By the way, is "Zero Homomorphism" the correct translation of the German word "Nullhomomorphismus" ? When I googled, I did not find the word Null Homomorphism in English
Let $K_1$ and $K_2$ be fields, and let $f: K_1 \to K_2$ be a Ring Homomorphism. Prove that $f$ is either injective or the Zero Homomorphism ($f(a) = 0_{K_2}$ for all $a \in K_1$)
My attempt: $$f: K_1 \to K_2$$ Some properties of Homomorphism are: $$f(a+b) = f(a) + f(b)$$ $$f(a \cdot b) = f(a) \cdot f(b)$$
We have two cases:
Case 1, $f$ is Zero Homomorphism:
$\exists k \neq 0$ such that $f(k) = 0$
$K$ \ {$0$} is an Abelian Group.
$a \neq 0 \implies a^{-1}$ exists.
So we have:
$$f(a) = f(a \cdot k^{-1}k) = f(a)f(k^{-1})f(k) = 0$$
Case 2, $f$ is injective:
$\forall k \neq 0$ we have $f(k) \neq 0$
$$f(a) = f(b) \implies f(a) - f(b) = 0 \implies f(a-b) = 0 \implies a - b = 0 \implies a = b$$
Is this correct ? Did I miss or forget something ?