Suppose $M$ and $N$ are free $R$-module($R$ is a commutative ring). The tensor product of $M\otimes_R N$ is free $R$-module? I know for projective modules it is true. How should we build its basis?
[Math] Tensor product of free modules
commutative-algebrafree-modulesmodulesprojective-moduletensor-products
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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).
The tensor product of two Noetherian modules is indeed Noetherian. Even better, if $L$ is finitely generated and $N$ is Noetherian, then $L \otimes_R N$ is Noetherian.
Because $L$ is finitely generated, there is an exact sequence $R^m \to L \to 0$ for some integern $m \geq 0$. Applying the right exact functor $-\otimes_R N$ yields an exact sequence $R^m \otimes_R N \to L \otimes_R N \to 0$. But $R^m \otimes_R N \cong N^m$ is Noetherian, so its homomorphic image $L \otimes_R N$ must also be Noetherian.
Best Answer
$$[\bigoplus\limits_{i} R] \otimes_R [ \bigoplus\limits_j R] \cong \bigoplus\limits_{i,j} R \otimes_R R \cong \bigoplus\limits_{i,j} R$$