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).

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