## Verify that the tensor product of two R-algebras is an R-algebra

Let $R$ be a commutative ring with 1 and let $A$ and $B$ be $R$-algebras via the ring homomorphisms $\alpha : R \rightarrow A$ and $\beta : R \rightarrow B$. Assuming that the multiplication $(a_1 \otimes b_1)(a_2 \otimes b_2) = a_1a_2 \otimes b_1b_2$ is well-defined, show that this operation makes $A \otimes_R B$ into an $R$-algebra.

We already know that $A \otimes_R B$ is an abelian group with respect to the usual addition. Moreover, in our proof (not given here) that this multiplication is well-defined, we found that multiplication is bilinear- that is, it distributes over addition from both sides. To show that $A \otimes_R B$ is a ring, it suffices to show that multiplication is associative. To that end, let $a_1, a_2, a_3 \in A$ and $b_1, b_2, b_3 \in B$.

Note that $(a_1 \otimes b_1)((a_2 \otimes b_2)(a_3 \otimes b_3)) = (a_1 \otimes b_1)(a_2a_3 \otimes b_2b_3)$ $= a_1(a_2a_3) \otimes b_1(b_2b_3)$ $= (a_1a_2)a_3 \otimes (b_1b_2)b_3$ $= (a_1a_2 \otimes b_1b_2)(a_3 \otimes b_3)$ $= ((a_1 \otimes b_1)(a_2 \otimes b_2))(a_3 \otimes b_3)$, so that multiplication is associative.

Now note that $(1 \otimes 1)(a \otimes b) = a \otimes b = (a \otimes b)(1 \otimes 1)$, so that in fact $A \otimes_R B$ is a ring with 1. To make $A \otimes_R B$ into an $R$-algebra, we need a unital ring homomorphism $\gamma : R \rightarrow A \otimes_R B$ whose image is contained in the center of $A \otimes_R B$. To that end, define $\gamma(r) = \alpha(r) \otimes 1$. Since $\gamma(r+s) = \alpha(r+s) \otimes 1$ $= (\alpha(r) + \alpha(s)) \otimes 1$ $= \alpha(r) \otimes 1 + \alpha(s) \otimes 1$ $= \gamma(r) + \gamma(s)$ and $\gamma(rs) = \alpha(rs) \otimes 1$ $= \alpha(r)\alpha(s) \otimes 1$ $= (\alpha(r) \otimes 1)(\alpha(s) \otimes 1)$ $= \gamma(r)\gamma(s)$, $\gamma$ is a ring homomorphism. Moreover, $\gamma(1) = \alpha(1) \otimes 1 = 1 \otimes 1$, so that $\gamma$ is a unital homomorphism. Finally, note that $\gamma(r)(a \otimes b) = (\alpha(r) \otimes 1)(a \otimes b)$ $= \alpha(r)a \otimes b$ $= a \alpha(r) \otimes b$ $= (a \otimes b)(\alpha(r) \otimes 1)$, so that $\mathsf{im}\ \gamma$ is contained in the center of $A \otimes_R B$. Thus via $\gamma$, $A \otimes_R B$ is an $R$-algebra.