Equivalence of Definitions of Field of Quotients

Theorem
Let $D$ be an integral domain.

Let $F$ be a field.

1 implies 2
Let $K$ be a field such that:
 * $\iota \sqbrk D \subseteq K \subseteq F$

We show that $F \subseteq K$.

Let $f \in F$.

By assumption, there exist $x, y \in D$ with $y \ne 0$ such that $f = \dfrac {\map \iota x} {\map \iota y}$.

Because $K$ is a field containing $\iota \sqbrk D$, $K$ also contains $f = \dfrac {\map \iota x} {\map \iota y}$.

Thus $F \subseteq K$.

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

Let:
 * $\bar \phi: F \to E$ is such that $\phi = \bar \phi \circ \iota$

and:
 * $f \in F$ such that $f = \dfrac {\map \iota x} {\map \iota y}$ with $x, y \in D$.

Then:
 * $\map {\bar \phi} f = \dfrac {\map {\bar \phi} {\map \iota x} } {\map {\bar \phi} {\map \iota y} } = \dfrac {\map \phi x} {\map \phi y}$

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

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

2 implies 1
Let $K$ be the subset of elements of $F$ that are of the form $\dfrac {\map \iota x} {\map \iota y}$.

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

Because:
 * $\map \iota x = \dfrac {\map \iota x} {\map \iota 1} \in K$ for $x \in D$

we have that:
 * $\iota \sqbrk D \subseteq H$

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

Let $\dfrac {\map \iota x} {\map \iota y}, \dfrac {\map \iota z} {\map \iota w} \in K$.

Then:

and:

Let $\dfrac {\map \iota x} {\map \iota y} \in K^\times$.

Then:
 * $x \ne 0$

so:

By Subfield Test, $K$ is a field.

By assumption, $K = F$.

3 implies 2
Let $K$ be a field such that:
 * $\iota \sqbrk D \subseteq K \subseteq F$

We show that $F \subseteq K$.

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, $\bar {\iota_2} = I_F$ 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 $F = \Img {\iota_2} = \Img {\iota_1} \subseteq K$, we have $F \subseteq K$.

3 implies 4
Let $S = D_{\ne 0}$.

Because $\iota$ is a monomorphism:
 * $\iota \sqbrk S \subseteq F_{\ne 0}$

Because $F$ is a field:
 * $\iota \sqbrk S \subset F^\times$

It remains to verify the universal property of the localization.

Let $B$ be a ring with unity

Let $g: D \to B$ be a ring homomorphism such that $g \sqbrk S \subseteq B^\times$.

We show that there is a unique ring homomorphism $h: F \to B$ such that $g = h \circ \iota$.

This is done in exactly the same way as in the implication $1$ implies $3$.

4 implies 3
Follows immediately from the definition of localization.

Also see

 * Equivalence of Definitions of Localization of Ring