The Fundamental Theorem of Galois Theory

After meeting the term of a Galois extension and Artin’s lemma, we are ready to prove the fundamental theorem of galois theory!

Recall that when $K/F$ is a galois extension, we can say a lot about the extension:

$K/F$ is a normal and separable.
$K$ 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 $F=K^G$ .
$K^{\text{Gal}(K/F)}=F$
$|\text{Gal}(K/F)|=[K:F]$

Moreover, artin’s lemma stated that if $G$ is an automorphism group of some field $K$ . Then:

[K:K^G]\leq|G|

However, we also had one unsolved problem:

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

So that’s where the fundamental theorem comes to help. It states that if the extension is Galois, then the maps are indeed inverting each other!

The proof is really easy since we’ve already done most of the work:

The proof

Note that since $K/F$ is a galois extension, it is in particular a splitting field of a separable polynomial $f\in F[x]$ . But for every subfield $F\sube L\sube K$ we can also think of $f$ as a polynomial over it – $f\in L[x]$ . Therefore, $K/L$ is also a galois extension.

Great – now we can use the 4-th equivalent defintion to get:

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

And that exactly what we wanted – If we map a subfield to a subgroup, and then map the subgroup to a field again – we will return to the same field that we started with.

Now, what about subgroups? Suppose that $H\leq \text{Gal}(K/F)$ is a sub-group. We can use artin’s lemma to get that:

|H|\geq[K:K^H]

We already know that:

H\sube \text{Gal}(K/K^H) \\ \Downarrow \\ |H|\leq |\text{Gal}(K/K^H)|

Since if $\sigma\in H$ so every $a\in K^H=\{a\in K| \forall \sigma \in H : \sigma(a)=a\}$ in particular satisfies $\sigma(a)=a$ Thus, $\sigma\in \text{Gal}(K/K^H)$ . Now, using the fact that the number of the automorphisms of an extension is bounded by the degree of the extension yields:

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

Let’s use all of those result to get:

|H|\leq|\text{Gal}(K/K^H)|{\leq}[K:K^H]\leq |H| \\ \Downarrow \\ |H|=|\text{Gal}(K/K^H)|

But we know that $H\sube \text{Gal}(K/K^H)$ , and since they have the same order, we conclude that they are the same:

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

As we wanted.

Note that in order to prove the equality $H=H^{*\circ}$ , we haven’t used any property of a galois extension! So this part of the proof is always true – for any extension. If you recall, we’ve already met this kind of asymmertry between the maps (here) – so galois extension is so great since it basically fixes this asymnetry.

Some insights

Note what we just proved without even noticing – if $H$ is a subgroup then:

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

That is, the degree of $K$ over $K^H$ is exactly the order of $H$ .

And we can take it even one step-forward: What if we want to know the degree of the extension $[K^H:F]$ ? Well, we can use the multiplicative property to get:

[K:K^H][K^H:F]=[K:F] \\ \Downarrow \\ [K^H:F]=\frac{[K:F]}{[K:K^H]}

And we also know that:

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

And by Lagrange’s theorem, we know that:

|\text{Gal}(K/F)|=|H|\cdot\overbrace{[\text{Gal}(K/F):H]}^{\text{ The index of the subgroup}} \\ \Downarrow \\ \frac{|\text{Gal}(K/F)|}{|H|}=[\text{Gal}(K/F):H]

Therefore:

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

And that’s amazing! The degree of the extension of the fixed field of $H$ is exactly the index of the subgroup $H$ ! That explains why the index of a subgroup and the degree of an extesion are written in the same way!

The theorem in action

Recall that we’ve proved in this post that $\text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3}))\cong V_4$ . Where $V_4\sube S_4$ is klein’s four group:

V_4=\{id, (1\ 2), (3, 4), (1\ 2)(3\ 4)\}

Where we identified the roots as:

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

It’s non-trivial subgroups are:

\lang (1\ 2)\rang =\{id, (1\ 2)\} \\ \lang (3\ 4)\rang =\{id, (3\ 4)\} \\ \lang (1\ 2)(3\ 4)\rang =\{id, (1\ 2)(3\ 4)\}

So the fundamental theorem tells us that there are exactly 3 sub-fields in the extension $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ . And we can find them by calculating the fixed field of each subgroup. So, let’s do it:

In the subgroup $\lang (1\ 2)\rang$ , there are only two automorphisms – the idenitity and the one that maps 1 to 2 and vice versa. That is:

\sqrt{2}\mapsto -\sqrt{2} \\ -\sqrt{2} \mapsto \sqrt{2}

And it also fixes the roots $\pm \sqrt{3}$ . Therefore, $\mathbb{Q}(\sqrt{3})\sube \mathbb{Q}(\sqrt3,\sqrt2)^{\lang (1\ 2)\rang}$ . However:

[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2=[V_4:\langle(1\ 2)\rangle]=[\mathbb{Q}(\sqrt{2},\sqrt{3})^{\lang (1\ 2)\rang}:\mathbb{Q}]

Great, so we can conclude that $\mathbb{Q}(\sqrt{3})= \mathbb{Q}(\sqrt{2},\sqrt{3})^{\lang (1\ 2)\rang}$ .

Similarly, $\mathbb{Q}(\sqrt{2})=\mathbb{Q}(\sqrt{2},\sqrt{3})^{\lang (1\ 2)\rang}$ .

What about the subgroup $\lang (1\ 2)(3\ 4)\rang$ ? It’s not fixing any of the roots – so what can it possibly fix?

So the trick here to use some expression involving some roots together. Let’s try this element:

\sqrt{6}=\sqrt{2}\cdot \sqrt{3}

Let’s denote by $\sigma$ the automorphism that represents the permutation $(1\ 2)(3\ 4)$ , and apply it on this element:

\sigma(\sqrt{2}\cdot \sqrt{3})=\sigma(\sqrt{2})\cdot \sigma(\sqrt{3})=(-\sqrt{2})\cdot(-\sqrt{3})=\sqrt{2}\cdot \sqrt{3}

Yes! This element works, and since the minimal polynomial of $\sqrt{6}$ over $\mathbb{Q}$ is:

x^2-6

Which has degree 2, we conclude that $[\mathbb{Q}(\sqrt{6}):\mathbb{Q}]=2$ . Thus:

\mathbb{Q}(\sqrt{6})=\mathbb{Q}(\sqrt{2},\sqrt{3})^{\lang (1\ 2)(3\ 4)\rang}

And that’s it!

Let’s see those diagrams next to each other:

Note that the order is inverted – in the diagrams, the biggest field is on the top and the biggest group is on the bottom.

Is that everything the fundamental theorem says?

So we’ve proved the fundametal theorem and we’ve seen a pretty cool example, though, we still have a major unanswered question:

If $K/F$ is galois, we know that for every sub-field, $K/L$ is galois as well. However, when it’s true that $L/F$ is galois?

To answer this question, I shall present part 2 of the fundamental theorem! Yes, there is a part 2! But, before I’ll present it – I want to have a quick discussion:

Automorphism on a fixed field

Suppose that $K/F$ is a galois extension and $H\leq \text{Gal}(K/F)$ . Moreover, let $\tau\in \text{Gal}(K/F)$ be some automorphism. Now consider the set:

\tau(K^H)=\{\tau(b):b\in K^H\}

And by the definition of the fixed field, we know that it is exactly the set:

\{\tau(b)\ |\ \forall \sigma\in H:\sigma(b)=b\}

However, $\tau$ is an automorphism, which is in particular onto and one-to-one – therefore, for every $b\in B$ there exists $a\in K$ such that $a=\tau(b)$ , which is the same as $\tau^{-1}(a)=b$ . Therefore, our set is:

\{a\in K |\ \forall \sigma\in H:\sigma(\tau^{-1}(a))=\tau^{-1}(a)\}

=\{a\in K |\ \forall \sigma\in H:\sigma\tau^{-1}(a)=\tau^{-1}(a)\}

=\{a\in K |\ \forall \sigma\in H:\tau\sigma\tau^{-1}(a)=a\}

But the elements of the form $\tau\sigma\tau^{-1}$ for some $\sigma\in H$ are exactly the elements of the subgroup $\tau H\tau^{-1}$ . So our set is in fact the fixed field $K^{\tau H\tau^{-1}}$ , and in total we’ve got that:

\tau(K^H)=K^{\tau H\tau^{-1}}

And that’s pretty interesting! Note that if $H$ is normal we get that:

\tau(K^H)=K^{\tau H\tau^{-1}}=K^H

This means that $K^H$ is closed under the action of $\tau$ . In other words, $\tau|_{K^H}$ is an automorphism of it.

And that’s true for every $\tau\in \text{Gal}(K/F)$ !

Can you see where I am going with it? If not, think about where a root $a\in K^H$ of an irreducible polymials is mapped to.

Ok, after this little discussion – let’s present the theorem:

The fundamental theorem of galois theory – part 2

Let $K/F$ be a galois extension, and suppose that $H\leq \text{Gal}(K/F)$ is a subgroup. So:

K^H/F\text{ is Galois} \iff H\text{ is normal: } H \vartriangleleft \text{Gal}(K/F)

Before I’ll prove it, I want to mention an important remark:

First, note that if $F\sube L\sube K$ and $K/F$ is separable, then so as $L/F$ . This follows from the simple fact that every element of $L$ is in particular an element of $K$ , hence, it’s separable.
The statment: “if $F\sube L\sube K$ and $K/F$ is normal, then so as $L/F$ ” is not always true! However, if it true – then the extension is automatically a Galois extension! Why? Since it’s a normal and separable extension.

Thus, $L/F$ is galois if and only if $L$ is a normal extesnion!

So what the theorem says is equivalent to:

H\text{ is normal} \iff K^H\text{ is normal }(\text{over } F)

What a great connection! It explains exactly why we chose this name for a normal extension.

The proof

$\Rightarrow$

First, Suppose that $K^H/F$ is Galois, so it’s in particular normal. Now, let $a\in K^H$ be some element and $\tau\in \text{Gal}(K/F)$ be some automorphism.

We know that $\tau(a)$ is a root of the minimal polynomial of $a$ over $F$ .

Since the extension is normal, we conclude that $\tau(a)\in K^H$ . Therefore:

\tau(K^H)= \{\tau(a):a\in K^H\}\sube K^H

However, we’ve seen that:

K^{\tau H\tau^{-1}}=\tau(K^H)

Therefore:

K^{\tau H\tau^{-1}}\sube K^H

However, we can use the first part of the fundamental theorem and the fact that:

|\tau H\tau^{-1}|=|H|

To get:

[K:K^H]=|H|=|\tau H\tau^{-1}|=[K:K^{\tau H\tau^{-1}}] \\ \Downarrow\\ K^H=K^{\tau H\tau^{-1}}

But since galois correspondence of a galois extension is invertible we conclude that:

H=\tau H\tau^{-1}

So $H$ is normal!

$\Leftarrow$

Suppose that $H$ is normal, and consider the restriction map:

\theta: \text{Gal}(K/F)\to\text{Gal}(K^H/F) \\ \tau\mapsto \tau|_{K^H}

this is a well defined homomorphism since the group is normal:

\tau(K^H)=K^{\tau H\tau^{-1}}\overset{H\vartriangleleft \text{Gal}(K/F)}{=}K^H \\ \Downarrow \\ \tau|_{K^H}=K^H \\ \Downarrow \\ \tau|_{K^H}\in\text{Gal}(K^H/F)

Let’s figure out what the kernel is:

\tau\in\ker \theta \iff \tau|_{K^H} =id \iff \tau\in \text{Gal}(K/K^H)

Thus $\ker\theta = \text{Gal}(K/K^H)$ . Now, by the first isomorphism theorem, we conclude that:

\text{Gal}(K/F)/\text{Gal}(K/K^H)\cong \text{Im }\theta\leq \text{Gal}(K^H/F)

But:

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

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

The first equality is just the size of the qutient group
The second follows from the fact that $K/F$ and $K/K^H$ are both galois.
The third follows from the mutliplicative property of extensions.
And finally, the inequality is due to the fact that the number of the automorphisms is bounded by the degree of the extension.

This result combined with the fact that we were able to embed the qutient group is $\text{Gal}(K^H/F)$ shows that $\theta$ is onto, and in particular:

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

So $K^H/F$ is galois, as we wanted. Moreover, we have the isomorphism

\text{Gal}(K/F)/\text{Gal}(K/K^H)\cong \text{Gal}(K^H/F)

And that’s pretty cool, not only we know that $K^H/F$ is Galois, but we also know exactly what the galois group is! It is a quotient group of the original galois group.

Note that by the first part of the fundametal theorem, we also know that $\text{Gal}(K/K^H)=H$ , so we can write the quality in the above a bit different:

\text{Gal}(K/F)/H\cong \text{Gal}(K^H/F)

Using the second part

Consider the polynomial $f(x)=x^3-2$ . It’s roots are:

\sqrt[3]{2}, \sqrt[3]{2}\rho, \sqrt[3]{2}\rho^2

Where $\rho=e^{\frac{2\pi i}{3}}$ is the root of unity of order 3. Let’s calculate it’s galois group (and by that I mean the galois group of it’s splitting field which we will denote by $E$ ):

Ok, so the polynomial is irreducible (use eisenstien criterion with $p=2$ , or just note that it has no rational roots…) hence separable as a polynomial over $\mathbb{Q}$ ( $\text{char}(\mathbb{Q})$ =0) so, the extension $E/\mathbb{Q}$ is galois as a splitting field of a separable polynomial.

Recall that since there are 3 roots, and the action of the galois group on the set of roots is faithul, we can embed the galois group in $S_3$ . Moreover, the action is also transitive since the polynomial is irreducible, so there is only one orbit of size $3$ that divides the order of the group – hence, by cauchy’s theorem there exists an element of order 3 – which is a 3-cycle.

Moreover the map that sends an element to it’s complex conjugate also forms an automorphism that fixes the root $\sqrt[3]{2}$ and maps $\sqrt[3]{2}\rho$ to $\sqrt[3]{2}\rho^2$ and vice versa. Hence, this automorphism is of order 2. That is, a 2-cycle

But 2-cycle and 3-cycle generate the whole group $S_3$ , so:

\text{Gal}_f\cong S_3

Great, now, what are the normal subgroups of $S_3$ ? There is only one:

A_3 =\{id, (1\ 2\ 3), (1\ 3\ 2)\}

So we know that the field:

E^{A_3} \\

is normal. Let’s try to figure out what is this field – First, we know it’s degree by the first part of the fundamtel theorem (here) :

[E^{A_3}:\mathbb{Q}]=[S_3:A_3]=2

So that’s a good start. Now let’s identify the roots as:

1\leftrightarrow \sqrt[3]{2} \\ 2 \leftrightarrow \sqrt[3]{2}\rho \\ 3\leftrightarrow \sqrt[3]{2}\rho^2 \\

And denote / identify the automorphisms as:

\sigma\leftrightarrow (1\ 2\ 3) \\ \tau\leftrightarrow (1\ 3\ 2)

Let’s see where $\rho$ is being mapped under those automorphisms:

\sigma(\rho)=\sigma(\frac{\sqrt[3]{2}\rho}{\sqrt[3]{2}})=\frac{\sigma(\sqrt[3]{2}\rho)}{\sigma(\sqrt[3]{2})}=\frac{\sqrt[3]{2}\rho^2}{\sqrt[3]{2}\rho}=\rho

Great! In a similar way we can show that $\tau$ also fixes $\rho$ and conclude that:

\mathbb{Q}(\rho)\sube E^{A_3}

However, $\rho$ is not a rational, so:

1<[\mathbb{Q}(\rho):\mathbb{Q}]\leq [E^{A_3}:\mathbb{Q}]=2 \\ \Downarrow \\ [\mathbb{Q}(\rho):\mathbb{Q}]=2

So, in fact: $E^{A_3}=\mathbb{Q}(\rho)$ , And we’ve found the only normal subfield without even finding the splitting field! (If you are curious, the splitting field is $\mathbb{Q}(\sqrt[3]{2},\rho)$ – try to prove it, it’s not hard at all – indeed, it has degree 6 – the order of it’s galois group: $S_3$ ).

The numbers represent the degrees of the extensions / index of the subgroup

If you want, you can try to calculate the rest of the subfields, and complete the subgroups / subfields lattices:

Note that by the second part of the theorem, the extensions:

\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q} \\ \mathbb{Q}(\sqrt[3]{2}\rho)/\mathbb{Q} \\ \mathbb{Q}(\sqrt[3]{2}\rho^2)/\mathbb{Q}

are not normal. Indeed, in those extensions, the polynomial has only one root, so we can only map it to itself – hence, the only automorphism is the identity. If we’ll focus on $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ , we’ll see that everything there falls apart, for example:

$|\text{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})|=1<2=[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$
$\mathbb{Q}(\sqrt[3]{2})^{\text{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})}=\mathbb{Q}(\sqrt[3]{2})\neq \mathbb{Q}$

Summary

So we’ve finally proved the fundamental theorem of galois theory, and I think that you can see it’s strength just by the examples I gave here! If we knowt he structure of the group – we know the structure of the field, and vice versa!

Now, after all the hard work, it’s time to have some fun and see some great results – Some are related to galois theory, and some are really unexpected!

The Fundamental Theorem of Galois Theory

The proof

Some insights

The theorem in action

Is that everything the fundamental theorem says?

Automorphism on a fixed field

The fundamental theorem of galois theory – part 2

The proof

$\Rightarrow$

$\Leftarrow$

Using the second part

Summary

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The proof

Some insights

The theorem in action

Is that everything the fundamental theorem says?

Automorphism on a fixed field

The fundamental theorem of galois theory – part 2

The proof

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Using the second part

Summary

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$\Rightarrow$

$\Leftarrow$