# Equivalence of Definitions of Quotient Field

## Theorem

Let $D$ be an integral domain.

Let $F$ be a field.

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

### Definition 1

A quotient field of $D$ is a pair $(F,\iota)$ where:

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

### Definition 2

A quotient field of $D$ is a pair $(F,\iota)$ where:

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

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

### Definition 3

A quotient field of $D$ is a pair $(F,\iota)$ 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 $\varphi = \bar\varphi \circ \iota$

### Definition 4

A quotient field of $D$ is a pair $(F,\iota)$ 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 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 $F\subset K$.

$\Box$

### 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

 $\displaystyle \frac{\iota(x)}{\iota(y)} \cdot \frac{\iota(z)}{\iota(w)}$ $=$ $\displaystyle \frac{\iota(x)\iota(z)}{\iota(y)\iota(w)}$ $\displaystyle$ $=$ $\displaystyle \frac{\iota(xz)}{\iota(yw)}$ $\iota$ is a ring homomorphism $\displaystyle$ $\in$ $\displaystyle K$

and

 $\displaystyle \frac{\iota(x)}{\iota(y)} - \frac{\iota(z)}{\iota(w)}$ $=$ $\displaystyle \frac{\iota(x)\iota(w) - \iota(z)\iota(y)}{\iota(y)\iota(w)}$ $\displaystyle$ $=$ $\displaystyle \frac{\iota(xw-zy)}{\iota(yw)}$ $\iota$ is a ring homomorphism $\displaystyle$ $\in$ $\displaystyle K$

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

 $\displaystyle \left(\frac{\iota(x)}{\iota(y)}\right)^{-1}$ $=$ $\displaystyle \frac{\iota(y)}{\iota(x)}$ $\displaystyle$ $\in$ $\displaystyle K$

By Subfield Test, $K$ is a field.

By assumption, $K=F$.

$\Box$

### 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 $F\subset K$.

$\Box$

### 3 implies 4

Let $S = D_{\neq0}$.

Because $\iota$ is a monomorphism, $\iota(S) \subset F_{\neq0}$.

Because $F$ is a field, $\iota(S) \subset F^\times$.

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 $g(S) \subset B^\times$.

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.

$\Box$

$\blacksquare$