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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    $\begingroup$ Atiyah-MacDonald appears to be wrong. The correct map is indeed $\alpha \mapsto f(\alpha) \otimes 1_C = 1_B \otimes g(\alpha)$. $\endgroup$ May 4, 2012 at 15:05
  • $\begingroup$ Very strange. I would have expected $f \otimes g$ myself. Georges Elencwajg points out the same error here. $\endgroup$
    – Zhen Lin
    May 4, 2012 at 18:04
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    $\begingroup$ @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. $\endgroup$ Jun 8, 2013 at 11:22

1 Answer 1


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


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