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 . So far, all we know is that .
- We don’t know if we can that and ? 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 with degree is separable if it has different roots. That is, we can represent in it’s splitting field as the product where the -s are all different.

Now, if is a field extension, we will say that an element is **separable** if it’s minimal polynomial over is separable. If every is separable, then the extension is called a** separable extension**.

Let’s see some examples:

- We’ve proved that if is a field with characteristic 0, then every irreducible polynomial over it is separable. Therefore, every extension where is separable.
- An immidieate conclusion from this fact, is that every extension of the form is separable (since ).
- Recall that we’ve seen an example of a non-separable polynomial. We took the field to be and the polynomial was . It’s splitting field was , and in this field: . Thus, the extension 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 is one with the following property:

If is irrecducible, and we’ve found a root for it in then all the roots are in . That is, one root is in all roots are in .

Formally, an extension is called **normal** if for every , the minimal polynomial of over , splits in .

Let’s see some examples:

Consider the extension . Note that is the root of the irreducible polynomial:

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

Over 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 is **irreducible** it’s roots are:

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

Where is the **root of unity** of order 5. Those roots are not elements of the field . Hence, the extension is **not** normal.

However, if we’ll add the root of unity to , we will get the field where the polynimoial 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 is a (finite) field extension. Then the following are equivalent:

- is a galois extension.
- is a splitting field of a separable polynomial.
- is a
**fixed field**, that is, there is a group of automorphisms of such that .

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 is the splitting field of . Moreover, this polynomial is **separable**. Hence, by second definition, we know that 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 is galois and is a subfield, then is galois as well. Why? since is a splitting field of a separable polynomial over . This polynomial is in particular in , and 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 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 , where is prime, then we know the galois group is of order . And the only group of order is the cyclic group .

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:

. And by doing so, we will prove that the first three are equivalent.

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

### Proving the equivalence

Since the extension is finite, we can write as . Now let be the minimal polynomials of each respectivly. Since the extension is galois it is:

**Normal**– so the -s split in ( has a root in , who? of course).**Separable**– so the -s are separable.

We can now define:

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

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

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

Let be the separable polynomial that 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 . 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 is the inclusion embedding and (since the extension is finite) is an **automorphism** that fix (in other words, it extends ).

The theorem stated that if is irreducible, is separable and is it’s splitting field (genarated by it’s roots), then there are exaclty extensions.

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

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

Moreover, , thus, is also the splitting field of over , so we can apply the same theorem and conclude that the number of automorphism of the extension is exaclty the degree – :

|\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 is a vector space of of degree 1, hence, it must be .

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

We now assume that where is an automorphism group of . We need to show that is galois. That is, separable and normal.

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

Let be the roots of in (different roots), and define:

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

Obviously, **divides** in and it is separable since the roots are different. Now, let’s pick some and apply it on the polynomial . But since roots of in must be sent to roots of it in , we know that 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 fixes the coefficients of , but that’s true for all the automorphisms in , thus: ! Therefore, divides over , but is irreducible over , thus and we conclude that are **all** the roots of . So is separable (since 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!

This one is trivial – statement 4 is stronger than statement 3. We can just pick to be .

Assume that where is an automorphism group of . We need to show that 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.

We’ve proved it during the proof

Suppose that . We want to show that .

Note that is a fixed field, therefore, the extension 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 , and this implies as we wanted.

## Summary

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

- is a galois extension.
- is a splitting field of a separable polynomial.
- is a
**fixed field**, that is, there is a group of automorphisms of such that .

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 and . 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.