Tensor product of quotient rings $A$ is a commutative ring with unit and $\mathfrak a$, $\mathfrak b$ ideals. I have to show that $$A/\mathfrak a \otimes_{A} A/\mathfrak b \cong A/(\mathfrak{a+b}).$$ Any hint ?
 A: It may be easier to prove the more general fact: if $R$ is any $A$-ring (i.e. we have a chosen homomorphism $\varphi : A \to R$), then we have
$$ A/I \otimes_A R \cong R / IR $$

In the following, $a,a'$ are variables denoting elements of $A$, $i$ an element of $I$, and $r,r'$ elements of $R$.
This can be done by explicitly writing down the maps: in the forward direction, it's enough to define it on pure tensors
$$ a \otimes r \mapsto \varphi(a) r $$
and in the reverse direction
$$ r \mapsto 1 \otimes r $$
All that's left is to show the maps are well-defined, are homomorphisms, and are actually inverses.
e.g. the forward map followed by the reverse map is the map
$$ a \otimes r \mapsto 1 \otimes \varphi(a) r = a \otimes r $$
and is this the identity (the equality is one of the arithmetic properties of the tensor product).
To show the map in the forward direction is well-defined, we need to show four things:


*

*$ i \otimes r$ maps to the zero element of $R / IR$

*$a a' \otimes r$ and $a \otimes \varphi(a') r$ map to the same element of $R/IR$.

*$(a+a') \otimes r$ maps to the sum of the images of $a \otimes r$ and $a' \otimes r$

*$a \otimes (r+r')$ maps to the sum of the images of $a \otimes r$ and $a \otimes r'$


I'l let you work out what else needs to be shown.
A: Define 
$$\phi : A/a \times A/b \rightarrow A/(a+b),\; \phi(x+a,y+b) = xy + (a+b).$$ 
Can you show $\phi$ is a well-defined bilinear map? 
Now if you use the universal property of tensor products to get a map 
$$\psi : A/a \otimes_A A/b \rightarrow A/(a+b),$$ can you show $\psi$ is injective? To show surjectivity, it will probably help you to show that an arbitrary simple tensor in $A/a \otimes_A A/b$ can be written in the form $(1+a)\otimes(x+b)$...
