Back to Algebra Page

On this page:

• Introduction
• Automorphisms of Field Extensions
• Galois Extensions
• Finite Fields
• Worked Exercises

Introduction

On this page, we are going to build up to the fundamental theorem of Galois theory. The idea that we’re going to be chasing around is that given a polynomial $f(x)$, its splitting field extension $K/F$ can be understood by understanding the permutation of the roots of $f(x)$.

…or at least that’s what I told when I first learned Galois theory. I personally think this is such a stupid way to motivate Galois theory because why the hell would we care about permutations of roots and where does it even come from? I think that Evan Chen has a much better way of motivating Galois theory by instead thinking of terms of embeddings. (Something I want to make very clear is that this introduction is basically the exact same thing Evan writes in the Napkin so none of this is mine or original, these are just me writing down my notes.)

Given a number field (an extension of $\QQ$) $K$, an embedding is a homomorphism of $K$ into $\CC$ (recall that homomorphisms of fields are injective).

[example] [extitle]Example %counter% ($\QQ(i) \injto \CC$)[/extitle]

It’s pretty clear that $\QQ(i)$ is not algebraically closed. That being said, we know that we can embed a copy of it into $\CC$ to fully factor everything.

There are two canonical choices of embeddings: $z \mapsto z$ (the identity) or $z \mapsto \conj{z}$ (conjugation). It’s also pretty obvious that the images of $\QQ(i)$ under these maps are isomorphic.

As it turns out, these are the only embeddings! Why? [/example]

[solution] Because if $\sigma:K \to \CC$ is an embedding, then we can take any (nonzero) $q \in \QQ$ to obtain

where $a, b\in \QQ$. Thus,

So then $b = 0$ which implies that

Now write $q = r/s$ where $r, s \in \ZZ$ are relatively prime. Then

Since $\sigma(s) = s$, it follows that $\frac{r}{a} = s$ which implies that $a = r/s = q$. So $\sigma$ fixes every element of $\QQ$. Similarly, using $f(i^2) = -1$ and $f(i\overline{i})$ to form a system of equations shows that the above are the only embeddings. [/solution]

[example] [extitle]Example %counter% ($\QQ(\sqrt{2}) \injto \CC$)[/extitle]

The two canonical choices of embedding here are $a + b\sqrt{2}$ gets sent to itself or its radical conjugate.

Again, it turns out that these are the only two possible embeddings. [/example]

[solution] The previous example logic can be used to show an embedding must fix $\QQ$ element-wise. So it remains to show where $\sqrt{2}$ gets sent.

It’s actually even easier to show that these are the two possible embeddings.

So $\sigma(\sqrt{2}) = \pm \sqrt{2}$ as desired. [/solution]

[example] [extitle]Example %counter% ($\QQ(\sqrt[3]{2}) \injto \CC$)[/extitle]

The three canonical choices of embedding here are:

1. The identity embedding.
2. The embedding that sends $a + b\sqrt[3]{2} + c\sqrt[3]{2}^2$ to $a + b\omega\sqrt[3]{2} + c(\omega\sqrt[3]{2})^2$.
3. The embedding that sends $a + b\sqrt[3]{2} + c\sqrt[3]{2}^2$ to $a + b\omega^2\sqrt[3]{2} + c(\omega^2\sqrt[3]{2})^2$.

Again, we can show that these are the only embeddings. [/example]

[solution] The previous example logic can be used to show an embedding must fix $\QQ$ element-wise. So it remains to show where $\sqrt[3]{2}$ gets sent.

Since

it follows that $\sigma(\sqrt[3]{2})$ must be sent to a cube root of $2$. [/solution]

[example] [extitle]Example %counter% (The punchline)[/extitle]

More generally, if we have a simple extension $\QQ(\alpha)/\QQ$ and we want to embed it into $\CC$, then taking the minimal polynomial $m_{\alpha,\QQ}(x)$ for $\alpha$ over $\QQ$ gives

Applying the embedding map to both sides gives

So this means that an embedding map must send roots of minimal polynomials to other roots of the same minimal polynomial.

Notice that this does not mean that every permutation of roots is admissible — there are $3! = 6$ ways to permutate three objects but there are only three embeddings of $\QQ(\sqrt[3]{2})$ into $\CC$. What is true, however, is that the degree of the field $K$ is equal to the number of embeddings. [/example]

Something to also note is that in the examples above, the first two actually embed $K$ back into $K$ inside of $\CC$ but the third has two embeddings that cause $K$ to leave itself inside $\CC$. The first two examples will be what we call Galois extensions (which, in our case, means that the roots are actually all inside $K$).

Now that we have this build-up of natural examples to look at, it is obvious that there is some sort of relationship between roots (namely, conjugation) of minimal polynomials which imposes relations on the permutations allowed. So it is very natural to explore that connection!

Automorphisms of Field Extensions

[definition] [deftitle]Definition %counter%[/deftitle]

Let $K/F$ be an extension of fields. Then $\Aut(K/F)$ is the set of automorphisms of $K$ that fix $F$ element-wise. [/definition]

Something that is very strongly related to what we were describing in the introduction is how prime subfields relate to automorphisms. Recall that a prime subfield of $K$ is the field $F$ generated by $1$ in $K$. So if $K = \CC$, then $F$ is just $\QQ$. More generally, the prime subfield is fixed by any automorphism of $K$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $K/F$ be an extension of fields. Then $\Aut(K/F)$ is a subgroup of $\Aut(K)$. [/theorem]

[proof] Clearly the identity map on $K$ fixes $F$. On the other hand, if $\sigma,\tau \in \Aut(K/F)$ and $f\in F$, then

So $\sigma,\tau\in \Aut(K/F)$. [/proof]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $K/F$ be a field extension and let $\alpha \in K$ be algebraic over $F$. Then for any $\sigma\in\Aut(K/F)$, $\sigma\alpha$ is a root of the minimal polynomial for $\alpha$ over $F$. [/theorem]

[proof] Let $m_(\alpha,F)(x)$ be the minimal polynomial for $\alpha$ over $F$. Then

So $\sigma\alpha$ is a root as desired. [/proof]

Now we show how to associate a group with a subfield of $K$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $K$ be a field and $H \le \Aut(K)$. Then the set

is a subfield of $K$, called the fixed field of $H$. [/theorem]

[proof] Follows immediately from the fact that the elements of $\Aut(K)$ are ring homomorphisms. [/proof]

Later on, what we want to show is that the subgroups of $\Aut(K)$ (of the form $\Aut(F/K)$) and $K^H$ are in bijection (provided certain other assumptions).

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The correspondences between subfields of $K$ and subgroups of $\Aut(K)$ are inclusion-reversing. That is:

1. If $F \subseteq E \subseteq K$, then $\Aut(K/E) \subseteq \Aut(K/F)$.
2. If $G \le H \Aut(K)$, then $K^H \subseteq K^G$. [/theorem]

[proof] (1) Let $\sigma \in \Aut(K/E)$. Then $\sigma$ fixes $E$. But since $F \subseteq F$, $\sigma$ necessarily fixes $F$ so $\sigma \in \Aut(K/F)$.

(2) Let $x \in K^H$. Since every $\sigma$ in $G$ is also an element of $H$, it follows that $\sigma(x) = x$. Thus, $x \in K^G$ and we are done. [/proof]

Galois Extensions

[definition] [deftitle]Definition %counter%[/deftitle]

The finite field extension $K/F$ is called a Galois extension if $[K:F] = |\Aut(K/F)|$. If $K/F$ is Galois, then $\Aut(K/F)$ is called the Galois group of $K/F$ and is denoted $\Gal(K/F)$. [/definition]

So, in some sense, another way we could say this is that the degree of the extension is equal to the number of ways to permutate the roots of a minimal polynomial. In any case, some examples:

[example] [extitle]Example %counter%[/extitle]

The extension $\QQ(\sqrt{2})/\QQ$ is Galois. The degree of $\QQ(\sqrt{2})$ over $\QQ$ is obviously $2$ and $\Gal(\QQ(\sqrt{2})/\QQ)$ has order 2. To see this, it is straightforward that $a + b\sqrt{2} \mapsto a + b\sqrt{2}$ and $a + b\sqrt{2} \mapsto a - b\sqrt{2}$ are automorphisms. These are also the only automorphisms fixing $\QQ$ as

So the automorphism is fully determined by where it sends $\sqrt{2}$. Since

it follows that $\sigma(\sqrt{2}) = \pm \sqrt{2}$ as desired. [/example]

[example] [extitle]Example %counter%[/extitle]

The extension $\QQ(\sqrt[3]{2})/\QQ$ is not Galois. This is because the group of automorphisms is trivial since

implies that $\sigma(\sqrt[3]{2})$ is a cube root of $2$. Since two of these are nonreal (and those aren’t in $\QQ(\sqrt[3]{2})$), it follows that

[/example]

There are several different ways to characterize Galois extensions.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $K/F$ be a finite field extension. The following are equivalent:

1. $K/F$ is Galois.
2. $K$ is the splitting field of some separable polynomial $g(x) \in F[x]$
3. $F$ is the fixed field of some subgroup of $\Aut(K)$.
4. If $p(x) \in F[x]$ is irreducible and has a root in $K$, then $p(x)$ is separable and splits over $K$. [/theorem]

Finally, and the must important theorem:

[theorem] [thmtitle]Theorem %counter% (Fundamental Theorem of Galois Theory)[/thmtitle]

Let $K/F$ be a Galois extension and $G = \Gal(K/F)$. Then there is a bijection

given by maps

which are inverse to each other. This bijection is order-reversing, $[K:E] = |H|$ and $[E:F] = |G:H|$. We also have that $E/F$ is Galois if and only if $H$ is normal in $G$. [/theorem]

We will omit the rather annoyingly technical proofs of both of these theorems.

Finite Fields

In this section, we will finally discuss finite fields a little bit more appropriately and we will start referring to $\ZZ/p\ZZ$ as just $\FF_p$.

Consider a field $F$ that has characteristic $p$. Accordingly, the prime subfield of $F$ is just $\FF_p$ so such a field is an $F$-vector space of degree $n\coloneqq [F:\FF_p]$. This tells us that $F$ has $p^n$ elements which we will denote $\FF_{p^n}$. As it turns out, we can fully classify the finite fields and doing so is quite easy.

Consider the splitting field of $x^{p^n} - x$ over $\FF_p$. Since $F^{\times}$ is a group of order $p^n - 1$, Lagrange’s theorem implies that

for every $a \in \FF$. This implies that all elements of $\FF_p$ are roots of $x^{p^n} - x$ which means we can factor this polynomial as

This shows that $x^{p^n} - x$ is separable and also that $F$ is the splitting field of $x^{p^n} - x$ over $\FF_p$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

There is exactly one field $\FF_{p^n}$ of cardinality $p^n$ up to isomorphism. In particular, $\FF_{p^n}$ is the splitting field of $x^{p^n} - x$ over $\FF_p$. Moreover, $\FF_{p^n}/\FF_{p}$ is separable. [/theorem]

Accordingly, this tells us that $\FF_{p^n}/\FF_p$ is Galois. Moreover, since

it follows that the Galois group is order $n$. Now there are many groups of order $n$, but it turns out that the Galois group is actually nothing more than just $\ZZ/n\ZZ$ and the key to seeing this is the Frobenius automorphism.

[definition] [deftitle]Definition %counter%[/deftitle]

The Frobenius automorphism of a field $F$ of characteristic $p$ is the map $\sigma_p:F \to F$ defined by $a\mapsto a^p$. [/definition]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The Frobenius automorphism is actually an automorphism. [/theorem]

[proof] Multiplication is obviously preserved by the Frobenius automorphism. For addition, we apply the binomial theorem to $(a + b)^p$ and the fact that $F$ is characteristic $p$ to kill everything but $a^p + b^p$. The unit and additive identity are obviously preserved. Thus, the Frobenius automorphism is a ring homomorphism.

Ring homomorphisms into a field are necessarily injective so it remains to show that the Frobenius automorphism is surjective. However, since $F$ is a finite set, any injection is also a surjection. [/proof]

Now, here’s the claim that we will try to justify:

Clearly, $\sigma_p$ fixes $\FF_p$. It is also to see that $\sigma_p^n$ is just the identity as

This implies that the order of $\sigma_p$ divides $n$. To see that it is exactly $n$, suppose that $d \mid n$ and $1 < d < n$. Then $\sigma^d$ is the identity mapping. That is,

This violates the separability $x^{p^n} - x$ this implies that $x^{p^n} - x$ contains repeated linear factors. So it follows that $\sigma_p$ has order $n$ and the claim follows.

Worked Exercises

[example] [extitle]TODO: Cite[/extitle]

This problem is to determine the group $\Aut(\RR/\QQ)$.

1. Prove that any $\sigma \in \Aut(\RR/\QQ)$ takes positive reals to positive reals.
2. Show that $a < b$ implies $\sigma(a) < \sigma(b)$ for every $a, b\in \RR$.
3. Prove that $\sigma$ is continuous.
4. Prove that $\Aut(\RR/\QQ)$ is the trivial group. [/example]

[proof] (1) Let $r \in \RR_{>0}$. Then we know that $\sqrt{r} \in \RR_{>0}$ and so it follows that

(2) If $a < b$, then $b - a > 0$. So it follows that $\sigma(b - a) > 0$ by Part (1) and the claim follows.

(3) Let $(x_n)$ be a convergent sequence in $\RR$ (in particular, say $x_n \to x \in \RR$). Pick $\eps > 0$. If $\eps$ is irrational, then pick some $\eps’$ that is rational and $0 < \eps’ < \eps$. If $\eps$ is rational, set $\eps’$ to $\eps$.

Then there is some $N$ for which

Now as

applying $\sigma$ to the inequality gives

so it follows that

Now since $\sigma(\eps’) \le \eps$, it follows that $\sigma$ is sequentially continuous. For functions on $\RR$, this is equivalent to continuity.

(4) Note that $\sigma$ is constant on $\QQ$. If we let $r \in \RR\setminus\QQ$, then there is a sequence of rational numbers $q_n \to r$. Thus, $\sigma(q_n) \to \sigma(r)$. But since $\sigma(q_n) = q_n$, it follows that $\sigma(r) = r$ (since $\RR$ is Hausdorff). [/proof]

[example] [extitle]TODO: Cite[/extitle]

Let $K/F$ be a field extension, $G = \Aut(K/F)$, $H\le G$ a subgroup, and $\sigma \in G$. Prove that

[/example]

[proof] ($\subseteq$) Let $y\in K^{\sigma H \sigma^{-1}}$. For any $\tau \in H$, we have

Thus,

which implies that $\sigma^{-1}(y) \in K^H$. So then $y \in \sigma(K^H)$ as desired.

($\supseteq$) If $y \in \sigma(K^H)$, then $y = \sigma(x)$ where $x \in K^H$. So if $\tau \in H$, then $\tau(x) = x$. Thus,

which implies that $\sigma\tau\sigma^{-1}\in K^{\sigma H \sigma^{-1}}$. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Prove that if the Galois group of the splitting field of a cubic $f(x)$ over $\QQ$ is cyclic of order 3, then all the roots of the cubic are real. [/example]

[proof] As $f(x)$ has rational coefficients, any nonreal roots of $f(x)$ must occur in conjugate pairs. So the only possibilies for roots of $f(x)$ are:

• One real root, two nonreal occuring in conjugate pairs.
• All three roots are real. Suppose that $f(x)$ has roots $\alpha, \beta, \conj{\beta}$ with $\beta \in \CC\setminus\RR$. Now since $\beta$ is nonreal the minimal polynomial of $\beta$ over $\QQ(\alpha)$ has degree two (namely, it is $x^2 - (\beta + \conj{\beta})x + |\beta|^2$). By the tower law, we have that

But since

we arrive at a contradiction (as $\QQ(\alpha, \beta)$ is the splitting field of $f(x)$ over $\QQ$) as this would imply that $2\mid 3$. [/proof]

[example] [extitle]KU Algebra Qual August 2024 Problem 2[/extitle]

Set $f(x) \coloneqq x^{17} - 2 \in \QQ[x]$ and write $K$ for the splitting field of $f(x)$ over $\QQ$.

1. Calculate $[K:\QQ]$.
2. For $G$, the Galois group of $K$ over $\QQ$, show that $G$ is not abelian but $G$ has a cyclic subgroup of order $17$. [/example]

[proof] (1) Let $\omega$ be a primitive 17th root of unity. Then we claim that $K = \QQ(\sqrt[17]{2}, \omega)$. Obviously the $\subseteq$ claim holds. To justify that $\omega \in K$, we note that

Thus, $\omega \in K$. So then we obtain the $\supseteq$ direction.

The polynomial $f(x) = x^{17} - 2$ is irreducible over $\QQ$ by Eisenstein so it is the minimal polynomial for $\sqrt[17]{2}$ over $\QQ$. This implies that

Now since $(x^{17} - 1)/(x - 1) = x^{16} + x^{15} + \cdots + x + 1$ is the minimal polynomial for $\omega$ over $\QQ$. Now, since $K$ is the composite field of $\QQ(\sqrt[17]{2})$ and $\QQ(\omega)$, noting that these have degrees 17 and 16, respectively, over $\QQ$ implies that they have relatively prime degrees and so their composite field has degree $17\cdot 16 = 17\cdot 2^4$.

(2) The existence of a cyclic group of order 17 is given by applying Sylow’s theorem to $G$. By Sylow, there is a 17-Sylow subgroup and since $[K:\QQ] = |G|$, it follows that said subgroup has order 17. Since groups of prime order are always cyclic, it follows that $G$ has a cyclic subgroup of order $17$. To obtain the nonabelian assertion, consider the field automorphisms

and

Then we see that $\sigma\tau(\sqrt[17]{2}) \neq \tau\sigma(\sqrt[17]{2})$. [/proof]

[example] [extitle]KU Algebra Qual January 2025 Problem 6[/extitle]

Construct (with details) a Galois extension of fields $F \subseteq K$ such that $[K:F] = 11$. Give an explicit description of the corresponding Galois group. [/example]

[proof] We will let $F = \FF_p$ and we will define $L$ to be the splitting field $f(x) = x^{p^{11}} - x$ over $F$ which we will denote $\FF_{p^{11}}$. Let $K$ be the subfield of $L$ consisting of the roots of $f(x)$. The characteristic $p$ assumption ensures $K$ actually is a subfield. Furthermore, $K$ contains all the roots of $L$ so $K$ is actually the splitting field itself. Consider the fact that

Thus, $f(x)$ and $Df(x)$ are relatively prime so $f(x)$ is separable. This implies that $K$ has $p^{11}$ elements. Since $K$ is an $F$-vector space, and $F$ itself has $p$ elements, this implies that $[K:F] = 11$.

For the Galois group, we make note that since $K$ is the splitting field of the separable $f(x) \in F[x]$, it follows that $K$ is Galois. Thus,

Now we claim that the map $\sigma:K \to K$ defined by $\sigma(x) = x^p$ is an automorphism. It is clear by the Binomial theorem that $\sigma$ is a ring homomorphism. Furthermore, any ring homomorphism into a field is injective and $K$ is a finite set which implies that $\sigma$ is surjective. It is also clear that $\sigma^{11}$ is the identity map since

Notice that the order of $\sigma^{11}$ cannot be less than $11$ for if it was, then every element of $K$ would be the root of a polynomial that is degree less than $p^n$ which would contradict the separability of $f(x)$. [/proof]

[example] [extitle]KU Algebra Qual August 2016 Problem 2[/extitle]

Let $F = \ZZ/2\ZZ$.

1. Prove that $K = F[t]/(t^3 + t + 1)$ is a field of order $8$
2. Prove that $K$ is a splitting field for $f(x) = x^7 - 1$ over $F$. [/example]

[proof] (1) We know that $t^3 + t + 1$ is irreducible if and only if it has a root (because said polynomial is degree 3) in $F$. Testing both $t = 0$ and $t = 1$ shows that this polynomial has no roots in $F$ and so the polynomial is irreducible. By polynomial long division, we know that each distinct coset in $K$ is represented by

Thus, $K$ contains exactly $8$ elements.

(2) Note that $K^\times$ is a group of 7 elements so it is cyclic. Say that $\alpha$ is a generator of $K^\times$. Then we know that $(\alpha^k)^7 = 1$ by Lagrange and it follows immediately that every element of $K^\times$ is a root of $f(x)$. So $f(x)$ certainly splits over $K$, but also since $K$ is separable and none of the roots are $0$, it follows that $K$ is the smallest such field. [/proof]

[example] [extitle]KU Algebra Qual August 2021 Problem 6[/extitle]

Construct a field $F$ with nine elements with full justification. Then prove that the map $f:F \to F$ defined by $f(\alpha) = \alpha^3$ for all $\alpha \in F$ is an automorphism. [/example]

[proof] The polynomial $x^2 + 1$ is clearly irreducible over $\FF_3$. So setting

gives us a field $F$ over $\FF_3$. In the quotient, polynomial long division tells us that each distinct coset is represented by

So $F$ has $9$ elements.

Clearly $f$ preserves the unit and additive identity. The characteristic $3$-property of $F$ guarantees $f$ is a ring homomorphism. Since $F$ is a homomorphism from a field to a field, it is injective and injective maps on a finite set are surjective. Thus, $F$ is an automorphism and we are done. [/proof]

[example] [extitle]KU Algebra Qual August 2019 Problem 3[/extitle]

Consider a field $K$, and an irreducible polynomial $f(x) \in K[x]$. Prove that if $L$ is a field extension of $K$ such that $[L:K]$ is relatively prime to the degree of $f(x)$, then $f(x)$ remains irreducible in $L[x]$. [/example]

[proof] Let $p(x)$ be an irreducible factor of $f(x)$ in $L$ and let $\alpha$ be a root of $p(x)$. The consider the extensions

The tower law implies that

and

Now, since $f(x)$ is irreducible in $K$, it follows that $f(x)$ is the minimal polynomial (up to a scalar) of $\alpha$ over $F$. Similarly, since $p(x)$ is the minimal polynomial (up to a scalar) of $\alpha$ over $L$, the coprimality of $[L:K]$ and $\deg f = [K(\alpha):K]$ implies that $\deg f \mid \deg p = [L(\alpha):L]$. This means that $\deg f \le \deg p$. So then $f = p$ (up to a scalar) and so $f$ is irreducible. [/proof]

[example] [extitle]KU Algebra Qual January 2018 Problem 3[/extitle]

Let the complex number $\eps$ be a primitive $5^\text{th}$ root of unity. Set $F \coloneqq \QQ(\eps)$.

1. Find $[F:\QQ]$.
2. Determine (with proof) whether or not there exists a field $K$ such that $\QQ \subsetneq K \subsetneq F$. [/example]

[proof] (1) We claim that

Consider

Then $f(x+1)$ is irreducible by Eisenstein which implies that $f(x)$ is also irredicible. Thus, $f(x)$ is the minimal polynomial for $\eps$ over $\QQ$ which implies that $[F:\QQ] = 4$.

(2) Because $\QQ(\eps)$ is the splitting field for $f(x)$ (which is separable), $\QQ(\eps)/\QQ$ is Galois. Thus

So then $\Gal(F/\QQ)$ is a group of order $4$. Since groups of orders that are squares of a prime are abelian, the classification of finite abelian groups tells us that $\Gal(F/\QQ)$ is either $\ZZ/4\ZZ$ or $\ZZ/2\ZZ \times \ZZ/2\ZZ$. In the former either case, we can find a subgroup of order $2$ and the Fundamental Theorem of Galois Theorem tells us that such a subgroup would correspond to an intermediate field between $\QQ$ and $F$. [/proof]