Prove the fractional field of an integral domain is the smallest field containing the integral domain

Question

I have two questions about the fractional field of an integral domain.

Given an integral domain $D$:

1. Is there a difference between saying "the fractional field of $D$ is the smallest field containing $D$" or "the fractional field of $D$ is the smallest field containing an embedding of $D$"?

2. How do you prove that the fractional field is the smallest field containing $D$ (or an embedding of $D$, if there is a difference...)? Specifically, I want to show that if $F$ is any field containing $D$, then $F$ must contain the fractional field of $D$.

Thanks for your help.

Answer 1

Let $F'$ be a smallest field containing an embedding of $D$ ($f:D\to F'$), and $F$ a field of fraction of $D$.

We can extend $f$ to a morphism of the field $\tilde f:F\to F'$ by $\tilde f(a/b)=f(a)/f(b)$.

Now we have that $\tilde f(F)\subseteq F'$ and $\tilde f(F)$ containing an embedding of $D$ , by smallest property we have $\tilde f(F)=F'$.

So the two fields $F$ and $F'$ are isomorphic.

Edit: If $F$ is any field containing $D$. And denote $K$ the field of fraction of $D$.

Let $a/b\in K$, $a\in D$ and $0\neq b\in D$, hence $a,b\in F$, it follow that $a$ and $1/b $ are in $F$ so $a. (1/b)=a/b\in F$. Thus $K\subseteq F$.

Answer 2

The right (i.e. categorical) way to say this (without the ambiguities of words like "smallest", "containing", etc.) ought to be that the inclusion $\iota: D\to Q(D)$ has the following universal property:

If $K$ is a field, and $f: D\to K$ is any morphism of rings, then there is a unique morphism of fields $g : Q(D) \to K$ such that $f = g \circ \iota$.

(In particular, $Q(D)$ embeds into any field that $D$ embeds into.)

This property uniquely determines (up to isomorphism) not only $Q(D)$, but $\iota$ as well.

And it's easily proved, since $g(1/b)g(b)=g(1)$ forces $g(a/b) = f(a)/f(b)$, so this amounts to checking that $a/b \mapsto f(a)/f(b)$ is actually a homomorphism.