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 $f: R \to S$ and $g: R \to T$ be two $R$-algebras. To show that $S \otimes_R T$ is an $R$-algebra I need to define a ring structure (multiplication) on it and a ring homomorphism $h : R \to S \otimes_R T$.

Using the universal property of the (multi-)tensor product, defining multiplication is clear to me. What I'm confused about is the map $h : R \to S \otimes_R T$. According to this answer here, $r \mapsto 1 \otimes g(r) = f(r) \otimes 1$.

How are they the same?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

The basic fact is that for two $R$-modules $M,N$ you have:
$$r\cdot m\otimes _R n=m\otimes_R r\cdot n \quad (\text {for all} \quad r\in R,\; m\in M,\; n\in N) \quad (*)$$
In particular, in your case, you have: $$r\cdot 1_S\otimes _R 1_T=1_S \otimes_R r\cdot 1_T \quad (**)$$

In order to conclude, you just have to remember that built into the notion of algebra is the equality $r\cdot s=f(r)s$ where on the right hand side of the equality $f(r)s$ means the product of the elements $f(r),s$ in the ring $S$.
In particular $r\cdot 1_S=f(r)1_S=f(r)$ and similarly $r\cdot 1_T=g(r)$.
Transporting these last equalities into $(**)$, you get the required equation $$ f(r) \otimes_R 1_T= 1_S \otimes_R g(r) \quad (***) $$

share|improve this answer
    
Dear @Georges. Now it's clear. Thank you. I'm pleased to get an answer by the other of the post I linked to. If I may I'd like to point out a tiny, what I believe to be a, typo in your answer to the linked MO post: isn't the second author of the book spelled "Macdonald"? –  Rudy the Reindeer Jun 12 '12 at 9:57
1  
Dear Matt, you have sharp eyes! Actually I had thought about the question and I don't know the answer.The book has that author's name spelled in capitals, which leaves the problem open. There are all sorts of variants: the chain of fast food restaurants is called McDonald's, and was founded by the brothers Roger and Maurice McDonald. The logic of a capital letter inside a proper name is that Mac stands for "son" in Scottish. (to be continued) –  Georges Elencwajg Jun 12 '12 at 11:07
1  
(continued) Similarly, the physicist who demonstrated that there is no ether was called FitzGerald, where Fitz in Ireland again stands for "son" and, amazingly, is derived from the French "fils" which still exactly means "son". To come back to your original remark, if I see a trustworthy source proving that Atiyah's co-author is named Macdonald, I'll be happy to correct my post accordingly. –  Georges Elencwajg Jun 12 '12 at 11:09
    
Dear Georges, actually I think you spelled it correctly. Looking at the book cover on Amazon suggests that the correct spelling is MacDonald. The reason why I thought it was spelled Macdonald is because I remember reading a post (I believe on MO) pointing out that the correct spelling was Macdonald. I should have checked the (latest) book cover on Amazon first before writing a presumptuous comment. My apologies! –  Rudy the Reindeer Jun 12 '12 at 11:18
    
Thank you for your explanation of the etymology of these names! –  Rudy the Reindeer Jun 12 '12 at 11:19

I believe your ring $R$ is supposed to be commutative, otherwise see here for example http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring.

The answer is hidden in the what are the actions of $R$ on the two algebras. An $R$-algebra is a ring with an action of $R$ (which is a left and right action since $R$ is commutative), and the maps $f \colon R \to S$ and $f \colon R \to T$ give you the actions.

They are $R \times S \to S \colon (r,s) \mapsto r \ast s := f(r) s$, with the multiplication of $S$, and similarly for $T$.

Therefore, in the tensor product $S \otimes_{R} T$, the left action on $T$ and the right action on $S$ are needed in order to define the tensor product. Moreover, the left action on $S$ and the right action on $T$ will give the additional structure of $R$-algebra, and these are the ones we use.

To come back to your question, the equality is proved by

$f(r) \otimes 1_T = (f(r) \cdot 1_{S}) \otimes 1_T = (r \ast 1_S) \otimes 1_T = 1_S \otimes (1_T \ast r) = 1_S \otimes (1_T \cdot g(r)) = 1_S \otimes g(r)$, where the essential step is the use of the relation $r a \otimes b = a \otimes br$.

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.