## Over a commutative ring, the tensor product of two free modules is free

Let $R$ be a commutative ring with 1.

1. Prove that the tensor product of two free $R$-modules is free.
2. Prove that the tensor product of two projective $R$-modules is projective.

Suppose $A$ and $B$ are free $R$-modules; we can assume (since $R$ is commutative) that $A = \bigoplus_I R$ and $B = \bigoplus_J R$ for some nonempty index sets $I$ and $J$. By this previous exercise, binary tensor products essentially commute with direct sums. Thus, $A \otimes_R B = (\bigoplus_I R) \otimes (\bigoplus_J R)$ $\cong_R \bigoplus_I (R \otimes_R (\bigoplus_J R))$ $\cong_R \bigoplus_I \bigoplus_J (R \otimes_R R)$ $\cong_R \bigoplus_{I \times J} R$. Thus $A \otimes_R B$ is free.

Now suppose $A$ and $B$ are projective. Then there exist $M$ and $N$ so that $A \oplus M$ and $B \oplus N$ are free. By the previous argument, $(A \oplus M) \otimes_R (B \oplus N)$ is free. Now $(A \oplus M) \otimes_R (B \oplus N) \cong_R A \otimes_R (B \oplus N) \oplus M \otimes_R (B \oplus N)$ $\cong_R (A \otimes_R B) \oplus (A \otimes_R N) \oplus (M \otimes_R (B \oplus N))$ is free, so that $A \otimes_R B$ is projective.

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