Tensor product of quotient rings [duplicate]

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.

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 $(1+a)\otimes(x+b)$...