Let $K$ and $K'$ be two fields and let $f : K \to K'$ a homomorphism (we do not include $f(1) = 1$ in the definition like many notes do). I want to show that $f(1_K) = 1_{K'}$ and I would like to receive many opinions about the truth of my two differents proofs.
- If $G$ and $G'$ are two groups with unit elements $e$ and $e'$, an homomorphism $G \to G'$ maps $e$ to $e'$, so, since $(K \setminus \{0_K\} , {\cdot}_K)$ and $(K' \setminus \{0_{K'}\} , {\cdot}_{K'})$ are two groups, then $f$ maps $1_K$ to $1_{K'}$.
- Assuming that $Im f$ is a subfield in $K'$, we can prove that $f$ is inyective using indeed the First Isomorphy Theorem; in fact, by FIT, $\ker f \trianglelefteq K$, so we have two options: $\ker f = \{0_K\}$ or $\ker f = K$ (since $K$ is a field, the only ideals of $K$ are these). If $\ker f = K$, we would obtain that $Im f = \{0_{K'}\}$, so $Im f$ would not be a subfield in $K'$ because a field has at least two elements: $0$ and $1$. This imply that $f$ is injective. Thus, $f(1_K) = 1_{K'}$ because we have that $$ f(1_K) = f(1_K {\cdot}_K 1_K) = f(1_K) {\cdot}_{K'} f(1_K) $$ and now we can remove $f(1_K)$ from the two sides of the equality (we can do because $f(1_K) \neq 0_{K'}$. It is just because $\ker f = \{0_K\}$, as $f$ is injective).
Both your proofs look fine to me.
I do it like this:
$1_{K'}f(1_{K}) = 1_{K'}f(1_{K}^2) = 1_{K'}(f(1_{K}))^2 \Longrightarrow 1_{K'} = f(1_{K}), \tag 1$
since we assume
$f(1_{K'}) \ne 0. \tag 2$