# If B and C are A -algebras with ring morphisms f : A &#x2192;<!-- → --> B and

Lillianna Andersen 2022-07-15 Answered
If $B$ and $C$ are $A$-algebras with ring morphisms $f:A\to B$ and $g:A\to C$, and $D=B{\otimes }_{A}C$ is an $A$-algebra with morphism $a↦f\left(a\right)\otimes g\left(a\right)$, then $uf=vg$, where $u:B\to D$ is $u\left(b\right)=b\otimes 1$.The map $v:C\to D$ is not defined in the text, but my guess is it's $v\left(c\right)=1\otimes c$.I don't understand why the diagram is commutative though. That would imply $f\left(a\right)\otimes 1=1\otimes g\left(a\right)$ for all $a\in A$. Is that true, or is v something else?Added: On second thought, does this follow since $f\left(a\right)\otimes 1=a\cdot \left(1\otimes 1\right)$ and $1\otimes g\left(a\right)=a\cdot \left(1\otimes 1\right)$ where ⋅ is the $A$-module structure on $D$?
$\begin{array}{ccc}A& \stackrel{f}{\to }& B\\ g↓& \mathrm{#}& ↓u& \\ C& \underset{v}{\overset{\phantom{\rule{2.75em}{0ex}}}{\to }}& D\end{array}$
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## Answers (1)

diamondogsaz
Answered 2022-07-16 Author has 12 answers
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 $ab=f\left(a\right)b$.
So we have
$u\circ f\left(a\right)=f\left(a\right)\otimes {1}_{C}=f\left(a\right)\cdot {1}_{B}\otimes {1}_{C}=a\cdot {1}_{B}\otimes {1}_{C}=a\left({1}_{B}\otimes {1}_{C}\right)$
and
$v\circ g\left(a\right)={1}_{B}\otimes g\left(a\right)={1}_{B}\otimes {1}_{C}\cdot g\left(a\right)={1}_{B}\otimes {1}_{C}\cdot a=\left({1}_{B}\otimes {1}_{C}\right)\cdot a$
but $D=B{\otimes }_{R}C$ is a commutative ring with identity ${1}_{B}\otimes {1}_{C}$ so you can conclude that $a\left({1}_{B}\otimes {1}_{C}\right)=\left({1}_{B}\otimes {1}_{C}\right)a$ and hence $uf=vg$ as desired.
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