Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial

Theorem

Let $F$ be a field.

Let $\map P x$ be a polynomial in $F$.

Let $z$ be a root of $\map P x$.

Then $\map P x$ is a multiple of the minimal polynomial $\map m x$ in $z$ over $F$.

Proof

For $z$ to be a root of $F$, $z$ must be algebraic over $F$.

Let us write:

$\map P x = \map m x \, \map q x + \map r x$

where $\map q x$ and $\map r x$ are polynomials in $F$.

Then either $\map r x = 0$ or $\map \deg {\map r x} < \map \deg {\map m x}$.

Then:

$\map P z = \map m z \, \map q z + \map r z$

But as $z$ is a root of both $\map P x$ and $\map m x$, we have that:

$\map P z = \map m z = 0$

and so:

$\map r z = 0$

So if $\map r x \ne 0$ we have that $\map r x$ is a polynomial of smaller degree than $\map m x$.

This contradicts the minimality of $\map m x$.

Thus $\map r x = 0$ and so $\map P x$ is a multiple of $\map m x$.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $71 \ \text{(ii)}$