Minimal Polynomial is Unique
Theorem
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial of $\alpha$ over $K$ is unique.
Proof 1
Let $f$ be a minimal polynomial of $\alpha$ over $K$ according to definition 1.
By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$.
Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$.
By the definition of minimal polynomial, $\map \deg f \le \map \deg g$, where $\deg$ denotes degree.
By Division Theorem for Polynomial Forms over Field, there exist polynomials $q, r \in K \sqbrk x$ such that:
- $g = q f + r$ and $\map \deg r < \map \deg f$.
Suppose $\map \deg r > 0$.
Then evaluating both sides of the equation above at $\alpha$, we obtain $\map r \alpha = 0$.
This contradicts the minimality of the degree of $f$.
Thus, $r$ is constant and equal to $0$.
We have now shown that $f$ divides all polynomials in $K \sqbrk x$ which vanish at $\alpha$.
By the monic restriction, it then follows that $f$ is unique.
$\blacksquare$
Proof 2
Let $f, g \in K \sqbrk x$ be minimal polynomials for $\alpha$ according to definition 2.
That is, let $f$ and $g$ be irreducible monic polynomials in $K \sqbrk x$ with $\map f \alpha = \map g \alpha = 0$.
Suppose $f$ and $g$ are distinct.
Then $f$ and $g$ are coprime.
Thus there exist polynomials $a, b \in K \sqbrk x$ with $a f + b g = 1$.
Taking the evaluation homomorphism in $\alpha$, we obtain the contradiction that $0=1$.
$\blacksquare$