It can be proved that if $K$ a field, then $f \in K[x]$ has $a$ as a multiple root if and only if $f(a) = f'(a) = 0$.
And as a corollary, if $K$ has characteristic $0$, then irreducible polynomials do not have multiple roots.
The proof is that $f$ would be the minimal polynomial of $a$, so $f$ clearly doesn't have $a$ as a multiple root. For $\text{deg}(f)\geq 2$, so $f'$ would be a non-zero polynomial with a lower degree than $f$ that also has $a$ as a root, giving a contradiction.
So I have been thinking that the following statement should be true:
Suppose $K$ is an arbitrary field (not necessarily of characteristic $0$). Suppose that $f \in K[x]$ is irreducible. If $f' \neq 0$, then $f$ does not have multiple roots.
This is because we can make the same argument as in the proof of the previous corollary.
The question is whether this is correct. It seems to me that it is, but I am afraid I may have made some incorrect assumptions. If that is the case, then what would be a counterexample to the proposition?
Best Answer
Your proposition is true and your proof is correct.
If $f \in K[x]$ is irreducible with $a$ as a root, then $f$ is the minimal polynomial for $a$. If $a$ is a multiple root of $f$, then $a$ is also a root of $f'$. If $f' \neq 0$, then $f'$ is a non-zero polynomial of lower degree than $f$ that has $a$ as a root. This contradicts the fact that $f$ is the minimal polynomial of $a$. This argument is valid for $K$ of any characteristic.
Perhaps it might be instructive to exhibit an example of a field $K$ and an irreducible $f \in K[x]$ that does has a multiple root. By your proposition, this $f$ must be chosen such that $f' = 0$. The classic example is to take $K = \mathbb F_p (t)$ and $f(x) = x^p - t$. (You can show that this $f$ is irreducible in $K[x]$ by applying Gauss' lemma and Eisenstein's criterion - see here.) Anyway, if $a$ is an element of an extension field of $K$ such that $a$ is root of $f$, then $a^p = t$, so $f(x) = (x - a)^p$. Thus $a$ is a multiple root of $f$. However, $f'(x) = px^{p-1} = 0$, so the example I've given you is not a counterexample to your proposition.