The kernel of a ring homomorphism is an ideal - that is the whole motivation of the definition of an idea.
Any polynomial in a polynomial ring of the form $F[x]$ generates an ideal. It turns out that with $F$ a field, any ideal in $F[x]$ is principal - that is, it is generated by a single polynomial, which can, in fact be taken monic (leading coefficient $1$) and is then unique.
The kernel of a ring homomorphism is a prime ideal if and only if the image is a domain (has no non-trivial zero divisors). The kernel is maximal if and only if the image is a field.
In a polynomial ring, if $p(x)=q(x)r(x)$ factorises, and $I$ is the ideal generated by $p$ then $(q+I)(r+I)\in I$, and I is therefore not prime, and certainly can't be maximal. So the image could not be a field - hence not an extension field.
So, answering your last two questions, most of this happens simply because the ring you are dealing with is a polynomial ring over a field, and does not depend on anything being finite.
The fact that the ideal is generated by the minimal polynomial for $\alpha$ can be established by showing that $\alpha$ satisfies the polynomial which generates the kernel of the homomorphism, and then showing that if $\alpha$ satisfied a polynomial of lower degree that would be in the kernel too.