Field of fractions of $R = \{ \frac{a}{b} | \ a \in \mathbb{Z}, b \in \mathbb{Z} \setminus p\mathbb{Z} \}$.

Question

Let $p$ be a prime number. Define the subring $R = \{ \frac{a}{b} | \ a \in \mathbb{Z}, b \in \mathbb{Z} \setminus p\mathbb{Z} \}$ of $\mathbb{Q}$. Can someone explain to me why the set of rationals $\mathbb{Q}$ is the field of fractions of $R$ ? And also, why does $p$ have to be a prime number ? I haven't studied alot of ring theory yet so an explanation based on only the definition of field of fractions would be useful.

Thanks in advance !

Answer 1

The field of fractions of $R$ is the field $\{\frac{a}{b}\mid a,b\in R, b\neq 0\}$ under the relation of equivalence of fractions, i.e. $\frac{a}{b}=\frac{c}{d}$ iff $ad=bc$ (this products in $R$), with the usual operations of addition and multiplication of fractions.

Let $a=\frac{x}{y}$ and $b=\frac{z}{t}\in R$. We can see that the fraction $\frac{a}{b}$ is equivalent to the fraction $\frac{\frac{xt}{y}}{\frac{z}{1}}\in Frac(R)$, which is identified with the fraction $\frac{xt}{yz}\in \mathbb{Q}$. Indeed, we have that these two fractions are equivalent iff $a\frac{z}{1}=b\frac{xt}{y}$ in $R$, but with the usual operations we have $b\frac{xt}{y}=\frac{z}{t}\frac{xt}{y}$. As you might expect, $t$ cancels, but you can show this using the definition of equivalence of fractions, i.e. $\frac{zxt}{ty}$ is equivalent to $\frac{zx}{y}=a\frac{z}{1}$ in $R$.

This means that any fraction of elements of $R$ can be seen a a rational number after the identification. Therefore, $Frac(R)$ is a non-trival (find two fractions that are not equivalent to show this) subfield of the rationals, and you can use that the only such subfield is $\mathbb{Q}$ itself. Or artenatively, given a rational number $\frac{r}{s}$ you can find an element of $Frac(R)$ that can be identified with it, for instance $\frac{\frac{r}{1}}{\frac{s}{1}}$.

Answer 2

Note that the most you can say is that “the field of fractions of $R$ is isomorphic to $\mathbb{Q}$”.

Trying to look at the problem from a more general point of view may help.

You're given a subring $R$ of a field $F$. I want to show that $\operatorname{Frac}(R)$ can be naturally embedded in $F$ as a subfield containing $R$.

Let's take $a,b\in R$, with $b\ne0$; denoting as usual by $a/b$ the fraction in $\operatorname{Frac}(R)$, we can think to send it to $ab^{-1}\in F$.

Now, is $a/b\mapsto ab^{-1}$ a ring homomorphism? First we need to show it's well defined.

Saying that $a/b=c/d$ amounts to $ad=bc$. Then $$ ab^{-1}=add^{-1}b^{-1}=bcd^{-1}b^{-1}=cd^{-1} $$ and we're done. The other verifications are easy.

So we have $f\colon\operatorname{Frac}(R)\to F$ such that $f(a/1)=a$, for $a\in R$. Hence the image of $f$ is (a realization of) the field of fractions.

In your case, (this realization of) the field of fractions of $R$ is a subfield of $\mathbb{Q}$ and there are not many of them: $\mathbb{Q}$ has a single subfield, namely itself.