Equivalence of Definitions of Field of Quotients

== Theorem ==
Let $D$ be an integral domain.
Let $F$ be a field.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== 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
Category:Fields of Quotients