# There exists an injective ring homomorphism $\bar{\phi} : Q \to F$ such that $j \circ \phi = \bar{\phi} \circ j$.

Let $A$ and $B$ two integral domains and $Q$ and $F$ their fields of fractions, respectively. Consider the ring homomorphisms $i : A \to Q$ and $j:B \to F$, defined as $i(a)=\frac{a}{1_A}$ and $j(b)=\frac{b}{1_B}$. Show that if $\phi : A \to B$ is an injective ring homomorphism, then there existe another injective ring homomorphism $\bar{\phi} : Q \to F$ such that $j \circ \phi = \bar{\phi} \circ j$.

I thought $\bar{\phi}= j \circ \phi \circ i^{-1}$; the problem is $i$ and $j$ are not bijective.

Is anyone could give me a hint to solve this problem?

• Consider $\bar{\phi}\left( \frac{a_1}{a_2} \right) = \frac{\phi(a_1)}{\phi(a_2)}$. – Jendrik Stelzner Jan 20 '16 at 23:13

## 1 Answer

The following statement is the “universal property of the field of fractions”.

Let $A$ be an integral domain, with field of fractions $Q$ and canonical inclusion $i\colon A\to Q$. If $K$ is any field and $f\colon A\to K$ is an injective ring homomorphism, then there is a ring homomorphism $g\colon Q\to K$ such that $f=g\circ i$.

Proof. Define the map $g$ by $$g(a/t)=f(a)f(t)^{-1}$$ Note that, by injectivity of $f$, if $t\ne0$, then $f(t)\ne0$.

This is well defined, because, for any $s\in A$, we have $$f(as)f(ts)^{-1}=f(a)f(s)f(t)^{-1}f(s)^{-1}=f(a)f(t)^{-1}$$ Verifying that $g$ is a ring homomorphism is just a standard computation.

Apply this to $K=F$ and $f=j\circ\phi$ and set $\bar\phi=g$. Note that a ring homomorphism $Q\to F$ is necessarily injective.