Every field contains a unique smallest subfield

Let $F$ be a field. Prove that $F$ contains a unique smallest subfield $F_0$ and that $F_0$ is isomorphic to either $\mathbb{Q}$ or $\mathbb{Z}/(p)$ for some prime $p$. ($F_0$ is called the prime subfield of $F$.)

We begin with a lemma.

Lemma: Let $F$ be a field and let $\mathcal{C}$ be a set of subfields of $F$. Then $\bigcap \mathcal{C}$ is a subfield of $F$. Proof: $\bigcap \mathcal{C}$ is a subring of $F$, and contains 1 since $1 \in E$ for all $E \in \mathcal{C}$. Let $r \in F$. Then $r \in E$ for each $E \in \mathcal{C}$, and so $r^{-1} \in E$. Then $r^{-1} \in F$, and thus $F$ is a field. $\square$

Let $F$ be a field, and let $\mathcal{F}$ denote the set of all subfields of $F$. By the lemma, $F_0 = \bigcap \mathcal{F}$ is a subfield of $F$ and is contained in every other subfield.

Now consider the ring homomorphism $\varphi : \mathbb{Z} \rightarrow F_0$ with $\varphi(1) = 1$. The induced map $\psi : \mathbb{Z}/(n) \rightarrow F_0$ is an injective ring homomorphism, and by these two previous exercises, $n$ is a prime or 0. Certainly for all nonzero $a \in \mathbb{Z}/(n)$, $\psi(a)$ is a unit in $F_0$. Thus we have an injective homomorphism $\Phi : F(\mathbb{Z}/(n)) \rightarrow F_0$, where $F(\mathbb{Z}/(n))$ denotes the field of fractions. Note that $\mathsf{im}\ \Phi$ is a subring of $F$ which is a field, and thus $\mathsf{im}\ \Phi = F_0$. Thus $\Phi$ is an isomorphism. Now if $n = 0$, then $F_0 \cong \mathbb{Q}$, and if $n = p$ is prime, then $F_0 \cong \mathbb{Z}/(p)$.