Galois extensions

In the last post, we’ve met the Galois correspondence between sub-fields and sub-groups:

L^{\circ}:=\text{Gal}(K/L)\longleftarrow L

H\longmapsto K^{H}=:H^{*}

However, we had two major problems:

We don’t know if $F=\text{Gal}(K/F)^*$ . So far, all we know is that $F\sube \text{Gal}(K/F)^*$ .
We don’t know if we can that $H=H^{*\circ}$ and $L=L^{\circ *}$ ? In other words, we don’t know when those maps are inverse maps of each other!

We also concluded that the thing we want from an extension to fulfill, is that all the sub-fields / sub-groups will be images of some sub-groups / sub-fields. This follows from the facts we proved (the “back and forth” property) that were:

L^\circ=L^{\circ*\circ} \\ H^*=H^{*\circ*}

Which is the same:

\text{Gal}(K/L)=\text{Gal}(K/K^{\text{Gal}(K/L)}) \\

K^H=K^{\text{Gal}(K/K^H)}

In this post we’ll start solving those problems, we will present a special type of an extension which is called a Galois extension. As it turns out, those kind of extensions are exactly what we want!

Before I’ll define what is a galois extension, I have to introduce two new terms:

Separable extension

What is a separable extension? We already know the term of a separable polynomial. Those are polynomials that splits to linear factors in their splitting field. In other words, a polynomial $f$ with degree $n$ is separable if it has $n$ different roots. That is, we can represent $f$ in it’s splitting field as the product $\prod_{i=1}^n(x-\alpha_i)$ where the $\alpha_i$ -s are all different.

Now, if $K/F$ is a field extension, we will say that an element $a\in K$ is separable if it’s minimal polynomial over $F$ is separable. If every $a\in K$ is separable, then the extension is called a separable extension.

Let’s see some examples:

We’ve proved that if $F$ is a field with characteristic 0, then every irreducible polynomial over it is separable. Therefore, every extension $K/F$ where $\text{char}(F) = 0$ is separable.
An immidieate conclusion from this fact, is that every extension of the form $K/\mathbb{Q}$ is separable (since $\text{char}(\mathbb{Q}) = 0$ ).
Recall that we’ve seen an example of a non-separable polynomial. We took the field to be $\mathbb{F}_p(t^p)$ and the polynomial was $x^p-t^p$ . It’s splitting field was $\mathbb{F}_p(t)$ , and in this field: $x^p-t^p=(x-t)^p$ . Thus, the extension $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$ is not separable.

That’s pretty much everything I want to say about separable extensions, so let’s move on to the second term.

Normal extension

Informally, a normal extension $K/F$ is one with the following property:

If $f\in F[x]$ is irrecducible, and we’ve found a root for it in $K$ then all the roots are in $K$ . That is, one root is in $K$ $\Rightarrow$ all roots are in $K$ .

Formally, an extension is called normal if for every $a\in K$ , the minimal polynomial of $a$ over $F$ , splits in $K$ .

Let’s see some examples:

Consider the extension $\mathbb{Q}(\sqrt[3]{5})/\mathbb{Q}$ . Note that $\sqrt[3]{5}$ is the root of the irreducible polynomial:

x^3-5\in\mathbb{Q}[x]

Over $\mathbb{Q}(\sqrt[3]{5})$ this polynomial is reducible:

x^3-5=(x-\sqrt[3]{5})(x^2+\sqrt[3]{5}x+(\sqrt[3]{5})^2)\in \mathbb{Q}(\sqrt[3]{5})[x]

However, the polynomial $(x^2+\sqrt[3]{5}x+(\sqrt[3]{5})^2)$ is irreducible it’s roots are:

\sqrt[3]{5}\rho_5,\sqrt[3]{5}\rho_5^2

Where $\rho_5=e^{{2\pi i}/{5}}$ is the root of unity of order 5. Those roots are not elements of the field $\mathbb{Q}(\sqrt[3]{5})$ . Hence, the extension $\mathbb{Q}(\sqrt[3]{5})/\mathbb{Q}$ is not normal.

However, if we’ll add the root of unity to $\mathbb{Q}(\sqrt[3]{5})$ , we will get the field $\mathbb{Q}(\sqrt[3]{5},\rho_5)$ where the polynimoial $x^3 - 5$ is indeed splitting in.

It turns out that this is indeed a normal extension, soon we’ll see why.

Now for what we’ve been waiting for:

Galois extension

A Galois extension is a normal and separable extension.

This simple definition is exactly what we need in order to solve our problems from the beggining.

Recall that the automorphisms of an extension must map a root of an irreducible polynomial to another root. By the strong assumption that the extension is normal, we are guranteed to have all the roots in the ‘big’ field. Moreover, since the extension is also separable, we know that all the roots are different. Hence, such extensions are not ‘missing opportunities’ of automorphisms.

This intuition already shows us the power of galois extension. Now, I want to present 4 more equivalent defintions for galois extension. Once I’ll present them, I think it will be really clear why those extensions are good.

Equivalent definitions

Suppose that $K/F$ is a (finite) field extension. Then the following are equivalent:

$K/F$ is a galois extension.
$K$ is a splitting field of a separable polynomial.
$F$ is a fixed field, that is, there is a group $G$ of automorphisms of $K$ such that $F=K^G$ .
$K^{\text{Gal}(K/F)}=F$
$|\text{Gal}(K/F)|=[K:F]$

Before I’ll prove that all of these are indeed equivalent, I want to talk about what can we learn from this equivalnce.

First, remember that in the example from before, we’ve seen that the field $\mathbb{Q}(\sqrt[3]{5},\rho_5)$ is the splitting field of $x^3-5$ . Moreover, this polynomial is separable. Hence, by second definition, we know that $\mathbb{Q}(\sqrt[3]{5},\rho_5)/\mathbb{Q}$ is galois and in particular normal – that was exactly what we wanted to prove there. This equivalence allows us to prove something that we didn’t even know how to approach before in two rows! That’s great!

In addition, we can also conclude that if $K/F$ is galois and $F\sube L\sube K$ is a subfield, then $K/L$ is galois as well. Why? since $K$ is a splitting field of a separable polynomial $f$ over $F$ . This polynomial is in particular in $L$ , and $K$ is still it’s splitting field…

Moreover, note that definitions 3 and 4 are exactly the one that solves one of our problems from the beginnig. The extensions that satisfy $F=\text{Gal}(K/F)^*$ are exactly the galois extensions!

Finally, definition 5 shows us that there are as many automorphisms of the extensions as the degree of the extension. In general, we know that the number of the automorphism is bounded by the degree, so, in fact, galois extension are such extension that have the maximal amount of automorphisms possible.

This definition also gives us a great tool for guessing what the galois group is. For example, if the extension is galois and of degree $p$ , where $p$ is prime, then we know the galois group is of order $p$ . And the only group of order $p$ is the cyclic group $\mathbb{Z}_p$ .

OK, now that we’ve seen the power of galois extension, it’s time to prove that all those definition are indeed equivalent.

My plan for this proof is to prove the statements in the following order:

$1\Rightarrow 2\Rightarrow 3\Rightarrow 1$ . And by doing so, we will prove that the first three are equivalent.

After that, I’ll show: $4\Rightarrow 3$ and then $3\Rightarrow 4$ , which will impliy that 4 is also equivalent to the first three. And finally, I’ll show that $2\Rightarrow 5$ and $5\Rightarrow 4$ to conclude that all of the statements are equivalent.

Proving the equivalence

$1\Rightarrow 2$

Since the extension is finite, we can write $K$ as $K=F[a_1,\dots,a_n]$ . Now let $f_1\dots,f_n$ be the minimal polynomials of each $a_1,\dots,a_n$ respectivly. Since the extension is galois it is:

Normal – so the $f_i$ -s split in $K$ ( $f_i$ has a root in $k$ , who? $a_i$ of course).
Separable – so the $f_i$ -s are separable.

We can now define:

f=\prod_{i=1}^nf_i\in F[x]

This is a separable polynomial and $K$ is it’s splitting field:

It is separable since all the $f_{i}$ ‘s are separable and differ by their roots (an element can’t be a root of two irreducible polynomials! The minimal polynomial is unique). $K$ is a splitting field since all the roots of the $f_{i}$ -s, are in $K$ and a subfield of $K$ won’t contain some $a_i$ , hence it’s not splitting $f$ , thus $K$ is the minimal field that splits $f$ .

$2\Rightarrow 3$

Let $f$ be the separable polynomial that $K$ is it’s splitting field. We’ll prove that:

F=K^{\text{Gal}(K/F)}

(This will also show that 2 implies 4). For comfort purposes let’s deonte $G=\text{Gal}(K/F)$ . Consider the following diagram:

\begin{array}{ccc} K & \overset{\sigma}{\to} & K\\ \cup & & \parallel\\ F & \overset{i}{\to} & K \end{array}

This fits perfectly to the theorem we proved in this post where $i$ is the inclusion embedding and (since the extension is finite) $\sigma$ is an automorphism that fix $F$ (in other words, it extends $i$ ).

The theorem stated that if $f\in F$ is irreducible, $i(f) = f$ is separable and $K$ is it’s splitting field (genarated by it’s roots), then there are exaclty $[K:F]$ extensions.

This theorem fits perfrectly to our case, therefore, the number of automorphism of the extension $K/F$ is exaclty the degree – $[K:F]$ (we just proved that 2 imples 5):

|G|=|\text{Gal}(K/F)|=[K:F]

Moreover, $F\sube F^G \sube K$ , thus, $K$ is also the splitting field of $f$ over $K^G$ , so we can apply the same theorem and conclude that the number of automorphism of the extension $K/K^G$ is exaclty the degree – $[K:K^G]$ :

|\text{Gal}(K/K^G)|=[K:K^G]

Now comes the punch line: I state that:

[K:K^G]=[K:F]

To see it, we’ll use the “back and forth” property:

[K:K^G]=|\text{Gal}(K/K^G)|\overset{\text{ back and forth}}{=}|G|=[K:F]

Ok, but how is that helpul? we can use the multiplicative property to get:

[K:F]=[K:K^G][K^G:F] \\ \Downarrow \\ [K^G:F] =1 \\ \Downarrow \\ K^G=F

The equality follows from the fact the $K^G$ is a vector space of $F$ of degree 1, hence, it must be $F$ .

This part reveals us a the stength of a splitting field of a separable polynominal.

$3\Rightarrow 1$

We now assume that $F=K^G$ where $G$ is an automorphism group of $K$ . We need to show that $K/F$ is galois. That is, separable and normal.

Let $a\in K$ be some element. Suppose that $f$ is the minimal polynomial of $a$ over $F$ . We shall prove that all of it’s roots are in $K$ (normal) and then show that they are different (separable).

Let $\alpha_1,\alpha_2,\dots,\alpha_n$ be the roots of $f$ in $K$ (different roots), and define:

g(x)=\prod_{i=1}^n (x-\alpha_i)

Obviously, $g$ divides $f$ in $K$ and it is separable since the roots are different. Now, let’s pick some $\sigma\in K^G$ and apply it on the polynomial $g$ . But since roots of $f$ in $K$ must be sent to roots of it in $K$ , we know that $\sigma$ is in fact a permutation on the roots – It is one-to-one as an automorphism. In other words:

\{\alpha_1,\dots,\alpha_n\}=\{\sigma(\alpha_1),\dots,\sigma(\alpha_n)\}

Therefore:

\sigma(g)=\sigma(\prod_{i=1}^n (x-\alpha_i))=\prod_{i=1}^n (x-\sigma(\alpha_i))=\prod_{i=1}^n (x-\alpha_i)=g

So $\sigma$ fixes the coefficients of $g$ , but that’s true for all the automorphisms in $G$ , thus: $g\in K^G[x] = F[x]$ ! Therefore, $g$ divides $f$ over $F$ , but $f$ is irreducible over $F$ , thus $g=f$ and we conclude that $\alpha_1,\dots,\alpha_n$ are all the roots of $f$ . So $f$ is separable (since $g$ is separable) and the extension is normal.

I want to mention that I think this method of proving – applying an automorphism on the polynomal – is just beautiful! And this is not the last time that I am going to use it!

$4\Rightarrow 3$

This one is trivial – statement 4 is stronger than statement 3. We can just pick $G$ to be $\text{Gal}(K/F)$ .

$3\Rightarrow 4$

Assume that $F=K^G$ where $G$ is an automorphism group of $K$ . We need to show that $K^{\text{Gal}(K/F)}=F$ By the “back and forth” property:

K^{\text{Gal}(K/K^G)}=K^G

Therefore:

K^{\text{Gal}(K/F)}=K^{\text{Gal}(K/K^G)}=K^G=F

As we wanted.

$2\Rightarrow 5$

We’ve proved it during the proof $2\Rightarrow 3$

$5\Rightarrow 4$

Suppose that $|\text{Gal}(K/F)|=[K:F]$ . We want to show that $K^{\text{Gal}(K/F)}=F$ .

Note that $K^{\text{Gal}(K/F)}$ is a fixed field, therefore, the extension $K/K^{\text{Gal}(K/F)}$ satisfies property 3, hence it is galois. So in particular it satisfies property 2 that implies property 5, which states that:

|\text{Gal}(K/K^{\text{Gal}(K/F)})|=[K:K^{\text{Gal}(K/F)}] \ \ \ \ \ \ \ \ \ \ \ (*)

Moreover, by the “back and forth” property:

\text{Gal}(K/F)=\text{Gal}(K/K^{\text{Gal}(K/F)})

Now:

[K:F]\overset{\text{property 5}}{=}|\text{Gal}(K/F)|=|\text{Gal}(K/K^{\text{Gal}(K/F)})|\overset{(*)}{=}[K:K^{\text{Gal}(K/F)}]

Same as before, by the multiplicative property we know that $[K^{\text{Gal}(K/F)}:F]=1$ , and this implies $K^{\text{Gal}(K/F)}=F$ as we wanted.

Summary

Great, now we have 5 different definitions for a galois extension. I am going to list them again:

$K/F$ is a galois extension.
$K$ is a splitting field of a separable polynomial.
$F$ is a fixed field, that is, there is a group $G$ of automorphisms of $K$ such that $F=K^G$ .
$K^{\text{Gal}(K/F)}=F$
$|\text{Gal}(K/F)|=[K:F]$

This is really important to know them all since we are going to use all of them, depends on the situation of course. Moreover, all those definition shows the ‘bright side’ of such extension. There are so many properties that only a galois extension satisfies.

However, there is still that second problem that I’ve mentioned in the beginning:

We don’t know if we can I say that $H=H^{*\circ}$ and $L=L^{\circ *}$ . In other words, we don’t know when those maps are inverse maps of each other!

As you may already guessed, the answer to the ‘when’ is exactly when the extension is galois. Before I’ll prove it, I’ll dedicate one post to an important lemma called Artin’s Lemma, after doing so, the proof will be really easy and straightforward.

Separable extension

Normal extension