In the one of my previous posts, we’ve discussed about **splitting fields** of a polynomial . Those are fields where splits in, and they are the minimal fields with this property.

We’ve also seen that the splitting field is **unique** up to isomorphism. We worked kind of hard trying to figure out In how many ways we can **extend** an embedding. That is, if are fields and is a field such that is an embedding. How many extensions of are there?

We also denoted the number of the extensions as . and we’ve seen that this number is bounded by the degree of the extension: .

We found out that if and is the minimal polynomial of , then the number of **extensions **is the number of **roots **of over .

On the other hand, if where are roots of such that **splits **in , then we are **guaranteed** to have extensions.

Moreover, if is **separable** in then we know the **exact **number of extensions – it is . This fact gives us motivation to understand and study **separable polynomials**.

## What’s next?

Why would I remind all those stuff when you can just go the last post and read them and their proof? Well, that follows from a simple reason – they are the **key** to everything we are doing. As you will see, I am going to use those conclusions multiple times. Thus, it is very important to **understand** them, and remember them.

## What is an automorphism?

Suppose that is a field extension. Of course, is a field, so we can define homomorphisms from it to itself:

\varphi:K\to K

In addition, we know that every field homomorphism is actually an embedding. If the embedding is also onto then it is an isomorphism.

Usually we call an isomorphism from an algebraic structure to itself an automorphism.

However, we still need to consider our base field, . So I’ll define an **automorphism of an** **extension** which is an automorphism such that the restriction to is the **identity**:

\text{for every } a\in F:\varphi(a)=a

That is – **Fixes **the elements of .

For example, Consider the extension . An automorphism of it should fix the elements of :

\varphi(a+bi)=\varphi(a)+\varphi(bi)=\varphi(a)+\varphi(b)\varphi(i)=a+b\cdot\varphi(i)

We can define and that would be a valid automorphism.

Wait a minute, I have already used this example before. It was an example of an embedding extension…

That should make you wonder – Automorphisms are in fact a special case of an embedding extension, right?

Well **of course** that’s right, they extend the **inclusion** embedding from to :

\begin{array}{ccc} K & \overset{\varphi}{\longrightarrow} & K\\ \cup & & \parallel\\ F & \overset{i}{\longrightarrow} & K \end{array}

And that’s such a great fact! Since we know a lot about embedding extensions! Let’s see some conclusions:

## Automorphisms as embeddings

Notice that if is finite, and is an **embedding** then is not just an embedding – it is an **isomorphism**. Why? we only need to show it is onto:

Since is an embedding, it is in particular a** one-to-one linear map** between vector spaces of the same dimension over – Thus it is an isomorphism of vector spaces. Which is in particular onto.

So if we are dealing with a **finite **extension we know that embedding extensions and automorphism are the **same**.

In that case, we have a **bound** for the number of the extensions – the degree of the extension – . Moreover, If is a **splitting field of a separable polynomial** , then there are exactly automorphisms.

But, we know exactly how such extensions behave – they map a root of to another root of – it **acts** as a **permutation** on the roots of , and even more than that – it is determined entirely by it’s action on the roots!

If – where the ‘s are irreducible, it has to map a root of to another root of it – since we create an extensions using the isomorphism:

F[\alpha_1]\cong F[x]/\langle f_i\rangle \cong F[a_2]

(Note that we **can’t** divide by since we won’t get a field, is not a maximal ideal!)

For example, consider the polynomial , it’s splitting field is and it has degree **4** over (I’ve showed it here). Since this polynomial is separable, we are guaranteed to have **4** extensions.

However, is reducible, it has two irreducible factors: with roots and with roots .

So every automorphism must acts as follows:

\sqrt{2}\mapsto \sqrt{2} \text{ and } \sqrt{3}\mapsto \sqrt{3} \\ -\\ \text{or} \\-\\ \sqrt{2}\mapsto -\sqrt{2} \text{ and } \sqrt{3}\mapsto \sqrt{3} \\-\\ \text{or} \\-\\ \sqrt{2}\mapsto \sqrt{2} \text{ and } \sqrt{3}\mapsto -\sqrt{3} \\-\\ \text{or} \\-\\ \sqrt{2}\mapsto -\sqrt{2} \text{ and } \sqrt{3}\mapsto -\sqrt{3}

So we’ve got exactly 4 extension, and that’s all of them!

## Properties of Automorphisms

Let’s try to prove some properties of automorphism.

First of all, every extension has at least one automorphism – the idenitity on . Let’s denote it by .

Since an automorphism is one-to-one and onto, it has an inverse function .

It’s also not hard to check that a composition of two automorphisms is an automorphism as well (that fixes the elements of ). Moreover, composition is an associative action.

So in total, the set of automrphisms of a field extension, equipped with the action of function composition is a **group**.

In fact – this is the most important group in this topic – all we are going to do from now on is related to this group, and it has a special name:

We call the group of automorphism of the extension the **Galois Group** of the extension, and denote it by:

\text{Gal}(K/F)

(FYI, some people may call this group the automorphisms group and save the name galois group for something ‘stronger’)

If is a finite extension, we know that the order of the group is bounded by the degree of the extension:

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

And if is a splitting field of a separable polynomial – then the order of the group is exactly the degree of the extension.

In the example of the extension we’ve already found 2 automorphisms – the identity, and the one that sends each element to it’s complex conjugate (denote it by ). The number of automorphisms is bounded by so that’s all the automorphisms. moreover, every group with two elements is isomorphic to

Therefore :

\text{Gal}(\mathbb{C}/\mathbb{R})=\{id_\mathbb{C},\varphi\}\cong \mathbb{Z}/2\mathbb{Z}

I want to give a bit more complicated example, but first, I want to mention some pretty great insights that we are going to use a lot:

## The action of the group of the set of roots

For any field extension such that is the splitting field of some polynomial , the group acts on the set of **roots** of – .

#### Faithful action

Since the automorphisms are defined **entirely **by their action on the roots, the group acts **faithfully** on the set of the roots. That is, if are two automorphism such that:

\forall i\in\{1,\dots, n\} : \sigma_1(\alpha_i)=\sigma_2(\alpha_i)

Then they must be the **same** automorphism!

How is that useful? well, because of the well known fact from group theory that states that if a group acts on a set faitfully, then we can **embed** it in the symmetric group , which is isomorphic to the group . A proof for that is not hard, we can easilly define the homomorphism:

\varphi:G\to S_{|A|} \\ \sigma\mapsto f_\sigma

Where defined as: . You can validate yourself that if acts faithfully, then is indeed an embedding (aka – a **monomorphism**).

Therefore, we can embed the galois group that acts on the set of roots of size in !

#### Sometimes – it might be transitive

Finally, if is irreducible, then the action is also **transitive** – that is, for every two roots there exists an automorphism such that . Why? For every two roots , we have the following isomorphism:

F(\alpha_i)\cong F[x]/\langle f\rangle \cong F(\alpha_j)\ (\sube K)

Let’s denote such an isomorphism from to as . This is in particular an **embedding** of to that **extends **the **inculsion** embedding . This yields the diagram:

\begin{array}{ccc} K & \overset{\sigma}{\longrightarrow} & K\\ \uparrow & & \shortparallel\\ F_{1} & \overset{\varphi_1}{\longrightarrow} & K\\ \uparrow & & \shortparallel\\ F & \overset{i}{\longrightarrow} & K \end{array}

And we already proved that we can extend to an **automorphism** . So we’ve found and automorphism that satisfies what we want!

Great, But what can we do with such information?

Well, we know by the **Orbit-stabilizer theorem** that the size of an orbit **divides** the order of the group. However a transitive action means that there is only **one **orbit – the whole set, which is of size . Therefore, we conclude that divides the order of the group:

n\ | \ |\text{Gal}(K/F)|

And that’s a pretty useful conclusion!

#### Back to the examples

##### A more interesting example

We’ve just calculated the automorphisms in , we have found **four** extensions! And we know that this is the splitting field of the polynomial whose roots are . There are 4 roots, and the action of the group on the set of the roots is faithful, therefore we can embed the group in . Let’s label the roots:

1\leftrightarrow \sqrt{2} \\ 2\leftrightarrow -\sqrt{2} \\ 3\leftrightarrow \sqrt{3} \\ 4\leftrightarrow -\sqrt{3} \\

So we can represent the automorphism that we’ve found before as the following permutations:

id \\ (1\ 2) \\ (3\ 4) \\ (12)(34)

So . And that’s exactly **4 Klein four-group**, which is also isomorphic to !

\text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})\cong V_4\cong \mathbb{Z}_2\times \mathbb{Z}_2

Pretty cool right? Let’s raise the level a bit:

##### Some more complicated example

Let’s calculate the galois group of the **splitting field** of the polynomial which we’ll denote as . It’s kind of hard calculating it’s roots. I mean, there is a formula for that but it’s a pretty long one…

Note that this polynomial is irreducible:

The only candinates of **rational** roots are . This folloes from the fact that for any polynomial with integer coefficients, a rational root must satisfy: (try to prove it yourself by substituting ), and we can check and see that those are not roots of . Hence, as a polynomial of degree **3**, it must be irreducible.

Moreover, note that is a monotonically increasing – it’s derivative is:

f^\prime(x)=3x^2+1>0

And we can use the **mean-value theorem** to conclude that it has only **one** real-valued root, which we’ll denote as .

The other two roots are **complex roots**, and one has to be a **conjucate** of the other (try to prove that if is a root of a polynomial then so as ). So let’s denote the other two roots as and .

Ok, the galois group acts faitfully on the set of roots (which is of size 3) and we can embed it in the group .

Moreover, since is irreducible, the action is transitive and we conclude that . By **Cauchy’s theorem**, we are guaranteed to have an element of order in the group – which has to correspond to a 3-cycle in .

Moreover, the map that sends an elements to it’s complex conjucate also forms an automorphism that maps to and vice versa, while fixing . This automorphism corresponds to a 2-cycle. So we now have a 2-cycle and a 3-cycle in the group, but together they form the whole group . Hence:

\text{Gal}_f\cong S_3

And that’s great! We have caluclated the galois group without even knowing what the roots are!

(Note that is separable as an irreducible polynomial over a field with characteristic zero), so we can conclude that the degree of the splitting field of over is the number of the automorphisms, which is the order of the group – **6**. You can try to find the splitting field yourself and see that the degree of the extension is indeed 6.

## Summary

So we’ve met the super important term of an **automorphism**, and also met the **galois group** and saw some pretty elegant ways of computing it.

However, there is much more than just computing the group, as it turns out, the galois group will give us an understanding of not only the extension – but it will also help us to find sub-fields and sub-extensions! How exactly? That’s what we are going to see in the next post – we’ll meet the **galois correspondence**.