Is $M\otimes_R N$ a projective $R$-module?

[Chris L T521]

Here's this week's problem.

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Problem: Let $R$ be a commutative ring, and let $M$ and $N$ be two projective $R$-modules. Prove that $M\otimes_R N$ is a projective $R$-module.

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[Chris L T521]

This week's problem was correctly answered by mathbalarka. You can find his solution below.

Spoiler

Suppose that $$F_1 = M \oplus A$$ and $$F_2 = N \oplus B$$ for free $$R$$-modules $$F_1, F_2$$ and some $$R$$-modules $$A, B$$. Then we have

$$F_1 \otimes_R F_2 = (M \oplus A) \otimes_R (N \oplus B) = (M \otimes_R N) \oplus (A \otimes_R B)$$

To prove that $$M \otimes_R N$$ is projective, we see that it is sufficient to prove that $$F_1 \otimes_R F_2$$ is free.

Suppose $$E_1$$ and $$E_2$$ are the basis of the free modules $$F_1$$ and $$F_2$$, respectively. Hence, $$F_1 = \bigoplus_{E_1} R$$ and $$F_2 = \bigoplus_{E_2} R$$. Applying tensor multiplication then gives

$$F_1 \otimes_R F_2 = \left ( \bigoplus_{E_1} R \right ) \otimes_R \left (\bigoplus_{E_2} R \right ) = \bigoplus_{E_1} \bigoplus_{E_2} (R \otimes_R R) = \bigoplus_{E_1 \times E_2} R$$

This concludes that $$F_1 \otimes_R F_2$$ is indeed free and thus $$M \otimes_R N$$ is projective.