Properties of the Tensor Product

Here are various results about the tensor product, one of my favorite mathematical constructions.

Lemma:

Let $A$ be a ring with elements $f, g$. Then $A/(f)\otimes_{A}A/(g)\cong A/(f, g)$, where we give the quotients the natural $A$-algebra structure.

Proof. We can always write an element $a\otimes b$ of $A/(f)\otimes_AA/(g)$ as $b'a\otimes 1$ for some lift $b'\in A$ of $b$, because the maps which we use to make $A/(f)$ and $A/(g)$ into $A$-algebras are surjective. Explicitly, we can lift $b$ to an element $b'$ of $A$ with $a\otimes b = a\otimes (b'\cdot 1) = (b'\cdot a)\otimes 1$ (where $\cdot$ is the action of $A$ on $A/(g)$ or $A/(f)$, as appropriate). From this, we see that the map $A\to A/(f)\otimes_AA/(g)$ by the rule $c\mapsto c\otimes 1$ is surjective. Suppose $c= af+bg$. Then $c\mapsto (af+bg)\otimes_A1=af\otimes_A1 + 1\otimes_Abg = 0$. Similarly, if $c\otimes 1 = 0$, then we must have $c = t\cdot 1\otimes_A 1+ 1\otimes v\cdot 1 = 0+0$ for $t,v\in A$; the only way for this to happen is if $t\in (f)$ and $v\in (g)$. Thus the ideal of this map is $(f, g)$, and we are done.