Am I interpreting this theorem correctly?

Question

Here is a theorem which my professor stated in Algebra class, in the chapter on Galois theory.

Let $f(x)=x^p+a_{p-1}x^{p-1}+...+a_0\in \mathbb{Q}[x]$, where $p$ is prime. Assume $f(x)$ is irreducible over $\mathbb{Q}$ and $f(x)$ has exactly two complex non-real roots. Then the Galois group of $f$ is isomorphic to $S_p$.

Here is where I believe there is some ambiguity: what does it mean for $f(x)$ to have "exactly two complex non-real roots?" I mean, does this count multiplicity? Even if it doesn't satisfy the other hypotheses, how many "complex non-real roots" does $x^4+2x^2+1$ have? Two or four?

Answer 1

As Morgan Rogers said in the comments, the roots will always be distinct anyway for an irreducible polynomial, so it won't make a difference.

In general with wording like this, context matters, but prima facie I'd assume multiplicity counts, so your (reducible) example would have 4 nonreal roots, not 2.

Answer 2

It means that $f$ has $2$ roots in $\mathbb{C} \setminus \mathbb{R}$, and $p-2$ in $\mathbb{R}$.

If $f$ is irreducible over a field $F$ of characteristic zero, it has no multiple roots in any field extension of $F$ (consider the GCD of $f$ and its formal derivative).

Finally, your example doesn't have prime degree, so the result doesn't apply to it anyway, but it has 4 (distinct) complex roots.