Field of Fractions of a Field

Question

I am trying to prove that if $R$ is a field, then its field of fractions is simply itself.
What would be the isomorphism that I would set up for this problem?
Any hints would be appreciated.

Answer 1

In this answer, I'll be assuming that all rings have a multiplicative identity, and that if $\phi:R\to S$ is a ring homomorphism, then $\phi(1_R)=1_S$

Let $\text{Frac}(R)$ denote the field of fractions for an integral domain $R$. Then $\text{Frac}(R)$ satisfies the following property: for any field $K$ and any injective ring homomorphism $h:R\to K$, there is a unique ring homomorphism $H:\text{Frac}(R)\to K$ that extends $h$.

Let $F$ be a field, and let $h:F\to F$ be the identity map. Then $h$ is an injective ring homomorphism from $F$ to a field. Hence it can be extended to ring homomorphism $H:\text{Frac}(F)\to F$. Note that $H$ is surjective since it extends $h$ which is surjective. Also $H$ is injective since $H$ is a ring homomorphism whose domain is a field. It follows that $H:\text{Frac}(F)\to F$ is an isomorphism.