I've read the following theorem:
And am trying to understand what was done there. I think it is the following: We want to prove that
$$\text{Non zero hom of field to ring} \implies \text{hom is injective}$$
I think they used the contrapositive:
$$\overbrace{\varphi(a)=0 \;\wedge \;a\neq0}^{\text{hom not injective}}\implies \overbrace{\varphi(b)=0}^{\text{zero hom}}$$
Is that correct? I believe it is, I just want to confirm.
On
What is not said, but should have already been proven, if the homomorphism is not injective, it has a non-trivial kernel.
I would call this a proof by contraction.
Suppose we have a homomorphism with a non-trivial kernel. Let $a$ be a non-zero member of that non-trival kernel. Since the domain is field, then $a$ has an inverse. Then by the algebra shown above, everything in the field must map to zero.
Yes, you've understood it correctly. It is perhaps clearer to see it in the following way, though. Given a ring map $\varphi:k\to A$ from a field $k$ to a ring $A$, we know that $\ker \varphi$ is an ideal of $k$. The ideals in a field are $(0)$ or $(1)$. So, $\varphi$ is either injective when $\ker \varphi=(0)$ or the zero map when $\ker\varphi=(1)=k$.