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Let $E$ be a field extension of $F$ and let $\alpha \in E$. Define $\phi_{\alpha}:F[x]\to F(\alpha)$ by $\phi_{\alpha}(f(x))=f(\alpha)$. Why is that the kernel of $\phi_{\alpha}$ is the principal ideal generated by the minimal polynomial $p_{\alpha,F}(x)$ for $\alpha$ over $F$?

Why first is it decided that the kernel is a principal ideal in $F[x]$? Is it because $F$ is finite?

Why second must the polynomial that generates the ideal be minimal? Is it because it is irreducible and zero at $\alpha$?

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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.

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