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 \(\displaystyle F_1 = M \oplus A\) and \(\displaystyle F_2 = N \oplus B\) for free \(\displaystyle R\)-modules \(\displaystyle F_1, F_2\) and some \(\displaystyle R\)-modules \(\displaystyle A, B\). Then we have

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

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

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

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

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