Galois Theory HT25, Computing the Galois group

• Course - Galois Theory HT25^U
  • Notes - Galois Theory HT25, Galois extensions^U
  • Notes - Galois Theory HT25, Main theorems of Galois theory^U

This page serves as a collection of useful results and examples of actually calculating the Galois group of a field extension or polynomial and working through the Galois correspondence.

Tips and tricks

Why isn’t the Galois group just all the permutations of the roots?

Suppose $f \in \mathbb Q[t]$ is an irreducible polynomial with roots $\alpha _ 1, \ldots, \alpha _ n$. The Galois group of $f$ is then

\[G = \text{Gal}(\mathbb Q(\alpha_1, \ldots, \alpha_n) / \mathbb Q) := \{\mathbb Q\text{-linear automorphisms of }\mathbb Q(\alpha_1, \ldots, \alpha_n)\}\]

It’s clear that $G$ can be identified with a subgroup of $\text{Sym}(\alpha _ 1, \ldots, \alpha _ n)$, since if $\sigma \in G$ is a $\mathbb Q$-linear automorphism, then

\[f(\sigma(\alpha_i)) = \sigma(f(\alpha_i)) = 0\]

and so $\sigma(\alpha _ i)$ is also a root of $f$.

But why isn’t the Galois group always exactly $\text{Sym}(\alpha _ 1, \ldots, \alpha _ n)$? This is because there might be non-trivial relationships between the roots that have to be preserved by any $\mathbb Q$-linear automorphism of $\mathbb Q(\alpha _ 1, \ldots, \alpha _ n)$.

For example, consider

\[f(t) := t^3 - 3t + 1\]

You can show that the roots of $f$ are:

\[\begin{aligned} \alpha_1 &= 2\cos(2\pi/9) \\ \alpha_2 &= 2\cos(4\pi / 9) \\ \alpha_3 &= 2\cos(8\pi / 9) \end{aligned}\]

None of these roots are in $\mathbb Q$ (otherwise $f$ wouldn’t be irreducible). Trigonometric identities give non-trivial relationship between roots. For example, since $\cos(4\pi / 9) = 2\cos^2(2\pi / 9) - 1$, it follows that

\[\alpha_2 = \alpha_1^2 - 2\]

and by a similar identity

\[\alpha_3 = \alpha_2^2 - 2\]

The important part is that any element $\sigma \in G$ must preserve these identities:

\[\begin{aligned} \alpha_2 = \alpha_1^2 - 2 &\iff \sigma(\alpha_2) = \sigma(\alpha_1^2 - 2) \\\\ &\iff \sigma(\alpha_2) = \sigma(\alpha_1)^2 - 2 \end{aligned}\]

For example, the permutation $(13)$ cannot be in $G$, since permuting the second identity gives:

\[\alpha_1 \ne \alpha_2^2 - 2\]

It turns out that the only permutations which work are those in $A _ 3$, and so $G = A _ 3$, not $S _ 3$.

You can explore this directly: pick a polynomial, choose a permutation of its roots, and watch which algebraic relations between them survive. The permutations that preserve every relation are exactly the Galois group (green), and the rest are not (red).

The whole group can also be drawn at once, with one small plot per automorphism showing how it permutes the roots in the complex plane, so the grid is the group drawn out. Cyclic and dihedral groups give tidy rotations and reflections, while a generic polynomial has the full symmetric group and so produces $n!$ of them (this gallery follows Chris Grossack, “Making Pretty Pictures for Galois Theory” (2021)).

Extending the identity

One very useful result is ∆extending-the-identity:

If

• $\varphi : F \to \tilde F$ is an isomorphism between two fields
• $K / F$ and $\tilde F / \tilde K$ are finite extensions
• $\alpha \in K$, $\tilde \alpha \in \tilde K$
• $\varphi(m _ {F, \alpha})(\tilde \alpha) = 0$

Then there is a unique extension of $\varphi$ to

\[\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha)\]

such that $\varphi^\ast (\alpha) = \tilde \alpha$.

(this is from Notes - Galois Theory HT25, Bounds on the size of the Galois group^U).

This lets you build up the Galois group one element at a time. Suppose that you want to determine the Galois group of the field extension $\mathbb Q(\sqrt[3]{2}, \omega) / \mathbb Q$ where $\omega$ is a primitive 3rd root of unity.

Start by taking $F = \tilde F = \mathbb Q$ and $K = \tilde K = \mathbb Q(\sqrt[3]{2}, \omega) / \mathbb Q$ in the above. Then:

• $\text{id} _ \mathbb Q : \mathbb Q \to \mathbb Q$ is an isomorphism between two fields
• $\mathbb Q(\sqrt[3]{2}, \omega) / \mathbb Q$ is a finite extension

Elements of the Galois group are isomorphisms $\sigma : \mathbb Q(\sqrt[3]{2}, \omega) \to \mathbb Q(\sqrt[3]{2}, \omega)$ which fix $\mathbb Q$. Any extension of $\text{id} _ \mathbb Q$ fixes $\mathbb Q$, so the idea is to repeatedly apply the result to extensions of $\text{id} _ \mathbb Q$ to yield different automorphisms.

For example, take $\alpha = \omega$ and $\tilde \alpha = \omega^2$. Since $m _ {\mathbb Q, \omega}(x) = 1 + x + x^2$ and $m _ {\mathbb Q, \omega}(\omega^2) = 0$, it follows there is a map

\[\sigma_1 : \mathbb Q(\omega) \to \mathbb Q(\omega^2) \]

such that $\sigma _ 1(\omega) = \omega^2$. But since $\mathbb Q(\omega^2) = \mathbb Q(\omega)$, $\sigma _ 1$ is an automorphism $\mathbb Q(\omega) \to \mathbb Q(\omega)$.

But this is not yet an element of the Galois group since it’s only over $\mathbb Q(\omega)$. So we now apply the result taking $F = \tilde F = \mathbb Q(\omega)$ and $K = \tilde K = \mathbb Q(\sqrt[3]{2}, \omega)$, note that $K/F$ is still a finite extension. Now pick $\alpha = \sqrt[3]{2}$ and by the above our choices for $\tilde \alpha$ are $\sqrt[3]{2}, \omega \sqrt[3]{2}, \omega^2 \sqrt[3]{2}$ since $m _ {\mathbb Q(\omega), \sqrt[3]{2}} = x^3 - 2 = (x - \sqrt[3]{2})(x - \omega \sqrt[3]{2})(x - \omega^2 \sqrt[3]{2})$. Take $\tilde \alpha = \omega^2 \sqrt[3]{2}$ to obtain

\[\sigma_2 : \mathbb Q(\omega, \sqrt[3]{2}) \to \mathbb Q(\omega, \omega^2 \sqrt[3]{2})\]

where $\sigma _ 2 (\omega) = \omega^2$ and $\sigma _ 2 (\sqrt[3]{2}) = \omega^2 \sqrt[3]{2}$.

Continuing this way, you can enumerate all the automorphisms and with careful choices of $\alpha$ and $\tilde \alpha$ you can do this quite quickly. The tree below traces exactly this: each tower step adjoins a generator that may map to any root of its minimal polynomial over the field built so far, so the branching factor at each step is that step’s degree, the leaves enumerate the automorphisms, and the degrees multiply to give $[K : F] = \vert \text{Gal}(K/F) \vert $.

The Galois group is often a transitive subgroup of $S _ n$

One useful result is that if $K$ is the splitting field of a separable, irreducible degree $n$ polynomial, then $\text{Gal}(K/F)$ is isomorphic to a transitive subgroup of $S _ n$. This follows from considering the injection

\[\rho : G \to \text{Sym}(V(f))\]

It is transitive because for any pair of generators in $K$, you can find at least one automorphism sending one to the other (consider the Extending the identity above applied to the minimal polynomials).

If the extension is simple and you know a primitive element, i.e. $K = F(\alpha)$ for some $\alpha \in K$, then you in fact have:

\[\begin{aligned} |\text{Gal}(K/F)| &= [K(\alpha) : F] \\\\ &= \deg m_{F, \alpha} \end{aligned}\]

The transitive subgroups of $S _ n$ are as follows:

• Transitive subgroups of $S _ 2$:
  • $C _ 2$ (order 2)
• Transitive subgroups of $S _ 3$:
  • $S _ 3$ (order 6)
  • $C _ 3$ (order 3)
• Transitive subgroups of $S _ 4$:
  • $S _ 4$ (order 24)
  • $A _ 4$ (order 12)
  • $D _ 8$ (order 8)
  • $V _ 4$ (order 4)
  • $C _ 4$ (order 4)
• Transitive subgroups of $S _ 5$:
  • $S _ 5$ (order 120)
  • $A _ 5$ (order 60)
  • $F _ {20} = \mathbb Z/5\mathbb Z \rtimes (\mathbb Z/5\mathbb Z)^\times$ (order 20)
  • $D _ {10}$ (order 10)
  • $C _ 5$ (order 5)

Giving generators

Often it’s possible to determine the Galois group up to isomorphism without actually working out any of its elements (e.g. by considering the determinant of a cubic polynomial). For example, suppose that you have identified that a Galois group $G$ is isomorphic to $S _ n$ for some $n$ and now have to find the generators.

One straightforward observation is that you can find the automorphisms corresponding to some generators of the group you’ve identified. For example:

• If $G \cong S _ n$, then it suffices to find a transposition and an $n$-cycle in $G$.
• If $G \cong C _ n$, then it suffices to find an $n$-cycle in $G$.

Galois groups of prime-degree polynomials over $\mathbb Q$ with mostly real roots

Another useful result says that if some polynomial $f \in \mathbb Q$ has a lot of real roots, then it is the full symmetric group:

Suppose:

• $f$ is a degree $p$ polynomial where $p$ prime
• $f$ has exactly $p - 2$ real roots

Then:

• $\text{Gal} _ \mathbb Q(f) \cong S _ p$

Galois groups of reducible polynomials and composite field extensions

Consider the following polynomial

\[f(x) = x^5 - 2x^3 - 2x^2 + 4 = (x^3 - 2)(x^2 - 2) \in \mathbb Q[x]\]

Most examples of calculating the Galois group consider irreducible polynomials. This polynomial is reducible, so how can we calculate the Galois group?

The definition is still exactly the same: $\text{Gal} _ \mathbb Q(f) = \text{Gal}(K/\mathbb Q)$ where $K$ is a splitting field for $f$ over $\mathbb Q$. Here

\[K = \mathbb Q(\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2 \sqrt[3]{2}, -\sqrt{2}, \sqrt{2}) = \mathbb Q(\omega, \sqrt[3]{2}, \sqrt 2)\]

and you can show (e.g. by Extending the identity) that $\text{Gal}(K/\mathbb Q)$ is isomorphic to $S _ 3 \times C _ 2$.

Some things to note:

• $S _ 3 \times C _ 2$ is not a transitive subgroup of $S _ 5$. This is not too surprising because the theorem stating the Galois group of a degree $n$ polynomial is a transitive subgroup of $S _ n$ requires $f$ to be irreducible.
• $\text{Gal} _ \mathbb Q(x^3 - 2) \cong S _ 3$
• $\text{Gal} _ \mathbb Q(x^2 - 2) \cong C _ 2$

From this example, it might seem sensible to suggest the following:

(Wrong) The Galois group of a reducible polynomial $f = g _ 1 \ldots g _ k$ where each $g _ i$ is irreducible is isomorphic to $\bigotimes^k _ {i = 1} \text{Gal} _ F(g _ i)$.

But this is not true in general. For example, consider

\[f(x) = (x^3 - 2)(x^3 - 1)\]

Then a splitting field is $\mathbb Q(\sqrt[3]{2}, \omega)$ and it’s straightforward to verify that $\text{Gal}(\mathbb Q(\sqrt[3]{2}, \omega) / \mathbb Q) \cong S _ 3$. But $\text{Gal} _ \mathbb Q(x^3 - 2) \cong S _ 3$ and $\text{Gal} _ \mathbb Q(x^3 - 1) \cong C _ 3$ and so $\text{Gal} _ \mathbb Q(f) \not\cong \text{Gal} _ \mathbb Q(x^3 - 2) \times \text{Gal} _ \mathbb Q(x^3 - 1)$.

What went wrong this time compared to before? The problem is now that the splitting fields for each of the reducible factors now overlap as they both contain $\omega$, and so you can’t guarantee it’s possible to shuffle parts of one of the splitting fields around without shuffling the others around. However, if we add an additional condition on the splitting fields of each of the factors, we have the following result:

(Right) The Galois group of a reducible polynomial $f = g _ 1 \ldots g _ k$ where for each $i$ and $j$, $g _ i$ is irreducible and $K _ i \cap K _ j = F$ where $K _ i$ is the splitting field of $g _ i$ is isomorphic to $\bigotimes^k _ {i = 1} \text{Gal} _ F(g _ i)$.

Or, more generally we have the result:

Suppose:

• $K \supseteq L _ 1, L _ 2 \supseteq F$
• $L _ 1/F$ and $L _ 2/F$ are Galois extensions
• $L$ is the smallest subfield of $K$ containing $L _ 1$ and $L _ 2$ ($L = L _ 1 L _ 2$, their “compositum”)

Then:

• $L/F$ is also Galois
• There is an injective homomorphism $\phi : \text{Gal}(L/F) \to \text{Gal}(L _ 1/F) \times \text{Gal}(L _ 2 / F)$
• If $L _ 1 \cap L _ 2 = F$ (“linearly disjoint”), then this homomorphism is an isomorphism

Checking irreducibility

When calculating the Galois group of a polynomial, a lot of useful results depend on the polynomial being irreducible (e.g. that the Galois group is a transitive subgroup of $S _ n$). Here are some general techniques for checking that a polynomial is irreducible:

• Polynomials $f \in F[t]$ of degree three and less are irreducible iff they have no roots in the base field $F$
  • Notes - Galois Theory HT25, Rational root test^U can be used to check whether there are any rational roots.
• Eisenstein’s criterion:
  • Notes - Rings and Modules HT24, Eisenstein’s criterion^U
  • Applies more generally than just $\mathbb Z[x]$
  • Sometimes useful to make a substitution
• Reducing modulo some prime
• $f$ is irreducible iff $G$ acts transitively on the roots of $f$
  • Useful if you already know an element of the Galois group, e.g. the Frobenius endomorphism

When the extension isn’t Galois

Consider the finite extension $K/F$. It’s still possible to talk about $\text{Aut} _ F(K)$, which is still well-defined even when $K/F$ is not Galois. The important results in this context are:

Suppose:

• $K / F$ is a finite field extension
• $G = \text{Aut} _ K(F)$

Then:

\[ \vert G \vert \le [K : F]\]

So you know that the group is still finite.

Suppose

• $F(\alpha)/F$ is a simple algebraic extension
• $K/F$ is another extension

Then:

• The $F$-linear homomorphisms $\varphi : F(\alpha) \to K$ are in one-to-one correspondence with the roots of $m _ {F, \alpha}$ in $K$.

This result is useful in cases like where $F = \mathbb Q$, $K = \mathbb Q(\sqrt[4]{2})$ and $\alpha = \sqrt[4]{2}$, since it tells you that the automorphisms are just $\lbrace \text{id}, \tau \rbrace$ where $\tau$ maps $\sqrt[4]{2} \mapsto -\sqrt[4]{2}$.

Examples

Galois groups of field extensions

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$\text{Gal}(S/\mathbb Q) \cong D _ 4$ where $S$ splitting field for minimal polynomial of $\sqrt{2 + \sqrt{5}}$ over $\mathbb Q$

Let $\alpha := \sqrt{2 + \sqrt 5}$.

Compute the minimal polynomial $m$ over $\mathbb Q$: One likely candidate is just

\[\begin{aligned} m(x) &= (x^2 - 2)^2 - 5 \\\\ &= x^4 - 4x^2 - 1 \end{aligned}\]

and in fact this factorises as

\[m(x) = (x - \sqrt{2 + \sqrt 5})(x + \sqrt{2 + \sqrt 5})(x - \sqrt{2 - \sqrt 5})(x + \sqrt{2 - \sqrt 5})\]

Since $m$ is a degree 4 polynomial and none of the pairs of factors multiply to give a polynomial in $\mathbb Q$, then $m$ must be the minimal polynomial of $\alpha$.

Calculate the splitting field and verify it’s Galois: Name the roots of the splitting field as

\[\alpha, -\alpha, \beta, -\beta\]

where $\beta := \sqrt{2 - \sqrt 5}$. Then the splitting field is

\[S := \mathbb Q(\alpha, \beta)\]

Note also that $S/\mathbb Q$ is Galois, as $S$ is the splitting field of a separable polynomial.

Calculate the size of the Galois group: Let $G := \text{Gal}(S/\mathbb Q)$ denote the Galois group. Since $S/\mathbb Q$ is Galois, we know

\[|G| = [S : \mathbb Q]\]

To calculate the degree of the extension, we use the tower law:

\[\begin{aligned} \text{}[S : \mathbb Q] &= [\mathbb Q(\alpha, \beta) : \mathbb Q] \\\\ &= [\mathbb Q(\alpha, \beta) : \mathbb Q(\alpha)] \cdot [\mathbb Q(\alpha) : \mathbb Q] \\\\ &= \deg m_{\beta,\mathbb Q(\alpha)} \cdot \deg m_{\alpha, \mathbb Q} \\\\ \end{aligned}\]

Hence we need to calculate $m _ {\beta, \mathbb Q(\alpha)}$. To do this, we can find some relationships between the roots. Note

\[\alpha \beta = \sqrt{2 + \sqrt 5} \sqrt{2 - \sqrt 5} = \sqrt{-1} = i\]

and so $\beta = i/\alpha$. Then $\beta^2 = -\alpha^{-2}$, so one candidate is

\[m_{\beta, \mathbb Q(\alpha)}(x) = x^2 + \alpha^{-2}\]

and since this is monic and irreducible, it is indeed the minimal polynomial. Hence

\[[S : \mathbb Q] = 2 \cdot 4 = 8\]

Find the automorphisms: Clearly the identity $\text{id} _ S : S \to S$ is included in the Galois group.

To find the other isomorphisms, by Extending the identity we know that we can then construct an isomorphism $\tilde \sigma : \mathbb Q(\alpha) \to \mathbb Q(\alpha)$ that maps $\alpha \mapsto \alpha$. We can then extend this again to an automorphism $\sigma : \mathbb Q(\alpha, \beta) \to \mathbb Q(\alpha, -\beta)$ (and so really $\sigma : S \to S$) by mapping $\beta \mapsto -\beta$; this is okay since $\tilde \sigma(m _ {\mathbb Q(\alpha), \beta})(x) = (x + \beta)(x - \beta)$, so can choose $\beta$ or $-\beta$ for the image. Note that $\sigma^2 = \text{id} _ S$, so we can’t find any additional automorphisms by taking powers of this one.

We could also have extended $\text{id} _ \mathbb{Q}$ to $\tilde \tau : \mathbb Q(\alpha) \to \mathbb Q(\beta)$ by mapping $\alpha \mapsto \beta$. Again by Extending the identity, $\beta$ must get mapped to a root of $\tilde\tau(m _ {\mathbb Q(\alpha), \beta})$. Since

\[\tilde\tau(m_{\mathbb Q(\alpha), \beta}) = \tilde\tau(x^2 + \alpha^{-2}) = x^2 + \beta^{-2} = (x - \alpha)(x+\alpha)\]

we can pick to extend to $\tau : S \to S$ where $\tau(\beta) = -\alpha$ (thinking ahead slightly, we can pick $-\alpha$ rather than $\alpha$ since then it will have a larger order).

We could continue this way, by making other choices for how we extend the identity. But in order to get a more natural description of the automorphisms, we can just take powers and products of the two automorphisms we already have:

• $\text{id} _ S : \alpha \to \alpha, \beta \to \beta$
• $\sigma : \alpha \mapsto \alpha, \beta \mapsto -\beta$
• $\tau : \alpha \mapsto \beta, \beta \mapsto -\alpha$
• …then…
• $\tau^2 : \alpha \mapsto -\alpha, \beta \mapsto - \beta$
• $\tau^3 : \alpha \mapsto -\beta, \beta \mapsto \alpha$
• $\tau \sigma : \alpha \mapsto \beta, \beta \to \alpha$
• $\sigma \tau : \alpha \mapsto -\beta, \beta \mapsto -\alpha$
• $\tau^2 \sigma : \alpha \to -\alpha, \beta \mapsto \beta$

Hence the Galois group is generated by $\sigma, \tau$ (and is in fact $D _ 4$).

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$\text{Gal}(\mathbb Q(\sqrt{d + \sqrt{b} }) / \mathbb Q) \cong C _ 4$ where $b$ is not square in $\mathbb Q$, $d$ not square in $\mathbb Q(\sqrt b)$ and $d^2 - b = bc^2$ for some integers $b,c,d$

Let $\alpha = \sqrt{d + \sqrt{b}}$ and $\beta = \sqrt{d - \sqrt b}$.

Verify the extension is Galois: To do this, we need to verify that $\mathbb Q(\alpha)$ is the splitting field of a separable polynomial. We can guess that such a polynomial is given by

\[f(x) := (x^2 - d)^2 - b = (x - \alpha)(x + \alpha)(x - \beta)(x + \beta)\]

$\alpha$ and $-\alpha$ are clearly in $\mathbb Q(\alpha)$. To verify $\beta$ and $-\beta$ in $\mathbb Q(\alpha)$, note that

\[\begin{aligned} \beta &= \sqrt{b - \sqrt d} \\\\ &= \frac{\sqrt{b - \sqrt d}\sqrt{b + \sqrt d}}{\sqrt{b + \sqrt d}} \\\\ &= \frac{\sqrt{b^2 - d}}{\sqrt{b + \sqrt d}} \\\\ &= \frac{c\sqrt b}{\alpha} \\\\ &= c(\alpha^2 - d) \cdot \frac{1}{\alpha} \\\\ &\in \mathbb Q(\alpha) \end{aligned}\]

Hence $\mathbb Q(\alpha)$ is the splitting field of a separable polynomial (all polynomials are separable in characteristic zero).

Calculate the size of the Galois group: Let $G = \text{Gal}(\mathbb Q(\alpha) / \mathbb Q)$. As the extension is Galois, $ \vert G \vert = [\mathbb Q(\alpha) : \mathbb Q]$. If you confirm that $f$ is the minimal polynomial e.g. by showing it is irreducible, then directly:

\[[\mathbb Q(\alpha) : \mathbb Q] = \deg m_{\alpha, \mathbb Q} = \deg f = 4\]

Alternatively, note that

\[\begin{aligned} \text{}[\mathbb Q(\alpha): \mathbb Q] &= [\mathbb Q(\alpha) : \mathbb Q(\sqrt b)] \cdot [\mathbb Q(\sqrt b) : \mathbb Q] \\\\ &= \deg m_{\alpha, \mathbb Q(\alpha)} \cdot \deg m_{\sqrt b, \mathbb Q} \\\\ &= 2 \cdot 2 \\\\ &= 4 \end{aligned}\]

(This also justifies that $f$ is indeed the minimal polynomial of $\alpha$).

Find the automorphisms: $G$ is either $C _ 4$ or $C _ 2 \times C _ 2$. To determine which, we can find the automorphisms. Since $G$ is a transitive subgroup of $S _ 4$, there must be at least one automorphism $\sigma : \mathbb Q(\alpha) \to \mathbb Q(\alpha)$ sending $\alpha$ to $\beta$, i.e. $\sigma(\sqrt{d + \sqrt{b}}) = \sqrt{d - \sqrt{b}}$. We can verify by direct calculation that $\sigma^2 \ne \text{id}$ as follows:

\[\begin{aligned} \sigma(\sqrt{d - \sqrt{b}}) &= \sigma\left( \sqrt{\frac{(d - \sqrt b)(d + \sqrt b)}{d + \sqrt b} } \right) \\\\ &= \sigma\left( \sqrt{\frac{d^2 - b}{d + \sqrt b} } \right) \\\\ &= \sigma\left( \frac{c\sqrt b}{\sqrt{d + \sqrt b} } \right) \\\\ &= \frac{-c\sqrt b}{\sqrt{d - \sqrt b} } \\\\ &= \frac{-c\sqrt b}{\frac{c\sqrt b}{\sqrt{d + \sqrt b}} } \\\\ &= -\sqrt{d + \sqrt b} \end{aligned}\]

Thus $G = \langle \sigma \rangle$ and $G \cong C _ 4$.

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$\text{Gal}(\mathbb F _ {p^n} / \mathbb F _ p) \cong C _ n$

This is also covered in Notes - Galois Theory HT25, Finite fields^U.

Verify the extension is Galois: $\mathbb F _ {p^n}$ is a splitting field of $x^{p^n} - x$, since by Fermat’s little theorem, $a^{p^n - 1} = 1$ for all $a \in \mathbb F _ {p^n}$ and so it splits as $\prod _ {a \in \mathbb F _ {p^n} } (x - a)$. It is also separable as all the roots are distinct.

Calculate the size of the Galois group: Let $G = \text{Gal}(\mathbb F _ {p^n} / \mathbb F _ p)$. Then as the extension is Galois,

\[\begin{aligned} |G| &= [\mathbb F_{p^n} : \mathbb F_p] \\\\ &= n \end{aligned}\]

so $G$ is a group of order $n$.

Find the automorphisms: Consider the Frobenius automorphism $\phi : \mathbb F _ {p^n} \to \mathbb F _ {p^n}$ given by $x \mapsto x^p$. This is indeed an automorphism in the Galois group as:

• It is additive: $\phi(x + y) = \phi(x) + \phi(y)$ by “Freshman’s dream”.
• It is $\mathbb F _ p$-linear: $\phi(a x) = a^p x^p = a x^p = a \phi(x)$ by Fermat’s little theorem.
• It is injective: if $\phi(x) = \phi(y)$, then $\phi(x - y) = 0$ and as $\mathbb F _ {p^n}$ is an integral domain, $(x - y)^p = 0$ implies $x - y = 0$ and hence $x = y$.
• It is surjective: it is a map between finite sets of equal size, so injectivity implies surjectivity.

It also has order exactly $n$ in $\text{Gal}(\mathbb F _ {p^n} / \mathbb F)$: since $a^{p^n} = a$ for all $a \in \mathbb F _ {p^n}$ by Fermat’s Little Theorem, it follows that $\phi^n = 1$. Now suppose that $\phi^m = 1$ for some $1 \le m \le n$. Then $a^{p^m} = a$ for all $a \in \mathbb F _ {p^n}$. But then $t^{p^m} - t$ has $ \vert \mathbb F _ {p^n} \vert $ distinct roots in $\mathbb F _ {p^n}$, which implies that $p^m \ge p^n$. Hence $m \ge n$, and the order of $\phi$ is precisely $n$.

Hence $\phi$ generates all of $G$, so $G = \langle \phi \rangle$ and therefore $G \cong C _ n$.

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$\text{Gal}(K/F) \hookrightarrow (\mathbb Z / n \mathbb Z)^\times$ where $K=F(\zeta _ n)$ is a cyclotomic extension and $n, \text{char}(F)$ coprime

Notes - Galois Theory HT25, Cyclotomic extensions^U covers the case where $[K : F] = p$ for some prime, and these lecture notes cover the general case in Section 5.1. The condition for this injection to also be a surjection is that $\Phi _ {n, F}$ is irreducible (since then the degrees match), which in particular is always true when $K = \mathbb Q$.

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$\text{Gal}(K/F) \hookrightarrow \mathbb Z / n\mathbb Z$ where $K = F(\alpha)$, $\alpha^n \in F$ is a Kummer extension and $n, \text{char}(F)$ coprime

Covered in Notes - Galois Theory HT25, Kummer extensions^U. There is also a partial converse in this case; namely that if $K/F$ is a Galois extension, $K$ contains an $n$-root of unity and $\text{Gal}(K/F)$ is a cyclic group of order $n$, then $K = F(\alpha)$ for some $\alpha$ where $\alpha^n \in F$.

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Both of these power-map families are interactive below. For a cyclotomic extension $\mathbb Q(\zeta _ n)/\mathbb Q$ the automorphisms are $\zeta \mapsto \zeta^k$ with $k$ coprime to $n$, giving $(\mathbb Z/n\mathbb Z)^\times$, and for a finite field the Frobenius $x \mapsto x^p$ cycles the roots of an irreducible polynomial.

Galois groups of polynomials

$\text{Gal} _ F(f) \cong \mathbb Z / p \mathbb Z$ where $f(x) = x^p - x + a$, $\text{char}(F) = p > 0$ and $a > 0$

Calculate the splitting field and verify it’s Galois: Recall that the Galois group of a polynomial $f$ is the Galois group of the splitting field $S$ of $f$.

Let $\alpha \in S$ be any root of $f$. Then note that $F(\alpha) = S$, since if $\alpha$ is a root of $f$, then so is $\alpha + k$ for $0 \le k \le p - 1$, since

\[\begin{aligned} f(\alpha + k) &= (\alpha + k)^p - (\alpha + k) + a \\\\ &= \alpha^p + k^p - \alpha - k + a \\\\ &= -a + k - k + a \\\\ &= 0 \end{aligned}\]

$f$ is separable as it has $p$ distinct roots in the splitting field (alternatively, since $f$ is irreducible, it’s separable as the derivative is non-zero). Then $S/F$ is Galois since $S$ is the splitting field of a separable polynomial.

Verify $f$ is irreducible: Let $G := \text{Gal}(S/F)$. $G$ contains the Frobenius endomorphism as for any root $\alpha$,

\[\begin{aligned} \alpha^p - \alpha + a = 0 &\implies \alpha^p = \alpha + a \\\\ &\implies \alpha^{p^k} = \alpha + ka \end{aligned}\]

and each $ka$ is distinct and so it runs over all the roots of $f$. Since any irreducible factor of $f$ must be closed under the action of the Frobenius endomorphism, and this already has orbit of size $p$, it follows that $f$ is irreducible (is this correct?).

Calculate the size of the Galois group: As $f$ is irreducible over $F$, $S = F(\alpha)$, and $S/F$ is Galois, it follows that

\[|G| = [S : F] = \deg f = p\]

Find the automorphisms: Since $ \vert G \vert = p$, we know that it’s going to be cyclic (and in fact isomorphic to $\mathbb Z / p \mathbb Z$). Hence consider the map $\sigma$ which maps $\alpha \mapsto \alpha + 1$ (this is definitely in the Galois group since we can always find at least one map that sends one root to another). Then this gives $p$ automorphisms $\text{id} _ S = \sigma^0, \sigma^1, \ldots, \sigma^{p-1}$.

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$\text{Gal} _ K(x^{25} - x - t) \cong C _ 5 \times C _ 5$ where $K = \mathbb F _ {5^2}(t)$

Let $f(x) = x^{25} - x - t \in K[x]$.

Calculate the splitting field: Recall that the Galois group of a polynomial $f$ is the Galois group of the splitting field $S$ of $f$.

Let $\alpha \in S$ be any root of $f$. Then note $K(\alpha) = S$, since if $\alpha$ is a root of $f$, then so is $\alpha + k$ for all $k \in \mathbb F _ {25}$, since

\[\begin{aligned} f(\alpha + k) &= (\alpha + k)^p - (\alpha + k) - t \\\\ &= \alpha^p + k^p - \alpha - k - t \\\\ &= t + k - k -t \\\\ &= 0 \end{aligned}\]

$f$ is separable as it is irreducible (consider Eisenstein’s criterion with $p = 3$) and has non-zero derivative. So $S = K(\alpha)$ and $S/K$ is Galois since $S$ is the splitting field of a separable polynomial.

Calculate the size of the Galois group: Let $G := \text{Gal}(S/K)$. Since the extension is Galois and $f$ is irreducible,

\[|G| = [S : K] = \deg f = 25\]

So $G$ has order 25. Thus either $G \cong C _ {25}$ or $G \cong C _ 5 \times C _ 5$.

Find the automorphisms: Since $f$ is irreducible, the Galois group must act transitively on the roots of $f$. Let $\\{1, u\\}$ be a basis for $\mathbb F _ {25}$ over $\mathbb F _ 5$. Then we have at least the two automorphisms:

\[\begin{aligned} \sigma &: \alpha \mapsto \alpha + 1 \\\\ \tau &: \alpha \mapsto \alpha + u \end{aligned}\]

Since both of these have order $5$ and $ \vert \langle \sigma, \tau \rangle \vert = 25$, it follows $G \cong C _ 5 \times C _ 5$.

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$\text{Gal} _ F(x^{124} - t) \cong C _ {124}$ where $F = \mathbb F _ {125}(t)$

Let $G := \text{Gal} _ F(x^{124} - t)$ and $f(x) := x^{124} - t \in F[x]$.

Find the splitting field: Note $f$ is irreducible and consider $E := \frac{K[x]}{\langle f \rangle}$. Then $E$ contains at least one root of $f$, say $\alpha := \langle f \rangle$. Note also that if $u$ is a unit in $F$, then $u \alpha$ is also a root. Since the units of $F$ are the elements of $\mathbb F _ {125}^\times$ and $\mathbb F _ {125}^\times \in F$, it follows all roots of $f$ are contained in $E$ and so it is a splitting field.

Calculate the size of the Galois group: Since the extension is Galois and $f$ is irreducible,

\[|G| = \deg f = 124\]

Find the automorphisms: Let $v$ be a generator of $\mathbb F^\times _ {125}$ ($\mathbb F _ {125}^\times$ is cyclic). Since $E = K(\alpha) = K(v\alpha)$ and all splitting extensions are isomorphic, $\sigma : \alpha \to v\alpha$ is an $F$-linear automorphism of $E$. Since $v$ is a generator of $\mathbb F _ {125}^\times$, $\sigma^i$ for $0 \le i \le 124$ are distinct, and hence $G = \langle \sigma \rangle$.

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$\text{Gal} _ {F}(x^{13} - x + 2) \cong C _ {13}$ where $F = \mathbb F _ {13}$

Find the splitting field: Pick some root $\alpha$ of $x^{13} - x + 2$. By properties of finite fields, we must have that

\[f(\alpha^{13}) = f(\alpha)^{13} = 0\]

and more generally $f(\alpha^{13k}) = 0$ also. It remains to check that each $\alpha^{13k}$ is distinct for $0, 1, \ldots, 12$. As $f()

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Galois group of a cubic polynomial

Covered in Notes - Galois Theory HT25, Cubic equations^U. The useful result is the following:

If:

• $F$ is a field
• $\text{char}(F) \ne 2, 3$
• $f := t^3 + pt + q \in F[t]$
• $f$ is irreducible
• $K$ is a splitting field for $f$
• $\Delta$ is the discriminant of $f$ (the square of the product of the differences of roots)
• $G = \text{Gal}(K/F)$

Then:

• If $\Delta$ is a square in $F$, then $G = A _ 3$
• If $\Delta$ is not a square in $F$, then $G = S _ 3$

This gives a complete characterisation of the Galois group in terms of expressions in the roots. Here $x^2 - \Delta$ is the resolvent quadratic of the cubic.

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Galois group of a quartic polynomial

Covered in Notes - Galois Theory HT25, Quartic equations^U.

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Galois group of a quintic polynomial

Covered in Notes - Galois Theory HT25, Quintic equations^U.

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The full Galois correspondence

By examples of the full Galois correspondence, I mean:

• Checking a field extension is Galois
• Determining the Galois group
• Pairing up subfields and subgroups

The two worked examples below are the presets in this visualisation. Click through the subgroup and field lattices to see the order-reversing bijection, with normal subgroups and their Galois fixed fields picked out.

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

Let $K = \mathbb Q(\sqrt 2, \sqrt 3)$.

Check it’s Galois

We show that $K$ is the splitting field of a separable polynomial. Let

\[\begin{aligned} f(x) &= x^4 - 10x^2 + 1 \\\\ &= (x - (\sqrt 2 + \sqrt 3))(x + (\sqrt 2 + \sqrt 3))(x - (\sqrt 2 - \sqrt 3))(x + (\sqrt 2 - \sqrt 3)) \end{aligned}\]

This polynomial is irreducible and separable.

Determine the Galois group

Let $G = \text{Gal}(K/\mathbb Q)$. Then

\[\begin{aligned} |G| &= [\mathbb Q(\sqrt 2, \sqrt 3) : \mathbb Q] \\\\ &= [\mathbb Q(\sqrt 2, \sqrt 3) : \mathbb Q(\sqrt 2)][\mathbb Q(\sqrt 2) : \mathbb Q] \\\\ &= \deg (x^2 - 3) \cdot \deg (x^2 - 2) \\\\ &= 4 \end{aligned}\]

(where the third equality is justified by the fact $x^2 - 3$ is irreducible over $\mathbb Q(\sqrt 2)$). Hence $G \cong \mathbb Z / 4\mathbb Z$ or $G \cong \mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z$.

$G$ must contain $\sigma$ and $\tau$ where:

• $\sigma : \sqrt 2 \mapsto -\sqrt 2, \sqrt 3 \mapsto \sqrt 3$
• $\tau : \sqrt 2 \mapsto \sqrt 2, \sqrt 3 \mapsto -\sqrt 3$

(this can be seen by Extending the identity), and hence it contains two subgroups of size 2. Thus $G \cong \mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z$.

Pair up subfields and subgroups

The Galois correspondence says that subfields of $K/\mathbb Q$ are in bijective correspondence with subgroups of $\mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z$. Then $K^{\langle \sigma \rangle} = \mathbb Q(\sqrt 3)$ and $K^{\langle \tau \rangle} = \mathbb Q(\sqrt 2)$ (consider that $1, \sqrt 2, \sqrt 3, \sqrt 6$ is a basis for $K/\mathbb Q$).

\[\begin{aligned} \mathbb Q(\sqrt 3) &\longleftrightarrow \langle (1, 0) \rangle \\\\ \mathbb Q(\sqrt 2) &\longleftrightarrow \langle (0, 1) \rangle \end{aligned}\]

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$\text{Gal}(\mathbb Q(\sqrt[3]{2}, e^{2\pi i / 3}) / \mathbb Q) \cong S _ 3$

See also:

• “FIT4.3. Galois Correspondence 1 - Examples” on YouTube
• “302.S9B: The Galois Correspondence” on YouTube

Let $\omega = e^{2\pi i / 3}$ and let $K = \mathbb Q(\sqrt[3]{2}, \omega)$.

Check it’s Galois

We show that $K$ is the splitting field of a separable polynomial. Let

\[f(x) = x^3 - 2 = (x - \sqrt[3]{2})(x - \omega\sqrt[3]{2})(x - \omega^2\sqrt[3]{2})\]

Hence the roots of $f$ are $\{\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}\}$. These roots are clearly contained in $K$, so the splitting field for $p$ is contained in $K$.

But also $K$ is contained in the splitting field, since $\sqrt[3]{2} \in K$ and $\omega = \frac{\omega \sqrt[3]{2}}{\sqrt[3]{2}} \in K$.

So $K$ is the splitting field of $f$.

Determine Galois group

Let $G = \text{Gal}(K/\mathbb Q)$. We know that

\[\begin{aligned} |G| &= [K : \mathbb Q] \\\\ &= [\mathbb Q(\sqrt[3]{2}, \omega): \mathbb Q(\omega)] \cdot [\mathbb Q(\omega) : \mathbb Q] \\\\ &= \deg (x^3 - 2) \cdot \deg(1 + x + x^2) \\\\ &= 6 \end{aligned}\]

So either $G \cong S _ 3$ or $G \cong \mathbb Z _ 6$.

Since there is already a natural injection $G \hookrightarrow S _ 3$ and the cardinalities of $G$ and $S _ 3$ match, it follows that this is an isomorphism and $G \cong S _ 3$.

Pair up subfields and subgroups

To do this, we need a more explicit description of the elements of the Galois group. Note that $S _ 3$ is generated by a transposition and a 3-cycle, i.e.

\[S_3 = \langle (123), (12)\rangle\]

Let $\sigma$ denote the automorphism such that

\[\begin{aligned} \sigma(\sqrt[3]{2}) &= \omega \sqrt[3]{2} \\\\ \sigma(\omega) &= \omega \end{aligned}\]

and $\tau$ the automorphism defined by

\[\begin{aligned} \tau(\sqrt[3]{2}) &= \sqrt[3]{2} \\\\ \tau(\omega) &= \omega^2 \end{aligned}\]

Then $\text{Gal}(K/\mathbb Q) = \langle \tau, \sigma \rangle$ and the correspondence is:

\[\begin{aligned} \mathbb Q(\sqrt[3]{2}) &\longleftrightarrow \langle \tau \rangle \\\\ \mathbb Q(\omega) &\longleftrightarrow \langle \sigma\rangle \\\\ \mathbb Q(\omega\sqrt[3]{2}) &\longleftrightarrow \langle \tau \sigma \rangle \\\\ \mathbb Q(\omega^2\sqrt[3]{2}) &\longleftrightarrow \langle \sigma \tau \rangle \end{aligned}\]

To verify this, you’d write down a basis of $K$ as a $\mathbb Q$-vector space ($1, \sqrt[3]{2}, \omega, \omega \sqrt[3]{2}, \omega^2 \sqrt[3]{2}, \omega^2$) and determine which ones were fixed by each subgroup.

Other examples

• Lecture notes
  • (Old) Example 4.16: $\text{Gal}(K/\mathbb Q)$ where $K$ is splitting field for $x^3 - 2$
  • (Old) Lemma 5.5: $\text{Gal}(K/\mathbb Q)$ where $K$ is splitting field for $f(t) = t^4 - a$ and $a$ is square-free
• Abstract Algebra, Judson^N
  • Example: $\text{Gal}(\mathbb Q[\sqrt 3, \sqrt 5] / \mathbb Q) \cong \mathbb Z _ 2 \times \mathbb Z _ 2$
  • Example: $f(x) = x^5 - 6x^3 - 27x - 3 \in \mathbb Q[x]$ has Galois group $S _ 5$. But $S _ 5$ is not solvable, so this polynomial is not solvable by radicals.
  • Exercises:
    • $\text{Gal}(\mathbb Q(\sqrt{30})/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt 6, i)/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt[4]{5})/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt{2}, \sqrt[3]{2}, i)/\mathbb Q)$
• Sheet 2
  • Q4a $\text{Gal}(\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q)$
  • Q4b $\text{Gal}(\mathbb Q(\sqrt[3]{2}, e^{2\pi i /3})/\mathbb Q)$
  • Q4c $\text{Gal}(\mathbb Q(e^{2\pi i / 7})/\mathbb Q)$
    • Proposition 16.10.2 of Artin’s Algebra
    • https://www.youtube.com/watch?v=SAV_6ZZCwGY
• Sheet 4
  • Q2a Galois group of $x^5 - 2x^3 - 2x^2 + 4$
  • Q2b Galois group of $x^5 - 2$
  • Q2c Galois group of $x^5 - 4x + 2$
• Other resources:
  • A Worked out Galois Group for the Classroom: Let $f(x) = x^6 - 3x^2 - 1 \in \mathbb Q[x]$ and let $L _ f$ be the splitting field of $f$ over $\mathbb Q$. Then $\text{Gal}(L _ f/\mathbb Q) \cong A _ 4$.
  • Determine the minimal polynomial of $\sqrt{2 + \sqrt 2}$ over $\mathbb Q$ and find its Galois group over $\mathbb Q$: Finding the Galois group for the minimal polynomial $\sqrt{2 + \sqrt{2}}$
  • How to find the Galois group of a polynomial?
  • Galois group of $x^5 - 2$: $F _ {20}$
  • Galois group of the splitting field of the polynomial $x^5 - 2$ over $\mathbb Q$
  • Galois group of $x^p - 2$ where $p$ is an odd prime
  • Galois group of a cubic polynomial
    • Example 1: $x^3 - 6x^2 + 11x - 6$ has trivial Galois group “$\text{Gal}(\mathbb Q / \mathbb Q)$” since it splits in $\mathbb Q$
    • Example 2: $x^3 - x^2 + x - 1$ has splitting field $\mathbb Q[i]$ and so has Galois group $S _ 2$
    • Example 3: $x^3 - 2$ has discriminant $-108$ and so has Galois group $S _ 3$
    • Example 4: $x^3 - 4x + 2$ has discriminant $202$ and so has Galois group $S _ 3$
    • Example 5: $x^3 - 3x + 1$ has discriminant $81$ and so the Galois group is $A _ 3$
  • Complete characterisation of Galois groups of biquadratic?