# 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. – Rudy the Reindeer 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. – Rudy the Reindeer 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.

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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 $f \in F[x]$. – Rudy the Reindeer Feb 17 '11 at 21:55
@Matt: Glad it was useful. In that setting, splitting fields are always Galois extensions, because you get separability from the fact that the base field is perfect and normality from being a splitting field. – Arturo Magidin Feb 17 '11 at 22:01
yes, very useful. I need to properly read Zev's answer tomorrow after the exam and then decide which of the two answers I accept as "the" answer. (Hope the professor won't ask me anything about Galois theory, really want to take a proper course about it and not just learn one theorem as the "end bit" of a general algebra course.) – Rudy the Reindeer Feb 17 '11 at 22:15
@Matt: Zev is answering what you actually wrote; I was commenting on a side issue that Zev had addressed in passing, about separability, and giving an explicit example of why it's needed. I think Zev did by far the better job. – Arturo Magidin Feb 17 '11 at 22:17
Thank you for your kind words, Arturo. – Zev Chonoles Feb 18 '11 at 17:37