Minimal Polynomial Exists
Theorem
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then there exists a minimal polynomial $f \in K \sqbrk x$ for $\alpha$ over $K$.
Proof 1
According to definition 1 of minimal polynomial, we ought to find $f \in K \sqbrk x$ such that:
- $f \in K \sqbrk x$ is a monic polynomial of smallest degree such that $\map f \alpha = 0$
Since $\alpha$ is algebraic over $K$, there is some $f \in K \sqbrk x$ such that $\map f \alpha = 0$.
Therefore we can define:
- $n = \min \set { \deg f : f \in K \sqbrk x, \map f \alpha = 0 }$
where $\deg f$ is the degree of $f$.
It follows that there is some $f \in K \sqbrk x$ such that $\deg f = n$.
Then $f$ is of smallest degree such that $\map f \alpha = 0$.
Let the first coefficient of $f$ be $a \in K$.
Since $K$ is a field, it follows that $g := f / a$ is monic.
Therefore $g$ is the sought monic polynomial of smallest degree such that $\map g \alpha = 0$.
$\blacksquare$
Proof 2
According to definition 2 of minimal polynomial, we ought to find $f \in K \sqbrk x$ such that:
- $f \in K \sqbrk x$ is an irreducible, monic polynomial such that $\map f \alpha = 0$
Since $\alpha$ is algebraic over $K$, there is some $f \in K \sqbrk x$ such that $\map f \alpha = 0$.
By Polynomial Forms over Field form Unique Factorization Domain, $\map f x$ has a complete factorization:
- $\map f x = a \cdot \map {g_1} x \cdots \map {g_n} x$
where $a \in K, a \ne 0$, and the $g_i$ are irreducible and monic.
Since $\map f \alpha = 0$, it follows that for some $g_i$:
- $\map {g_i} \alpha = 0$
as required.
$\blacksquare$
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