Isomorphisms between tensor products field of fractions of integral domain Let $D$ be an integral domain.  How can I show that there are isomorphisms
$$F(D) \otimes_D F(D) \to F(D) \otimes_{F(D)} F(D) \to F(D)$$
The identity map seems like the natural choice for the first one.  Does it have a nontrivial kernel? For the second one, I am stuck.
 A: This is a consequence of more general statements, which are easy to prove (just verify the universal properties):


*

*If $M$ is an $R$-module, then $R \otimes_R M \cong M$.

*If $M,N$ are $S$-modules and $R \to S$ is a homomorphism of commutative rings, there is a canonical homomorphism of $R$-modules $M \otimes_R N \to M \otimes_S N$. If $S$ is a localization of $R$ (or more general $R \to S$ is an epimorphism of commutative rings), it is an isomorphism. The reason is that a $R$-bilinear map on $M \times N$ is already $S$-bilinear.
A: Let me explain more as to why $F(D) \otimes_D F(D) \cong F(D) \otimes_{F(D)} F(D)$ as $D$ - modules. Now strictly speaking, the map 
$$p : F(D) \otimes_{F(D)} F(D) \to F(D) \otimes_D F(D)$$ 
that sends $m \otimes n \to m \otimes n$ is not the identity map because the two objects in question a priori are different! On the left we have the equalities
$$ma \otimes n = m \otimes an$$ with $m,n \in F(D), a \in D$ while on the right we have the further equalities
$$ m \left(\frac{j}{k} \right)\otimes n = m\otimes \left(\frac{j}{k} \right) n$$
where $m,n \in F(D)$ and $j,k \in D$, $ k \neq 0$.  The way to think about why it is an isomorphism is that those further equalities I wrote above also hold true in  $F(D) \otimes_D F(D)$. Using that $F(D)$ is an $F(D)$ - module, we have 
$$\begin{eqnarray*} \left(\frac{j}{k} m\right) \otimes n &=& (k^{-1}mj )\otimes n\\
 &=& k^{-1}m \otimes nj \\
&=& k^{-1}m \otimes (njk^{-1})k \\
&=& k^{-1}m k \otimes  njk^{-1} \\
&=& m\otimes \left(n\frac{j}{k}\right)\end{eqnarray*} $$
and boom! The $\frac{j}{k}$ has jumped over the other side. Thus we see that  $F(D) \otimes_D F(D)$ is actually a tensor product over the ring $F(D)$ and not just $D$.
A: You spotted the obvious choice of map
$$ F(D) \otimes_D F(D) \to F(D) \otimes_{F(D)} F(D) : a \otimes b \mapsto a \otimes b$$
The second choice should also be "obvious" once you're used to things: elements of $F(D) \otimes_{F(D)} F(D)$ are written as (sums of) (tensor) products of two ring elements. The only product available in $F(D)$, however, is ring multiplication, so the natural choice is
$$ F(D) \otimes_{F(D)} F(D) \to F(D) : a \otimes b \mapsto ab$$
The "obvious" choice of map $F(D) \to F(D) \otimes_D F(D)$ comes about by similar reasoning: to turn a single element into a product of two elements, you can use $1$ as the other element:
$$ F(D) \to F(D) \otimes_D F(D) : a \mapsto a \otimes 1 $$
Be sure to check all of these maps are well-defined!
Now, if you started with $\frac{a}{b} \otimes \frac{c}{d} \in F(D) \otimes_D F(D)$, going through all three maps results in $\frac{ac}{bd} \otimes 1$. We can simplify
$$ \frac{ac}{bd} \otimes 1 = \frac{a}{bd} \otimes c$$
but how can we get a $d$ in the denominator on the right hand side? Well is easy to do in $F(D)$: just put one there and put one in the numerator too to cancel it out
$$ \frac{a}{bd} \otimes c = \frac{a}{bd} \otimes \frac{cd}{d}
= \frac{a}{b} \otimes \frac{c}{d}$$
Okay, so the composite of all three maps starting from $F(D) \otimes_D F(D)$ is the identity. That was the only hard one: the triple composite starting at $F(D) \otimes_{F(D)} F(D)$ sends $x \otimes y \mapsto xy \otimes 1$, and
$$ xy \otimes 1 = x \otimes y $$
and if we start at $F(D)$, it sends $x \mapsto x$.
All three triple composites are identities, and so each of the individual maps are isomorphisms.
