# Why isn't the splitting field always Galois?

My question stems from the proof of the fundamental theorem of algebra in Dummit and Foote. There they claim that for any real polynomial $$f(x)$$ with roots $$\alpha_1, \dots, \alpha_n$$, $$\Bbb R(\alpha_1, \dots, \alpha_n, i) / \Bbb R$$ is Galois. I was under the impression that $$f$$ had to be separable in order to guarantee that the extension was Galois, but it seems that the Dummit and Foote claim ignores this requirement.

If we let $$\alpha_{n+1} = i$$, $$f(x) \mapsto f(x) (x ^2 + 1)$$, and replace $$\Bbb R$$ with any perfect field $$F$$ then we can assume that $$\alpha_i$$ are just roots of a polynomial $$f(x) \in F[x]$$. I can then rephrase my question as follows:

If $$\alpha_i$$ are the roots of a polynomial $$f(x) \in F[x]$$ for $$F$$ a perfect field, is $$F(\alpha_1, \dots, \alpha_N)/ F$$ Galois?

My attempt is as follows. Each $$\alpha_i$$ has a minimal polynomial $$m_i(x)$$ that divides $$f(x)$$ because they share a root (and thus all roots of $$m_i$$ are roots of $$f$$). Each of these is separable by virtue of being irreducible with coefficients in a perfect field, so letting $$g$$ be the square free product of the $$m_i$$'s, we have that $$g$$ is separable. Indeed, if $$m_i$$ and $$m_j$$ shared a root $$\alpha$$, then any extension $$K /F$$ splitting $$m_i$$ would be Galois because $$m_i$$ is separable, and $$m_j$$ having a root in $$K$$ implies all of its roots are in $$K$$. Since the roots of $$m_i$$ and $$m_j$$ are Galois conjugates of $$\alpha$$, we would have that $$m_i = m_j$$. In other words, we have that $$g$$ is a separable polynomial with the same roots as $$f$$ with splitting field $$F(\alpha_1, \dots, \alpha_N)$$.

Does this mean that the extension is indeed Galois or have I missed something, am I not proving that the splitting field of any polynomial (over a perfect field) is Galois?

• Every algebraic extension of a perfect field is separable, this is one of the equivalent definitions of perfect fields. Since every splitting field is normal it follows that every splitting field over a perfect field is Galois. Feb 12 at 7:03
• Extensions over fields of characteristic zero are always separable. The splitting field of $f$ is the same as the one you get by replacing $f$ with the product of its distinct irreducible factors, and that polynomial is separable. Feb 12 at 7:03
• "... and replace $\Bbb R$ with any perfect field $F$ ...". Why? $\mathbb{R}$ is already perfect. Feb 12 at 7:04