Equivalence of Definitions of Field of Quotients

Theorem
Let $D$ be an integral domain.

Let $F$ be a field.

The following definitions of quotient field are equivalent:

1 implies 2
Let $K$ be a field with $\iota \left({D}\right) \subset K \subset F$.

We show that $F\subset K$.

Let $f\in F$.

By assumption, there exist $x,y\in D$ with $y\neq0$ such that $f=\iota(x)/\iota(y)$.

Because $K$ is a field containing $\iota(D)$, $K$ also contains $f=\iota(x)/\iota(y)$.

Thus $F\subset K$.

1 implies 3
Let $E$ be a field and $\phi : D \to E$ a ring monomorphism.

If $\bar\phi : F \to E$ is such that $\phi = \bar\phi \circ \iota$ and $f \in F$ with $f = \iota(x)/\iota(y)$ with $x,y\in D$, then $\bar\phi(f) = \frac{\bar\phi(\iota(x))}{\bar\phi(\iota(y))} = \frac{\phi(x)}{\phi(y)}$.

Thus there is only one option for $\bar\phi$.

It remains to verify that the mapping which sends $f = \iota(x)/\iota(y)$ to $\frac{\phi(x)}{\phi(y)}$ is:
 * well-defined
 * a field homomorphism

2 implies 1
Let $K$ be the subset of elements of $F$ that are of the form $\iota(x)/\iota(y)$.

We show that $K$ is a field containing $\iota(D)$, which by assumption implies $K=F$.

$\iota(D)\subset H$, because $\iota(x) = \iota(x)/\iota(1) \in K$ for $x\in D$.

We use Subfield Test to show that $K$ is a field:

If $\frac{\iota(x)}{\iota(y)}, \frac{\iota(z)}{\iota(w)} \in K$, then

and

If $\frac{\iota(x)}{\iota(y)} \in K^\times$, then $x\neq0$, so

By Subfield Test to show that $K$ is a field.

Thus $K=F$.

3 implies 2
Let $K$ be a field with $\iota \left({D}\right) \subset K \subset F$.

We show that $F\subset K$.

We apply the universal property to $\iota : D \to K$ and $\iota : D \to F$.

By assumption, there exists:
 * a unique field homomorphism $\bar \iota_1 : F \to K$ such that $\iota = \bar\iota_1 \circ \iota$.
 * a unique field homomorphism $\bar \iota_2 : F \to F$ such that $\iota = \bar\iota_2 \circ \iota$.

By uniqueness, $\bar\iota_2 = \operatorname{id}_F$ is the identity mapping on $F$.

Because $K\subset F$, $\iota_1$ fulfills the second condition as well.

By uniqueness, $\iota_1 = \iota_2$.

Because $F = \operatorname{im}(\iota_2) = \operatorname{im}(\iota_1) \subset K$, we have $F\subset K$.

4 implies 3
Follows immediately from the definition of localisation.