Root of Polynomial iff Divisible by Minimal Polynomial

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Theorem

Let $K$ be a field.

Let $L / K$ be a field extension of $K$.

Let $\alpha \in L$ be algebraic over $K$.


Then there is a unique monic polynomial $\mu_\alpha \in K \left[{X}\right]$ of least degree such that $\mu_\alpha \left({\alpha}\right) = 0$.

Moreover $f \in K \left[{X}\right]$ is such that $f \left({\alpha}\right) = 0$ if and only if $\mu_\alpha$ divides $f$.



Proof

By definition, there is some polynomial $f \in K \left[{X}\right]$ such that $f \left({\alpha}\right) = 0$.

Let $\mu$ be such a polynomial of minimal degree, and define:

$\mu_\alpha = \dfrac 1 a \mu $

where $a$ is the leading coefficient of $\mu$.


Clearly this polynomial has the required properties, so we show uniqueness.

Let $\mu'$ be another such polynomial. Then

$\mu_\alpha \left({\alpha}\right) - \mu' \left({\alpha}\right) = 0 - 0 = 0$

Since both polynomials are monic, $\mu_\alpha - \mu'$ has degree strictly less than $\mu_\alpha$.

So we must have $\mu_\alpha - \mu' = 0$.


Let $\mu_\alpha \mathrel \backslash f$.

Then:

$f = g \mu_\alpha$ for some $g \in K \left[{X}\right]$

and:

$f \left({\alpha}\right) = 0 \cdot g \left({\alpha}\right) = 0$

Conversely, let $f \in K \left[{X}\right]$ be any polynomial such that $f \left({\alpha}\right) = 0$.

By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \left[{X}\right]$ such that:

$f = q \mu_\alpha + r$

and:

$\deg r < \deg \mu_\alpha$.

Evaluating this expression at $\alpha$ we find that:

$f \left({\alpha}\right) = q \left({\alpha}\right) \mu_\alpha \left({\alpha}\right) + r \left({\alpha}\right) \implies r \left({\alpha}\right) = 0$

since $\mu_\alpha \left({\alpha}\right) = f \left({\alpha}\right) = 0$.

But $\mu_\alpha$ has minimal degree among the non-zero polynomials that are zero at $\alpha$.

Therefore as $\deg r < \deg \mu_\alpha$ we must have $r = 0$.

Therefore:

$f = q \mu_\alpha$

That is, $\mu_\alpha$ divides $f$.

$\blacksquare$