Notes - Galois Theory HT25, Computing the Galois group

• [[Course - Galois Theory HT25]]^U
  • [[Notes - Galois Theory HT25, Galois groups and Galois extensions]]^U
  • [[Notes - Galois Theory HT25, Main theorems of Galois theory]]^U
• 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
  • Proposition 6.5: Prime degree irreducible polynomials with $p-2$ real roots has Galois group isomorphic to $S _ p$.
• [[Abstract Algebra, Judson]]^N
  • Example: $\text{Gal}(\mathbb Q[\sqrt 3, \sqrt 5] / \mathbb Q) \cong \mathbb Z _ 2 \times \mathbb Z _ 2$
  • Example: $\text{Gal}(\mathbb Q[\omega] : \mathbb Q) = \mathbb Z _ 4$ where $\omega$ is a root of $f(x) = x^4 + x^3 + x^2 + x + 1$.
  • 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:
    • $\text{Gal}(\mathbb Q(\sqrt{30})/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt 6, i)/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt[4]{5})/\mathbb Q)$
    • $\text{Gal}(\mathbb Q(\sqrt{2}, \sqrt[3]{2}, i)/\mathbb Q)$
• [[Algebra, Artin]]^N
  • Corollary 16.12.6: Let $f(x)$ be an irreducible polynomial of degree $5$ over $\mathbb Q$. If $f$ has exactly three real roots, its Galois group $G$ is the symmetric group $S _ 5$, and hence its roots are not solvable.
• Sheet 2
  • Q4a $\text{Gal}(\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q)$
  • Q4b $\text{Gal}(\mathbb Q(\sqrt[3]{2}, e^{2\pi i /3})/\mathbb Q)$
    • https://www.youtube.com/watch?v=SAV_6ZZCwGY
    • https://www.youtube.com/watch?v=P8__4fzMmsM&t=995s
  • 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 3
  • Q4 Galois group of $\text{Gal}(\mathbb Q(\alpha) / \mathbb Q)$ where $\alpha \in \mathbb C$ is a root of $f := t^4 - 4t^2 + 2$ is cyclic of order 4.
• 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 $A _ 3$
  • Galois group of reducible polynomial: What is the Galois group of $(x^3 - x + 1)(x^2 + 1)$? $S _ 3 \times C _ 2$
  • Example where the Galois group of a reducible polynomial is not just the Galois group of its irreducible factors?
  • Complete characterisation of Galois groups of biquadratic?

Why isn’t the Galois group just 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 $S _ 3$.