=== 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 $F \subseteq K$. {{qed|lemma}} === 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 {{finish}} === 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: {{begin-eqn}} {{eqn | l = \frac {\map \iota x} {\map \iota y} \cdot \frac {\map \iota z} {\map \iota w}       | r = \frac {\map \iota x \map \iota z} {\map \iota y \map \iota w}       | c =  }} {{eqn | r = \frac {\map \iota {x z} } {\map \iota {y w} }       | c = $\iota$ is a ring homomorphism }} {{eqn | r = K       | o = \in }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \frac {\map \iota x} {\map \iota y} - \frac {\map \iota z} {\map \iota w}       | r = \frac {\map \iota x \map \iota w - \map \iota z \map \iota y} {\map \iota y \map \iota w}       | c =  }} {{eqn | r = \frac {\map \iota {x w - z y} } {\map \iota {y w} }       | c = $\iota$ is a ring homomorphism }} {{eqn | r = K       | o = \in }} {{end-eqn}} Let $\dfrac {\map \iota x} {\map \iota y} \in K^\times$. Then: :$x \ne 0$ so: {{begin-eqn}} {{eqn | l = \paren {\frac {\map \iota x} {\map \iota y} }^{-1}       | r = \frac {\map \iota y} {\map \iota x}       | c =  }} {{eqn | o = \in       | r = K }} {{end-eqn}} By Subfield Test, $K$ is a field. By assumption, $K = F$. {{qed|lemma}} === 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 $F \subseteq K$. {{qed|lemma}} === 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 $3$. {{qed|lemma}} {{finish|maybe prove diretly that 1 implies 4 instead?}} === 4 implies 3 === Follows immediately from the definition of localization. {{qed|lemma}} {{qed}}