Determine minimal polynomial without arguing about irreducibility

Question

Let $E/K$ be a field extension and let $\alpha \in E$ be algebraic over $K$. Let $f \in K[x]$.

My question is, if I have the following:

• $f(\alpha)=0$
• $f$ is monic
• $\deg(f) = [E : K]$

Can I then conclude that

• $f$ is the minimal polynomial for $\alpha$ over $K$?

Then I don't have to prove that $f$ is irreducible over $K$.

Answer 1

Take any element $a$ in $K$ and $f(x)=(x-a)^n, n=[E:F]$ $f$ is not the minimal polynomial of $a$.