# Determine minimal polynomial without arguing about irreducibility

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 $$K$$.

Take any element $$a$$ in $$K$$ and $$f(x)=(x-a)^n, n=[E:F]$$ $$f$$ is not the minimal polynomial of $$a$$.
• Do you have a counter example if $a \in E$ is algebraic over $K$? – Johannes Jensen Dec 12 '19 at 0:41
• Take any $a$ with minimal polynomial $P$ of degree $l<n=[E:K]$, and consider $PX^{n-l}$. – Tsemo Aristide Dec 12 '19 at 0:46