0
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Feb 12, 2023 at 7:03
  • $\begingroup$ 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. $\endgroup$ Commented Feb 12, 2023 at 7:03
  • $\begingroup$ "... and replace $\Bbb R$ with any perfect field $F$ ...". Why? $\mathbb{R}$ is already perfect. $\endgroup$ Commented Feb 12, 2023 at 7:04

0

You must log in to answer this question.

Browse other questions tagged .