# Equivalence of Definitions of Field of Quotients

## Theorem

Let $D$ be an integral domain.

Let $F$ be a field.

The following definitions of the concept of Field of Quotients are equivalent:

### Definition 1

A field of quotients of $D$ is a pair $\struct {F, \iota}$ where:

$(1): \quad$ $F$ is a field
$(2): \quad$ $\iota : D \to F$ is a ring monomorphism
$(3): \quad \forall z \in F: \exists x \in D, y \in D_{\ne 0}: z = \dfrac {\map \iota x} {\map \iota y}$

### Definition 2

A field of quotients of $D$ is a pair $\struct {F, \iota}$ such that:

$(1): \quad F$ is a field
$(2): \quad \iota: D \to F$ is a ring monomorphism
$(3): \quad$ If $K$ is a field with $\iota \sqbrk D \subset K \subset F$, then $K = F$.

That is, the field of quotients of an integral domain $D$ is the smallest field containing $D$ as a subring.

### Definition 3

A field of quotients of $D$ is a pair $\struct {F, \iota}$ where:

$(1): \quad$ $F$ is a field
$(2): \quad$ $\iota : D \to F$ is a ring monomorphism
$(3): \quad$ it satisfies the following universal property:
For every field $E$ and for every ring monomorphism $\varphi: D \to E$, there exists a unique field homomorphism $\bar \varphi: F \to E$ such that $\varphi = \bar \varphi \circ \iota$
That is, the following diagram commutes:
$\begin{xy}\xymatrix@[email protected]+2px{D \ar[r]^\iota \ar[dr]_\varphi & F \ar[d]^{\exists_1 \bar \varphi} \\ & E}\end{xy}$

### Definition 4

A field of quotients of $D$ is a pair $\struct {F, \iota}$ which is its total ring of fractions, that is, the localization of $D$ at the nonzero elements $D_{\ne 0}$.

## 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 $F \subseteq K$.

$\Box$

### 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:

 $\ds \frac {\map \iota x} {\map \iota y} \cdot \frac {\map \iota z} {\map \iota w}$ $=$ $\ds \frac {\map \iota x \map \iota z} {\map \iota y \map \iota w}$ $\ds$ $=$ $\ds \frac {\map \iota {x z} } {\map \iota {y w} }$ $\iota$ is a ring homomorphism $\ds$ $\in$ $\ds K$

and:

 $\ds \frac {\map \iota x} {\map \iota y} - \frac {\map \iota z} {\map \iota w}$ $=$ $\ds \frac {\map \iota x \map \iota w - \map \iota z \map \iota y} {\map \iota y \map \iota w}$ $\ds$ $=$ $\ds \frac {\map \iota {x w - z y} } {\map \iota {y w} }$ $\iota$ is a ring homomorphism $\ds$ $\in$ $\ds K$

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

Then:

$x \ne 0$

so:

 $\ds \paren {\frac {\map \iota x} {\map \iota y} }^{-1}$ $=$ $\ds \frac {\map \iota y} {\map \iota x}$ $\ds$ $\in$ $\ds K$

By Subfield Test, $K$ is a field.

By assumption, $K = F$.

$\Box$

### 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 $F \subseteq K$.

$\Box$

### 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 $3$.

$\Box$

### 4 implies 3

Follows immediately from the definition of localization.

$\Box$

$\blacksquare$