Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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
1  
@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

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.