_p\otimes_{R_p} (R_p/pR_p)\ne 0$, hence $M\ne pM$. This condition for all maximal ideals $p$ plus flatness is equivalent to the faithfull flatness. $\endgroup$ – user18119 Nov 2 '12 at 21:39
  • $\begingroup$ @HectorPinedo: as $M$ is locally free, it is even easier to see the faithfull flatness just by applying the definition. $\endgroup$ – user18119 Nov 2 '12 at 23:04

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