Equivalence of Definitions of Minimal Polynomial

From ProofWiki
Jump to navigation Jump to search

Theorem

The following definitions of the concept of Minimal Polynomial are equivalent:

Let $L / K$ be a field extension.

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

Definition 1

The minimal polynomial of $\alpha$ over $K$ is the unique monic polynomial $f \in K \sqbrk x$ of smallest degree such that $\map f \alpha = 0$.

Definition 2

The minimal polynomial of $\alpha$ over $K$ is the unique irreducible, monic polynomial $f \in K \sqbrk x$ such that $\map f \alpha = 0$.

Definition 3

The minimal polynomial of $\alpha$ over $K$ is the unique monic polynomial $f \in K \sqbrk x$ that generates the kernel of the evaluation homomorphism $K \sqbrk x \to L$ at $\alpha$.

That is, such that for all $g \in K \sqbrk x$:

$\map g \alpha = 0$ if and only if $f$ divides $g$.


Proof

1 equals 2

By Minimal Polynomial is Irreducible, it follows that the two are equal.

$\Box$


1 equals 3

Let $f \in K \sqbrk x$ be the monic polynomial of smallest degree such that $\map f \alpha = 0$.

Let $g \in K \sqbrk x$ be a polynomial.

If $f \divides g$, then $g = q f$ for some $q \in K \sqbrk x$.

Thus $\map g \alpha = \map f \alpha \, \map q \alpha = 0$.

Conversely, suppose $\map g \alpha = 0$.

By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \sqbrk x$ such that:

$g = q f + r$

and:

$r = 0$ or $\deg r < \deg f$.

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

$\map g \alpha = \map q \alpha \, \map f \alpha + \map r \alpha \implies \map r \alpha = 0$

since $\map f \alpha = \map g \alpha = 0$.

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

Therefore $r = 0$.

Therefore:

$g = q f$

That is, $f$ divides $g$.

$\blacksquare$