Why does the minimal polynomial of α divide all polynomials for which α is a root?

Question

Suppose α ∈ E with E being a field extension of F. Let S be the set of all polynomials in F[x] for which α is a root. Prove that the minimal polynomial of α over F divides every polynomial in S.

Answer 1

Hint $\ S $ is an ideal of $\,F[x],\,$ which, having a Euclidean algorithm, is a PID. Hence the ideal $\,S\,$ is principal, generated by any element of minimal degree (= gcd of all elements of $S).$

Or, if ideals are unfamiliar, we can eliminate that language from the linked proof as follows. Note that $\,S\,$ is closed under gcd. Indeed, by Bezout, $\,\gcd(f,g) = a f + b g\,$ so if $\,\alpha\,$ is a root of $\,f,g\,$ then it is a root of their gcd. Thus any nonzero element $\,f\in S\,$ of minimal degree divides every $\,g\in S,\,$ else their gcd $\in S\,$ and has smaller degree than $\,f\,$ (or, equivalently, $\,0\neq g\bmod f = g - fh\in S\,$ has smaller degree than $\,f,\,$ contra minimality of $\,f)$.

Answer 2

Consider the evaluation homomorphism $ev_\alpha:F[x] \rightarrow F[\alpha]$ defined as follows: $g(x) \mapsto g(\alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.

As with any homomorphism, its kernel is an ideal (in $F[x]$), and it will be the set of all polynomials with $\alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $\ker(ev_\alpha)$ must be generated by a single element.

Now you just need to convince yourself that that element must be the minimal polynomial with $\alpha$ as a root. Finally, we can use this to conclude that the minimal polynomial will divide every polynomial with $\alpha$ as a root.