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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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    @QiaochuYuan Please consider converting your comment into an answer, so that it gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138/15416). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141/the-crusade-of-answers) 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](http://meta.stackexchange.com/q/143113/199957), [here](http://meta.math.stackexchange.com/q/1148/15416) or [here](http://meta.math.stackexchange.com/a/9868/15416).2013-06-08

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