# Turning the tensor product of algebras into an algebra

Let $B, C$ be $A$-algebras, where $A$ is a commutative ring, i.e. $B, C$ are rings and we have ring homomorphisms $f:A\rightarrow B, g:A \rightarrow C$. Since both $B, C$ are $A$-modules, we define $D=B \otimes_A C$. Now, D can be turned into a ring by defining multiplication $D \times D \rightarrow D$ by $(b \otimes c, b' \otimes c') \mapsto bb' \otimes cc'$. See for example "Intoduction to Commutative Algebra" by Atiyah and MacDonald, p. 30.

To turn $D$ into an $A$-algebra we need a ring homomorphism $A \rightarrow D$. One possibility is $\alpha \mapsto f(\alpha) \mapsto f(\alpha) \otimes 1_C$. It is mentioned in Atiyah in p. 31 that the map $\alpha \mapsto f(\alpha) \otimes g(\alpha)$ is a ring homomorphism $A \rightarrow D$. However, it seems to me that this map does not preserve addition, since $\alpha+\alpha' \mapsto f(\alpha+\alpha') \otimes g(\alpha+\alpha')$ and the latter quantity is not equal (seemingly) to $f(\alpha) \otimes g(\alpha) + f(\alpha') \otimes g(\alpha')$.

Am i missing something or is this a typo?

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Atiyah-MacDonald appears to be wrong. The correct map is indeed $\alpha \mapsto f(\alpha) \otimes 1_C = 1_B \otimes g(\alpha)$. –  Qiaochu Yuan May 4 '12 at 15:05
Very strange. I would have expected $f \otimes g$ myself. Georges Elencwajg points out the same error here. –  Zhen Lin May 4 '12 at 18:04
@QiaochuYuan Please consider converting your comment into an answer, so that it gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it, so that it gets an upvote. For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 8 '13 at 11:22

I think this is a typo. The correct map is $\alpha \mapsto f(\alpha) \otimes 1_C = 1_B \otimes g(\alpha)$.