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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?

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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 (***) $

  • 0
    Dear @Georges, after looking at [Wikipedia's entry about Ian G. Macdonald](http://en.wikipedia.org/wiki/Ian_G._Macdonald) (as they spell it) my confusion remains. Is it more likely that the publisher got his name right or Wikipedia? It seems to me that the answer to this is not immediately obvious.2012-06-30
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I believe your ring $R$ is supposed to be commutative, otherwise see here for example https://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 $r a \otimes b = a \otimes br$.