Polynomial Has Root If and Only If It Is Divisible by Minimal Polynomial

Theorem

A polynomial \(p \in \mathbb{F}[X]\) for some field \(\mathbb{F}\) has a root \(\alpha\) if and only if the minimal polynomial of \(\alpha\) over \(\mathbb{F}\) divides \(p\).

Proof

Consider the result of dividing \(p\) by \(f_{\alpha}\), the minimal polynomial of \(\alpha\):

\[ p = f_{\alpha}q + r.\]

Now, since \(\deg(r) < \deg(f_{\alpha})\) we have that either \(r = 0\), which gives the divisibility we desire, or we have a contradiction, as:

\[ p(\alpha) = f_{\alpha}(\alpha)q(\alpha) + r(\alpha)\]

which means since \(p(\alpha) = f_{\alpha}(\alpha) = 0\) that \(r(\alpha) = 0\), contradicting the minimality of \(f_{\alpha}\).