#because the image of F[X] under the homomorphism is an integral domain it must be that the kernel is a prime ideal

[bubbloquacious]

Jacobson's Basic algebra 1 gives such a nice argument for why algebraic numbers have minimum polymials, making it a corollary of the fact that polynomial rings over fields are principal ideal domains. Consider a field extension E/F (i.e. a field E that contains a field F). For any u ∈ E we have in E the subring F[u] and the subfield F(u). These are defined as the smallest subring and subfield of E respectively that contain both F and u. If E = F(u) for some u ∈ E, then E/F is called a simple field extension.

Let F(u)/F be an arbitrary simple field extension. Consider the polynomial ring F[X]. By the universal property of polynomial rings (which is essentially what one means when they say that X is an indeterminate), there is a unique ring homomorphism F[X] -> F(u) that sends any a ∈ F to itself and that sends X to u (you can think of this homomorphism as evaluation of a polynomial at u). If the kernel of this homomorphism is the zero ideal, then F[u] is isomorphic to F[X], so u is transcendental over F. If the kernel is non-zero, then by definition there are non-zero polynomials p ∈ F[X] such that p(u) = 0 in F(u), so u is algebraic over F. Because F[X] is a principal ideal domain, there is a polynomial p that generates the kernel. In other words, p divides all polynomials q such that q(u) = 0 in F(u). Two polynomials (over a field) generate the same ideal if and only if they differ by a constant factor, so there is a unique monic minimum polynomial in F[X] for u.