other roots of a minimal polynomial

Question

If $f(x)$ is the minimal polynomial of $\alpha$ over some field $K$, then is $f$ always the minimal polynomial of all of its other roots over $K$, or are there counterexamples?

Answer 1

This property is indeed always true. Recall that the minimal polynomial of an algebraic number $x$ must divide any other polynomial for which $x$ is a root.

The minimal polynomial of $\alpha$ is irreducible over $K$ (if $f$ were composite, say $f = gh$, then $f(\alpha)=0$ implies $g(\alpha)=0$ or $h(\alpha)=0$ so either $g$ or $h$ would be a smaller polynomial for which $\alpha$ is a root). Now if $\beta$ is another root of $f$, then the minimal polynomial of $\beta$ must divide $f$, and hence it has to be $f$ itself.

Answer 2

Yes. The other roots of the minimal polynomial of $f$ are all conjugate elements of $\alpha$. Let $\beta\neq \alpha$ be another root of $f$. Suppose there exists $g$ such that $\beta$ is a root of $g$ and $\text{deg}(g)<\text{deg}(f)$. Then all roots of $g$ are conjugate elements of $\beta$, which means that all roots of $g$ would be roots of $f$, which would imply that $g|f$. This contradicts the notion that $f$ is irreducible.