While reading, there is an isomorphism that I'm having trouble fulling seeing.
If you have two algebras $A$ and $B$ over a commutative ring $R$, with $I$ and $J$ two sided ideals in $A$ and $B$, then you should have an isomorphism
$$
(A/I)\otimes (B/J)\cong (A\otimes B)/(I\otimes 1+1\otimes J).
$$
Now there are bilinear maps from $A/I\times B/J\to (A\times B)/(I+J)$ and so the universal property of the tensor product gives unique maps $(A/I)\otimes (B/J)\to (A\times B)/(I+J)$. Does this somehow get to the isomorphism above, or am I completely off track? What is the quick way to see this? Thanks!
Best Answer
By the universal properties of quotient algebras, tensor products ( = coproducts of algebras), we have for every $R$-algebra $T$:
$\hom(A/I \otimes B/J,T) \cong \hom(A/I,T) \times \hom(B/J,T)$
$\cong \{f \in \hom(A,T),g \in \hom(B,T) : f|_I = 0, g|_J = 0\}$
$\cong \{h \in \hom(A \otimes_R B,T) : f:=h(- \otimes 1), g:=h(1 \otimes -) \text{ satisfy } f|_I = 0,~ g|_J = 0\}$
$\cong \{h \in \hom(A \otimes_R B,T) : h|_{I \otimes 1 + 1 \otimes J}=0\}$
$\cong \hom((A \otimes_R B)/(I \otimes 1 + 1 \otimes J),T)$
By the Yoneda lemma, we are done.
Remark: This is one of the thousands of trivial isomorphisms in basic algebra which are usually proved (in textbooks, lectures, etc.) in a too complicated way. Instead, you can always just use the Yoneda lemma and the involved universal properties. And then there is nothing to do at all ... By the way, this abstract approach is the only one which is applicable in more abstract contexts, where you can't use elements anyway.