Field Homomorphism Preserves Unity

Theorem
Let $\phi: \left({F_1, +_1, \times_1}\right) \to \left({F_2, +_2, \times_2}\right)$ be a field homomorphism.

Let:
 * $1_{F_1}$ be the unity of $F_1$
 * $1_{F_2}$ be the unity of $F_2$.

Then:
 * $\phi \left({1_{F_1}}\right) = 1_{F_2}$

Proof
By definition, if $\left({F_1, +_1, \times_1}\right)$ and $\left({F_2, +_2, \times_2}\right)$ are fields then $\left({F_1^*, \times_1}\right)$ and $\left({F_2^*, \times_2}\right)$ are groups.

Again by definition:
 * the unity of $\left({F_1, +_1, \times_1}\right)$ is the identity of $\left({F_1^*, \times_1}\right)$
 * the unity of $\left({F_2, +_2, \times_2}\right)$ is the identity of $\left({F_2^*, \times_2}\right)$.

The result follows from Group Homomorphism Preserves Identity.