Finite Fields, Field Extensions, and Minimal Polynomials

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 $$\alpha$$?

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.
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.
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.