Field Homomorphism Preserves Subfields

Theorem
Let $\left({F_1, +_1, \circ_1}\right)$ and $\left({F_2, +_2, \circ_2}\right)$ be fields.

Let $\phi: F_1 \to F_2$ be a field homomorphism such that $\phi$ is not the trivial homomorphism.

If $K$ is a subfield of $F_1$, then $\phi \left({K}\right)$ is a subfield of $F_2$.

Proof
First note that if $\phi$ is the trivial homomorphism then $\phi \left({K}\right) = 0_{F_2}$ and so is not a field.

Since $K$ is a field, we have that:
 * $0_{F_1} \in K$
 * $1_{F_1} \in K$

and so by Ring Homomorphism Preserves Zero and Field Homomorphism Preserves Unity:
 * $\phi \left({0_{F_1}}\right) = 0_{F_2} \in \phi \left({K}\right)$
 * $\phi \left({1_{F_1}}\right) = 1_{F_2} \in \phi \left({K}\right)$

So $\phi \left({K}\right)$ contains at least the zero and unity.

From Group Homomorphism Preserves Subgroups:
 * $\left({\phi \left({K}\right), +_2}\right)$ is a subgroup of $\left({F_2, +_2}\right)$
 * $\left({\phi \left({K}\right), \circ_2}\right)$ is a subgroup of $\left({F_2, \circ_2}\right)$.

Thus, as $\left({R_2, +_2}\right)$ and $\left({R_2, \circ_2}\right)$ are both groups, the result follows.