# A question regarding normal field extensions and Galois groups

The following is possibly true but I can't find a corresponding theorem:

If $E/F$ is the splitting field of some polynomial in $F$ and $F \subset K \subset E$ then:

$Gal(E/K)$ normal subgroup of $Gal(E/F)$ $\Leftrightarrow$ $K/F$ is a normal field extension

Is this true? I think saying that $E/F$ is the splitting field of some polynomial in $F$ is the same as saying $E/F$ is Galois and therefore the Galois correspondence theorem applies. But I'm not sure I see how the meaning of normality of a subgroup relates to the meaning of normality of a field extension. But maybe the above is wrong altogether?

In any case, many thanks for your help to clarify this.

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Two things. "K normal subgroup of E" should probably say $Gal(E/K)$ normal subgroup of $Gal(E/F)$ or something like that, and in order for $E/F$ to be Galois you need some other assumption like the polynomial is separable? In this case, I think many books include something like this in with the Fundamental Theorem of Galois Theory. –  Matt Feb 17 '11 at 17:48
Yes, of course! That was a typo, thanks for pointing it out. –  Matt N. Feb 17 '11 at 21:41
I'm not sure if the polynomial needs to be separable for the extension to be Galois. But for the fundamental theorem to apply it is enough to assume that $E$ is the splitting field of some polynomial in $F[x]$ and $char(F) = 0$ of $F$ finite. –  Matt N. Feb 17 '11 at 21:44
Let $E$ be the splitting field of $f\in F[x]$. Then $E/F$ is Galois iff $f$ is separable. Recall that the splitting field of $f\in F[x]$ is just $E=F(\alpha_1,\ldots,\alpha_n)$, where the $\alpha_i$ are the roots of $f$. By definition, the polynomial $f$ is separable iff each of its irreducible factors are, which is the case iff each of the $\alpha_i$ are separable. If any of the $\alpha_i$ are non-separable, then $E$ is non-separable because it has a non-separable element; if all of the $\alpha_i$ are separable, then the extension they generate, namely $E$, will also be separable. –  Zev Chonoles Feb 20 '11 at 20:25

You are confused about the Galois correspondence - normal subgroups $H$ of the Galois group $\text{Gal}(E/F)$ correspond to normal extensions $E^H/F$, where $E^H$ denotes the subfield of $E$ fixed by $H$. Note that $E$ being the splitting field of a polynomial in $F$ does not guarantee that $E/F$ is Galois. This is due to the fact that $E/F$ is Galois only when it is normal (i.e., is a compositum of some splitting fields) and separable.

However, I imagine the statement you intended was:

If $E/F$ is normal, then $$H\triangleleft \text{Aut}(E/F) \iff E^H/F \text{ is a normal field extension.}$$

This is actually still true. It can be regarded as a salvaging of the Fundamental Theorem of Galois Theory in the case that $E/F$ is not necessarily separable. Here is my reasoning: Let $E/F$ be normal and let $G=\text{Aut}(E/F)$. Then $E^G/F$ is purely inseparable, and $E/E^G$ is separable. We have that $\text{Aut}(E/E^G)=\text{Aut}(E/F)$. Because $E/F$ is normal, we have that $E/E^G$ is normal and hence $E/E^G$ is Galois, and therefore a normal subgroup $H\triangleleft\text{Aut}(E/E^G)=\text{Aut}(E/F)$ corresponds to a normal extension $E^H/E^G$. It is known that if $C\subseteq B\subseteq A$ is a tower of field extensions and $A/C$ is normal and $B/C$ is purely inseparable, then $A/B$ is normal. Thus $E^H/F$ is normal.

Conversely, given a normal subextension $L/F$ of $E/F$ that is the fixed field $L=E^H$ of some subgroup $H\subseteq\text{Aut}(E/F)=\text{Aut}(E/E^G)$, then $L$ in fact contains $E^G$, and $L/F$ normal implies $L/E^G$ normal, hence the subgroup of $G$ that fixes $L$, namely $H$, is normal in $G$.

Please take note of this question: it is not an obvious one!

However, constructing a counterexample is evading me. Here is a try:

Let $\mathbb{F}_p$ be a finite field where $3\nmid p-1$ (so that there are no cube roots of unity), let $F=\mathbb{F}_p(T)$, let $f=x^{3p}-T\in F[x]$, let $E$ be the splitting field of $f$ over $F$. Then $E=F(\sqrt[3p]{T},\sqrt[3]{1})$. This has as a subfield $M=F(\sqrt[3]{T})$, which is separable, but not normal, over $F$. EDIT: Nevermind, this doesn't work. $M$ isn't the fixed field of a normal subgroup of $\text{Aut}(E/F)$.

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I think your statement still holds. $\Longrightarrow$ is trivial, while $\Longleftarrow$ works as follows: Let $G=F^{\mathrm{Aut}\left(E/F\right)}$. Then, $E\subseteq G\subseteq F$ is a tower of fields, and $F$ is Galois over $G$. Thus, since you know that your equivalence holds for Galois extensions, you conclude that $E^H/G$ is normal (because $\mathrm{Aut}\left(E/F\right)=\mathrm{Aut}\left(E/G\right)$). On the other hand, $G/F$ is normal (easy to check), and normality is transitive, so you conclude that $E^H/F$ is normal. Where is my mistake? –  darij grinberg Feb 17 '11 at 19:00
Normality is not transitive, and $G$ is not an extension of $F$. I'm a bit confused by your notation, mainly due to the use of $G$ for a field. –  Zev Chonoles Feb 17 '11 at 19:07
Uhm, yeah. I was confused by $F$ being smaller than $E$ while standing farther in the alphabet. Also, I was completely wrong about normality, though it can be fixed - but your writeup is better anyway. –  darij grinberg Feb 17 '11 at 19:13
It is not true in general that saying that $E$ is a splitting field over $F$ implies that the extension is Galois: the missing ingredient is separability. The implication holds for characteristic zero and in more generality for perfect fields, but not always. For an example, take $F=\mathbb{F}_p(x)$, the field of rational functions over the field of $p$ elements, and let $E$ be the splitting field over $F$ of $t^p - x$. If $\alpha$ is a root of $t^p-x$ in $E$, then $t^p - x = t^p - \alpha^p = (t-\alpha)^p$ in $E$, since $E$ has characteristic $p$. So $E=F[\alpha]$ is a splitting field of $f(t)=t^p-x$ over $F$. But we also conclude that $\mathrm{Aut}(E/F)$ consists only of the identity (since $\sigma\in\mathrm{Aut}(E/F)$ is completely determined by its value in $\alpha$, but $\alpha$ must map to itself, since it must map to a root of $t^p - x$, and $\alpha$ is the only possibility). However, since $f(t)$ is irreducible over $F$ of degree $p$, then $[E:F]=p$. Hence, $|\mathrm{Aut}(E/F)|\lt[E:F]$, and the extension cannot be a Galois extension. The reason it fails to be a Galois extension is that the extension is not separable, since it is given by an irreducible polynomial with multiple roots.
thank you, you helped me resolve part of my confusion. In my book the fundamental theorem is stated for $F$ finite or $char(F) = 0$ and $E$ the splitting field of some $f \in F[x]$. –  Matt N. Feb 17 '11 at 21:55