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}

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

1 Answer 1


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.


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