Quotient Theorem for Monomorphisms

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Theorem

Let $K, L$ be quotient fields of integral domains $\left({R, +_R, \circ_R}\right), \left({S, +_S, \circ_S}\right)$ respectively.

Let $\phi: R \to S$ be a monomorphism.

Then there is one and only one monomorphism $\psi: K \to L$ extending $\phi$, and:

$\displaystyle \forall x \in R, y \in R^*: \psi \left({\frac x y}\right) = \frac {\phi \left({x}\right)} {\phi \left({y}\right)}$

Also, if $\phi$ is a ring isomorphism, then so is $\psi$.

Proof

By definition, $\left({K, \circ_R}\right)$ and $\left({L, \circ_S}\right)$ are inverse completions of $\left({R, \circ_R}\right)$ and $\left({S, \circ_S}\right)$ respectively.

So by the Extension Theorem for Homomorphisms, there is one and only one monomorphism $\psi: \left({K, \circ_R}\right) \to \left({L, \circ_S}\right)$ extending $\phi$.

Thus:

$\displaystyle \forall x \in R, y \in R^*: \psi \left({\frac x y}\right) = \frac {\phi \left({x}\right)} {\phi \left({y}\right)}$

By the Extension Theorem for Isomorphisms, $\psi$ is an isomorphism if $\phi$ is.

Thus, $\forall x, y \in R, z, w \in R^*$:

 $\displaystyle \psi \left({\frac x z +_R \frac y w}\right)$ $=$ $\displaystyle \psi \left({\frac {\left({x \circ_R w}\right) +_R \left({y \circ_R z}\right)} {z \circ_R w} }\right)$ Addition of Division Products $\displaystyle$ $=$ $\displaystyle \frac {\phi \left({\left({x \circ_R w}\right) +_R \left({y \circ_R z}\right)}\right)} {\phi \left({z \circ_R w}\right)}$ Definition of $\psi$ $\displaystyle$ $=$ $\displaystyle \frac {\left({\phi \left({x}\right) \circ_S \phi \left({w}\right)}\right) +_S \left({\phi \left({y}\right) \circ_S \phi \left({z}\right)}\right)} {\phi \left({z}\right) \circ_S \phi \left({w}\right)}$ Morphism Property $\displaystyle$ $=$ $\displaystyle \frac {\phi \left({x}\right)} {\phi \left({z}\right)} +_S \frac {\phi \left({y}\right)} {\phi \left({w}\right)}$ Addition of Division Products $\displaystyle$ $=$ $\displaystyle \psi \left({\frac x z}\right) +_S \psi \left({\frac y w}\right)$ Definition of $\psi$

Thus $\psi: K \to L$ is a monomorphism.

$\blacksquare$