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 $f$.
Link to an appropriate page defining divisibility among polynomials.
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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 $\mu_\alpha \divides f$.
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 $f$.
$\blacksquare$