# Automorphisms

In the one of my previous posts, we’ve discussed about splitting fields of a polynomial $f$. Those are fields where $f$ 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 $F\sub K$ are fields and $E$ is a field such that $\varphi:F\to E$ is an embedding. How many extensions $\varphi^\prime:K\to E$ of $\varphi$ are there?

We also denoted the number of the extensions as $n_{F\overset{\varphi}{\to} E}^K$. and we’ve seen that this number is bounded by the degree of the extension: $[K:F]$.

We found out that if $K=F[a]$ and $p\in F[x]$ is the minimal polynomial of $a$, then the number of extensions is the number of roots of $\varphi(p)$ over $E$.

On the other hand, if $K=F[a_1,\dots,a_n]$ where $a_i$ are roots of $f$ such that $\varphi(f)$ splits in $E$, then we are guaranteed to have extensions.

Moreover, if $\varphi(f)$ is separable in $E$ then we know the exact number of extensions – it is $[K:F]$. 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 $K/F$ is a field extension. Of course, $K$ 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, $F$. So I’ll define an automorphism of an extension which is an automorphism $\varphi:K\to K$ such that the restriction to $F$ is the identity:

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

That is – $\varphi$ Fixes the elements of $F$.

For example, Consider the extension $\mathbb{C}/\mathbb{R}$. An automorphism of it should fix the elements of $\mathbb{R}$:

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

We can define $\varphi(i)=-i$ 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 $F$ to $K$:

\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 $[K:F]$ is finite, and $\varphi:K\to K$ is an embedding then $\varphi$ is not just an embedding – it is an isomorphism. Why? we only need to show it is onto:

Since $\varphi$ is an embedding, it is in particular a one-to-one linear map between vector spaces of the same dimension over $F$ – 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 – $[K:F]$. Moreover, If $K$ is a splitting field of a separable polynomial $f$, then there are exactly $[K:F]$ automorphisms.

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

If $f=f_1\cdots f_k$ – where the $f_i$‘s are irreducible, it has to map a root of $f_i$ 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 $F[x]$ by $\langle f\rangle$ since we won’t get a field, $\langle f\rangle$ is not a maximal ideal!)

For example, consider the polynomial $f(x)=(x^2-2)(x^2-3)\in \mathbb{Q}[x]$, it’s splitting field is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ and it has degree 4 over $\mathbb{Q}$ (I’ve showed it here). Since this polynomial is separable, we are guaranteed to have 4 extensions.

However, $f$ is reducible, it has two irreducible factors: $(x^2-2)$ with roots $\pm \sqrt{2}$ and $(x^2-3)$ with roots $\pm \sqrt{3}$.

So every automorphism $\sigma:K\to K$ 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 $K$. Let’s denote it by $id_K$.

Since an automorphism $\varphi$ is one-to-one and onto, it has an inverse function $\varphi^{-1}$.

It’s also not hard to check that a composition of two automorphisms is an automorphism as well (that fixes the elements of $F$). 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 $K/F$ 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 $K/F$ 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 $K$ 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 $\mathbb{C}/\mathbb{R}$ we’ve already found 2 automorphisms – the identity, and the one that sends each element to it’s complex conjugate (denote it by $\varphi$). The number of automorphisms is bounded by $[\mathbb{C}:\mathbb{R}]=2$ so that’s all the automorphisms. moreover, every group with two elements is isomorphic to $\mathbb{Z}/2\mathbb{Z}$

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 $K/F$ such that $K$ is the splitting field of some polynomial $f\in F$, the group $\text{Gal}(K/F)$ acts on the set of roots of $f$$\{\alpha_1,\dots,\alpha_n\}$.

#### 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 $\sigma_1, \sigma_2$ 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 $G$ acts on a set $A$ faitfully, then we can embed it in the symmetric group $S_A$, which is isomorphic to the group $S_{|A|}$. A proof for that is not hard, we can easilly define the homomorphism:

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

Where $f_\sigma:A\to A$ defined as: $f_\sigma(a)=\sigma\cdot a$. You can validate yourself that if $G$ acts faithfully, then $\varphi$ is indeed an embedding (aka – a monomorphism).

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

#### Sometimes – it might be transitive

Finally, if $f$ is irreducible, then the action is also transitive – that is, for every two roots $\alpha_i,\alpha_j$ there exists an automorphism $\sigma$ such that $\sigma(a_i)=a_j$. Why? For every two roots $\alpha_i, \alpha_j$, 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 $F(\alpha_i)$ to $F(\alpha_j)$ as $\varphi_1$. This is in particular an embedding of $F(\alpha_i)$ to $K$ that extends the inculsion embedding $i:F\to K$. 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 $\varphi_1$ to an automorphism $\sigma$. 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 $n$. Therefore, we conclude that $n$ 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 $\text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$, we have found four extensions! And we know that this is the splitting field of the polynomial $f(x)=(x^2-2)(x^2-3)$ whose roots are $\pm 2, \pm 3$. 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 $S_4$. 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 $\text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})=\{id,(1\ 2),(3\ 4),(1\ 2)(3\ 4)\}$. And that’s exactly 4 Klein four-group, which is also isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$!

\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 $f(x)=x^3+x+1\in\mathbb{Q}[x]$ which we’ll denote as $\text{Gal}_f$. 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 $\pm 1$. This folloes from the fact that for any polynomial $a_nx^n+\cdots+a_0$ with integer coefficients, a rational root $\frac{p}{q}$ must satisfy: $p|a_0, q|a_n$ (try to prove it yourself by substituting $x=\frac{p}{q}$), and we can check and see that those are not roots of $f$. Hence, as a polynomial of degree 3, it must be irreducible.

Moreover, note that $f$ 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 $\theta$.

The other two roots are complex roots, and one has to be a conjucate of the other (try to prove that if $z$ is a root of a polynomial $f\in\mathbb{R}[x]$ then so as $\overline{z}$). So let’s denote the other two roots as $\rho$ and $\overline{\rho}$.

Ok, the galois group acts faitfully on the set of roots (which is of size 3) and we can embed it in the group $S_3$.

Moreover, since $f$ is irreducible, the action is transitive and we conclude that $3 \ |\ |\text{Gal}_f|$. By Cauchy’s theorem, we are guaranteed to have an element of order $3$ in the group – which has to correspond to a 3-cycle in $S_3$.

Moreover, the map that sends an elements to it’s complex conjucate also forms an automorphism that maps $\rho$ to $\overline{\rho}$ and vice versa, while fixing $\theta$. 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 $S_3$. 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 $f$ is separable as an irreducible polynomial over a field with characteristic zero), so we can conclude that the degree of the splitting field of $f$ over $\mathbb{Q}$ 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.