Can the minimal polynomial have multiple roots?

Question

Let $F$ be an extension field of $K$. I'm trying to think of an example of an element $\alpha$ which is algebraic over $K$ and whose minimal polynomial over $K$ has a root of multiplicity $> 1$ in some splitting field. Is this possible? Nothing comes to mind immediately, but I might have to looked hard enough.

Answer 1

I will explain a little bit the comment of @Lubin.

In your notations, let's say the minimal polynomial of an element $\alpha \in F$ is a polynomial $f \in K[x]$. Then $f$ is an irreducible polynomial over $K$: if $f = gh$ with $g, h \in K[x]$ of smaller degree than $f$, then we would have $g(\alpha)h(\alpha) = 0$, which implies that at least one of $g(\alpha)$ and $h(\alpha)$ is zero, contradicting the minimality of $f$.

On the other hand, if $f$ has multiple roots, then the $\gcd$ of $f$ and its derivative $f'$ is not equal to $1$ (as any multiple root is a root of both). Since $f$ is irreducible, we must have $\gcd(f, f') = f$, i.e. $f$ divides $f'$.

However, the degree of $f'$ is strictly smaller than that of $f$, thus the only possibility is $f' = 0$.

If $K$ has characteristic zero, then $f' = 0$ implies that $f$ is a constant, hence cannot be a minimal polynomial. This is the reason that you never find an example in characteristic zero.

In positive characteristic, we may have $f' = 0$ while $f$ is non-constant. The example provided by @Lubin in the comment is a typical one.