3
$\begingroup$

On page 31 of Atiyah and Macdonald, there is a commutative diagram. It essentially says that if $B$ and $C$ are $A$-algebras with ring morphisms $f:A\to B$ and $g\colon A\to C$, and $D=B\otimes_A C$ is an $A$-algebra with morphism $a\mapsto f(a)\otimes g(a)$, then $uf=vg$, where $u:B\to D$ is $u(b)=b\otimes 1$.

The map $v:C\to D$ is not defined in the text, but my guess is it's $v(c)=1\otimes c$.

I don't understand why the diagram is commutative though. That would imply $f(a)\otimes 1=1\otimes g(a)$ for all $a\in A$. Is that true, or is $v$ something else?

Added: On second thought, does this follow since $f(a)\otimes 1=a\cdot(1\otimes 1)$ and $1\otimes g(a)=a\cdot (1\otimes 1)$ where $\cdot$ is the $A$-module structure on $D$? $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V g V V\# @VV u V\\ C @>>v> D \end{CD}

$\endgroup$
1
  • $\begingroup$ Your second thought is indeed the answer. $\endgroup$ Oct 2, 2014 at 17:54

1 Answer 1

2
$\begingroup$

In their definition for algebras, they let $f : A \to B$ be a ring homomorphism; if $a \in A$, $b \in B$, define a product $ab = f(a)b$.

So we have $$u \circ f(a) = f(a) \otimes 1_C = f(a) \cdot 1_B \otimes 1_C = a \cdot 1_B \otimes 1_C = a (1_B \otimes 1_C)$$ and $$v \circ g(a) = 1_B \otimes g(a) = 1_B \otimes 1_C \cdot g(a) = 1_B \otimes 1_C \cdot a = (1_B \otimes 1_C) \cdot a$$ but $D = B \otimes_R C$ is a commutative ring with identity $1_B \otimes 1_C$ so you can conclude that $a (1_B \otimes 1_C) = (1_B \otimes 1_C)a$ and hence $uf = vg$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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