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.

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This is a consequence of more general statements, which are easy to prove (just verify the universal properties):

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

2. 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.

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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 $D$.

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$p$ is always surjective; the equations show that $p$ is invertible. – Martin Brandenburg Feb 13 '13 at 15:11
@MartinBrandenburg Sorry there was a typo which I have corrected. I used the word surjective twice :) – user38268 Feb 13 '13 at 15:13
@MartinBrandenburg I have edited my answer. – user38268 Feb 13 '13 at 15:32

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.

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