Algebraic Element of Field Extension is Root of Unique Monic Polynomial of Minimal Degree

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $L/K$ be a field extension.

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


Then there exists a unique monic polynomial $f \in K \left[{x}\right]$ of smallest degree such that $f \left({\alpha}\right) = 0$, called the minimal polynomial.


Proof

Existence

Follows from Well-Ordering Principle and because $K$ is a field


$\Box$

Uniqueness

Let $f$ and $g$ be two such polynomials.

Then $f \left({\alpha}\right) - g \left({\alpha}\right) = 0$.


Aiming for a contradiction, suppose $f - g \ne 0$.

Let $a$ be the leading coefficient of $f-g$.

Then $\dfrac 1 {a} \left({f -g}\right)$ is another such polynomial of strictly smaller degree.

This is a contradiction.

Thus $f - g = 0$.

$\blacksquare$