Some useful exercises and solutions: https://feog.github.io/chap4.pdf
-
Given \(x\coloneqq\sqrt{a} + \sqrt{b}\), to find a minimal polynomial consider \(x^2, x^3,\cdots\) and try to get a linear combination. Then check if its irreducible.
- General strategy here: try to isolate radicals on one side, then raise both sides to that power.
-
To find a minimal polynomial for an element \(\alpha\), figure out the dimension of \({\mathbf{Q}}(\alpha)/{\mathbf{Q}}\) – say it’s \(n\), then \(1, \alpha, \cdots, \alpha^n\) must be a \({\mathbf{Q}}{\hbox{-}}\)linearly dependent set, so you compute these powers and fiddle with \({\mathbf{Q}}\) coefficients (or invert a matrix).
-
Useful trick: for \(x \coloneqq\sqrt{a} + \sqrt{b}\), compute \(x,x^2, x^3, x^4\) and write them in terms of the basis \(\left\{{1, \sqrt{a}, \sqrt{b}, \sqrt{ab}}\right\}\). Then put this linear system into a matrix and invert: \begin{align*} A\mathbf{v} = \mathbf{c} \coloneqq A {\left[ {1, \sqrt a, \sqrt b, \sqrt{ab} } \right]} = {\left[ {x, x^2, x^3, x^4} \right]} .\end{align*} Once you get \(A^{-1}\mathbf{x} = \mathbf{b}\), read off the first row dotted against \(\mathbf{b}\) to get a polynomial in \(x\).
-
In general: take \(\alpha\), sort out the degree \(n\) of the extension \({\mathbf{Q}}(\alpha)/{\mathbf{Q}}\), and use the basis \(1,\alpha,\alpha^2,\cdots, \alpha^{n-1}\).
-
A trick to remember how degrees, indices and sizes match up: \(L/K/F\) corresponds to \(1/H/G\), and \([L:K] = [H:1] = {\sharp}H\), \([F:K] = [G:H]\), \([L:F] = [G:1] = {\sharp}G\), etc.
-
Trick: once you find \(\operatorname{SF}(f)/{\mathbf{Q}}\), if any subextension is not normal over \({\mathbf{Q}}\), then \(G\) can not be abelian.
- Example: \(f(x) = x^3-2\) splits in \({\mathbf{Q}}(\zeta_3, 2^{1\over 3})\) which has a non-normal subextension \({\mathbf{Q}}(2^{1\over 3})\), forcing \(G= S_3\).
-
If \(\alpha\beta \in {\mathbf{Q}}\), then \(\alpha \in {\mathbf{Q}}(\beta)\) and vice-versa (I think).
-
Checking subgroup lattices: https://hobbes.la.asu.edu/groups/groups.html
-
De-nesting radicals:
Assume all extensions here are algebraic and finite. Let \(f\in {\mathbf{Q}}[x]\) with \(n \coloneqq\deg f\).
-
Show your extension is Galois (normal and separable)
- Show \(f\) is irreducible and separable.
-
Find the degree of the extension \(d\), since then \({\sharp}G = d\).
- Note that in general, \(G\leq S_n\) and \(n\neq d\), \({\sharp}G\neq n\).
- Obtain \(n\divides d \coloneqq{\sharp}G \divides n!\) and \(G\leq S_n\) is a transitive subgroup, list possibilities.
- Rule out cases or determine the group completely by finding cycle types.
Consider \(f(x) \coloneqq x^5-9x+3\), let \(L\coloneqq\operatorname{SF}(f)/{\mathbf{Q}}\).
- \(f\) is irreducible: Apply Eisenstein with \(p=3\).
-
\(f\) is separable:
- \({\mathbf{Q}}\) is perfect, so irreducible implies separable.
-
\(L\) is Galois:
- \(L/{\mathbf{Q}}\) is a finite extension over a perfect field and thus automatically separable.
- \(L/{\mathbf{Q}}\) is the splitting field of a separable polynomial, and thus normal.
-
Since \(L\) is Galois, \({\sharp}G = d \coloneqq[L: {\mathbf{Q}}]\), so try to compute the degree by computing the splitting field (and its degree) explicitly.
- Here: difficult! The roots are complicated.
- Since \(L\) is Galois, \(G\leq S_5\) is a transitive subgroup. Possibilities: \begin{align*} S_5, A_5 , F_5\cong C_5\rtimes C_4, D_5, C_5 .\end{align*}
-
Claim: \(G = S_5\).
-
Reduce mod 2: \((x^2 + x + 1) (x^3 + x^2 + 1)\), yielding a cycle type \((2, 3)\). This rules out
- \(C_5, D_5\) since \(3\notdivides 5, 10\).
- \(A_5\), since this is an odd number of even length cycles.
- \(F_5\) since \(3\notdivides 20\).
- So this only leaves \(S_5\).
-
Reduce mod 2: \((x^2 + x + 1) (x^3 + x^2 + 1)\), yielding a cycle type \((2, 3)\). This rules out
Showing Extensions are Galois
Showing your polynomial is irreducible:
- Eisenstein (including shifting/inverting tricks, see section below)
- To show \(f\) is irreducible, it suffices to show it is irreducible over any \({ \mathbf{F} }_p[x]\).
- A quadratic with no real roots is irreducible.
Showing your polynomial is separable:
- Show directly that \(f\) has distinct roots in \(\overline{k}\) by factoring it.
- For perfect fields, irreducibles are automatically separable.
- For \(f\) irreducible, \(f\) is separable iff \(f'(x) \not\equiv 0\).
Showing your extension is separable:
- Splitting fields of separable polynomials are automatically separable and normal (and thus Galois).
-
Algebraic extensions of a perfect field are automatically separable.
- In particular, extensions over \({\mathbf{Q}}\) or any \(\operatorname{ch}k = 0\) are separable, and one only needs to show normality.
- (Hard) Show \([L:k]_s = [L:k]\).
- (Hard) Use that separability is a distinguished class.
Showing your extension is normal:
- Show that \(L/k\) is finite and the splitting field of some separable polynomial.
Showing your extension \(K/k\) is Galois:
- Show normality and separability.
- Show \(K\) is the splitting field of a separable polynomial (“separable splitting field”)
- Automatic when \(K/k\) is algebraic and a finite field, since it’s the splitting field of \(x^{p^n}-x\).
Irreducibility
If \(f\in {\mathbf{Z}}[x]\) is monic and there exists any prime \(p\) such that \(f\operatorname{mod}p\) is irreducible in \({ \mathbf{F} }_p[x]\), then \(f\) irreducible in \({\mathbf{Q}}[x]\).
Finding a good prime for this is hard, but irreducibility can be checked exhaustively in small fields: just enumerate all polynomials and try polynomial long division.
\(f(x) \coloneqq x^4 + x + 1\) is irreducible in \({\mathbf{Z}}[x]\), since checking manually in \({ \mathbf{F} }_2[x]\) shows that \(0, 1\) are not roots \(\operatorname{mod}2\) so there is not linear factor. Manually dividing \(a_1 x^2 + a_2 x + a_3\) for \(a_i\in 0, 1\) leaves remainders, so there are no quadratic factors.
If \begin{align*} f(x) = \sum_{i=0}^n \alpha_i x^i = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in {\mathbf{Q}}[x] .\end{align*} then \(f\) will be irreducible over \({\mathbf{Q}}[x]\) (and thus over \({\mathbf{Z}}[x]\) by Gauss’ lemma) if \(\exists p\) such that
- \(p\) divides every coefficient except \(a_n\) and
- \(p^2\) does not divide \(a_0\).
Note that if \(f\) is monic, it suffices to find any prime dividing all of the non-leading terms.
If \(f(x+a)\) satisfies Eisenstein for any \(p\), then \(f\) is irreducible. This is generally because \(\Delta_{f(x)} = \Delta_{f(x + a)}\), and if \(p\) works for Eisenstein on any \(f\) then \(p\divides \Delta_f\).
Set \(f(x) \coloneqq x^2+x+2\), then \(f(x+3) = x^2 + 7x + 14\) and Eisenstein applies at \(p=7\).
If \(n \coloneqq\deg(f)\) and \(x^n f(1/x)\) is irreducible, then \(f\) is irreducible. Note that this is just reversing the coefficients.
Take \(f(x) \coloneqq 2x^5 -4x^2-3\), then for \(g(x) \coloneqq 3x^5 +4x^2 - 2\) Eisenstein applies with \(p=2\).
If \(f(x) \equiv b(x+a)^n\operatorname{mod}p\) for some \(p\) where \(n\coloneqq\deg f\), then Eisenstein may work on \(f(x-a)\) using the prime \(p\). Note the change in signs/reverse translation.
In other words, reduce mod \(p\) for various \(p\), and if any \(p\) collapses \(f\) to a power of a linear factor, use that \(p\) for Eisenstein.
Check \begin{align*} f(x) \coloneqq x^3 + x^2 -48 x + 128 \leadsto f(x)\equiv (x-3)^3 \operatorname{mod}5 ,\end{align*} and Eisenstein on \(f(x+3)\) with \(p=5\) works.
Computing
Misc Useful Facts
Once you’ve confirmed that you have a Galois extension, some useful tricks are available:
- The size \({\sharp}G(f)\) is the degree \([\operatorname{SF}(f) : {\mathbf{Q}}]\).
-
The degree of \([{\mathbf{Q}}(\alpha): {\mathbf{Q}}]\) is the degree of \(\min_{\alpha}(x)\), or any irreducible polynomial with \(\alpha\) as a root.
- Note that \({\mathbf{Q}}(\alpha)\neq \operatorname{SF}(f)\) in general.
-
If \(f=\prod(x-r_i)\), then \(\operatorname{SF}(f)\) contains every \({\mathbf{Q}}(r_i)\). Thus
\begin{align*}
[{\mathbf{Q}}(r_i) : {\mathbf{Q}}] = d \implies
d\divides [\operatorname{SF}(f): {\mathbf{Q}}]
.\end{align*}
- Note that \(d\neq \deg f\) in general!
- Reminder of rational roots test: for \(f(x) = a_nx^n + \cdots + a_0\), rational roots are of the form \(p_0/p_n\) where \(p_i \divides a_i\).
- \({ \mathsf{Gal}}(L/k)\) permutes the roots of any irreducible polynomial in \(k[x]\). In particular, if \(L=\operatorname{SF}(f)\) with \(f\) reducible, then \(G\) must send roots of irreducible factors to conjugates of the same factor.
- \({\mathbf{Q}}(\zeta_a) = {\mathbf{Q}}(\zeta_b) \iff a = 2b\) and \(b\) is odd.
- If there are \(k\) complex conjugate pairs (accounting for \(2k\) roots) then \(G\) contains a cycle \((1,2)(3,4)\cdots(2k-1,2k)\).
-
If all exponents are even, \(f(r) = 0 \iff f(-r) = 0\), so roots occur in pairs \((r, -r)\).
- Pairs are preserved by \(G\) in the sense that every \(\sigma\in G\) satisfies either \(\left\{{r, -r}\right\}\mapsto \left\{{r, -r}\right\}\) or \(\left\{{r, -r}\right\} \mapsto \left\{{s, -s}\right\}\) for another pair.
- Example: \(x^4-5x^2+5\) has two pairs.
Transitive Subgroups
If \(f\in k[x]\) is irreducible, then \({ \mathsf{Gal}}(\operatorname{SF}(f)/k) \leq S_n\) is always a transitive subgroup, i.e. it acts transitively on the set of roots.
\begin{align*} n\divides {\sharp}{ \mathsf{Gal}}(K/{\mathbf{Q}}) \divides n! .\end{align*}
Why: \({ \mathsf{Gal}}(K/{\mathbf{Q}}) \cong G\leq S_n\), and Lagrange yields \({\sharp}H \divides n!\). Note that \(G\) acts on \(R\) the set of \(n\) roots, and since it acts transitively, \(R\) is a single orbit. By orbit stabilizer, \({\mathcal{O}}_r \cong G/{\operatorname{Stab}}_G(r)\) and thus \begin{align*} {\sharp}G = {\sharp}{\mathcal{O}}_r \cdot {\sharp}{\operatorname{Stab}}_G(r) ,\end{align*} so both terms on the right-hand side patently divide \({\sharp}G\)
Write \(C_n\) for the cyclic group of order \(n\). The following are transitive subgroups of \(S_n\) for small \(n\), where blue groups are nonabelian:
| \(n \text{ in }S_n\) | Transitive Subgroups | Sizes |
|---|---|---|
| 1 | 1 | 1 |
| 2 | \(S_2 \cong C_2\) | 2 |
| 3 | \({\color{blue}S_3\cong D_3}, A_3 \cong C_3\) | 6,3 |
| 4 | \({\color{blue}S_4, A_4, D_4}, C_4, C_2^2\) | 24,12,8,4,4 |
| 5 | \({\color{blue}S_5, A_5 , F_5\cong C_5\rtimes C_4, D_5}, C_5\) | 120,60,20,10,5 |
Other useful facts:
-
\({\sharp}D_n = 2n\), \({\sharp}S_n = n!, {\sharp}A_n = n!/2\), and \({\sharp}F_5 = 20\).
-
For degree 8 extensions (which sometimes arise as quadratic extensions of degree 4 extensions): \(Q_8 \leq S_8\) is transitive and nonabelian of order 8, and has presentation \begin{align*} Q_{8}=\left\langle\alpha, \beta \mathrel{\Big|}\alpha^{4}=\beta^{4}=1, \alpha \beta \alpha=\beta, \beta^{2}=\alpha^{2}\right\rangle .\end{align*} Note that \(Q_8 \leq S_8\) but \(Q_8\not\leq S_{<7}\).
-
\(F_5\) has presentation \begin{align*} \left\langle{a,b {~\mathrel{\Big\vert}~}a^5, b^4, bab^{-1}= a^2}\right\rangle .\end{align*}
Distinguishing Groups
Material borrowed from https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisSnAn.pdf
By \(n\) in \(G\leq S_n\):
\(n=4\):
-
\(C_2^2\) vs \(C_4\):
- \(C_2^2\) has two elements of order 2, the latter does not. So a cycle of type \((2, 2)\) forces \(C_2^2\).
-
\(S_4\) vs \(A_4\):
- \(S_4\) contains a Sylow-2 subgroup of order 8 (which divides \(4! = 24\)) but \(A_4\) does not since it’s of order \(4!/2 = 12\) and \(8\not\divides 12\).
- If \(G\) contains a transposition, then \(G= S_4\) or \(D_4\), since \(A_4\) doesn’t contain a transposition.
-
\(D_4\) vs \(Q_8\):
-
5 roots:
- \(S_5\) is generated by any transposition and any 5-cycle.
- \(S_n\) is generated by \((a,b)\) and \((1,2,\cdots,n) \iff \gcd(b-a, n) = 1\). In particular, the \((1, 2)\) and any length \(n\) cycle works.
The following are the cycle types that can occur:
Useful fact: if \(G \leq S_n\) for \(n\) prime contains a 2-cycle and a \(p{\hbox{-}}\)cycle, then \(G\cong S_n\). Note that for \(n\) not prime, a transposition and an \(n{\hbox{-}}\)cycle isn’t enough, since one needs the specific \(n{\hbox{-}}\)cycle \((1,2,\cdots,n)\) in general.
If \(n>2\) and \(G\) contains a 3-cycle and an \(n{\hbox{-}}\)cycle, then \(G = A_n\) or \(S_n\). Note that by Orbit-Stabilizer \(n\divides {\sharp}G\), and if \(n\) is prime then by Cauchy there is an \(n{\hbox{-}}\)cycle (but this is not always the case). In fact, it suffices to find a \(k{\hbox{-}}\)cycle for any \(k\geq n/2\), which can be found by reducing mod \(p\) and examining cycle types.
Moreover, if \(G\) contains a 2-cycle (transposition), then \(G = S_n\).
- Alternating groups have even numbers of cycles of even length.
- Elements in \(A_n\) either have cycle type with an even number of even lengths (including 0).
- \(A_4\) does not contain a subgroup isomorphic to \(C_2^2\).
Some useful generating sets: see https://kconrad.math.uconn.edu/blurbs/grouptheory/genset.pdf
Density: Cycle Types
For any \(p\not\divides \Delta\), writing \(f(x) = \prod_{i=1}^m f_i(x) \operatorname{mod}p\), \(G\) contains a cycle of type \((\deg f_1, \deg f_2, \cdots, \deg f_m)\). Equivalently, if \(\tilde f \coloneqq f\operatorname{mod}p\), then \(G(\tilde f) \leq G(f)\) is a subgroup.
Warning: this only works if the \(f_i\) are distinct, i.e. there are no repeated factors in the factorization \(\operatorname{mod}p\).
You can use this to rule out types of groups using Lagrange’s theorem: if you find a cycle of length \(m\) which doesn’t divide \({\sharp}H\), then \(H\) isn’t a possibility! Example: \(\deg f=5\) with exactly one conjugate pair of roots, then there is a 5-cycle \(\sigma \coloneqq(1,2,3,4,5)\) because \(5\divides {\sharp}G\) and a 2-cycle \(\tau \coloneqq(1,2)\) coming from complex conjugation. There check that \(a_1 \coloneqq\sigma \tau \sigma^{-1}= (1, 5)\) and \(a_1 \tau = (1, 5, 2)\) is a 3-cycle, so \(3\divides {\sharp}G\). This rules out \(F_5\) which is of order 20, since \(3\not\divides 20\).
Consider \(f(x) = x^5+2x+1\). Reducing to \({ \mathbf{F} }_3\) yields no roots, so \(f\) is irreducible, and moreover \(G(f)\) contains a 5-cycle. Reducing to \({ \mathbf{F} }_7\) splits \(f\) as \((x+2)(x+3)(x^3+2x^2+5x+5)\), so \(G(f)\) contains a 3-cycle.
Take \(f(x) \coloneqq x^6 + x^4 + x + 3\), then
\begin{align*} f(x) &\equiv (x+1)(x^2 + \cdots)(x^3 + \cdots) \operatorname{mod}2 &\implies \text{type } (1,2,3) \in G \\ f(x) &\equiv x(x+2)(x^4 + \cdots) \operatorname{mod}3 &\implies \text{type } (1,1,4) \in G \\ .\end{align*}
Take \(f(x) \coloneqq x^4+x+1\), then \begin{align*} f(x) &\equiv x^4+x+1 \operatorname{mod}2 &\implies \text{type } (4) \\ f(x) &\equiv (x-1)(x^3+x^2+x-1) \operatorname{mod}3 &\implies \text{type } (1,3) \\ .\end{align*}
So \(G\) contains a 4-cycle and a 3-cycle. This is enough to show \(G = A_4\).
Let \(f(x) = x^6 + x^4 + x + 3\), reduce \(\operatorname{mod}11\) to get a cycle type \((1, 5)\). So \(G\leq S_6\) contains a 5-cycle, where \(5>n/2 \coloneqq 6/2=3\), meaning \(G = A_n, S_n\). Now reduce \(\operatorname{mod}p\) for various \(p\) to look for a cycle type of the form \((2,1,1,\cdots)\) or \((3,1,1,\cdots)\). This is hard, but \(f\operatorname{mod}2\) has type \((1,2,3)\) and \(\qty{ (a,b)(c,d,e) }^3 = (a, b)\), so \(G\) contains a transposition and thus \(G = S_n = S_6\).
Let \(f(x) = x^7-x-1\), reduce \(\operatorname{mod}2\) to get a 7-cycle, and \(\operatorname{mod}3\) to get \((2, 5)\). Then use \((2, 5)^5 = (2,1,1,\cdots)\) to get a transposition, So \(G = S_7\).
Let \(f(x) \coloneqq x^7-7x+10\). Reducing \(\operatorname{mod}3\) yields \((2, 5)\) and \((2, 5)^5 = (2, \cdots)\) and have a transposition. Since \(5>n/2=7/2\), \(G = S_7\).
Discriminants
For \(f = \sum a_k x^k\) monic, \begin{align*} \Delta_f = \prod_{i < j} (r_i - r_j)^2 .\end{align*}
Note that \(\Delta = 0\) when \(f\) has a repeated root.
For cubics,
- \(\Delta > 0 \implies 3\) distinct real roots
- \(\Delta < 0 \implies 1\) real root and 1 conjugate pair.
For \(f\) a cubic: \begin{align*} \Delta_f &= (r_1 - r_2)^2 (r_1 - r_3 )^2 \\ &\quad (r_2 - r_3)^2 .\end{align*}
For \(f\) a quartic: \begin{align*} \Delta_f &= (r_1 - r_2)^2(r_1 - r_3)^2 (r_1 - r_4)^2 \\ & \quad (r_2 - r_3)^2 (r_2 - r_4)^2 \\ & \quad (r_3 - r_4)^2 .\end{align*}
In general, for a degree \(n\) polynomial this will have \(n(n-1)/2\) terms.
In general, \begin{align*} G \subseteq A_n \iff \sqrt{\Delta} \in k ,\end{align*} i.e. \(\Delta\) is a perfect square in the ground field \(k\).
Some special cases of discriminant values:
- Quadratics: \begin{align*} f(x) = ax^2 + bx + c \implies \Delta = b^2-4ac .\end{align*}
-
Cubics:
-
General:
\begin{align*}
f(x) = ax^3 + bx^2 + cx + d \implies \Delta = b²c² - 4ac³ - 4b³d - 27a²d² + 18abc
.\end{align*}
- Note that you can depress a general cubic by substituting \(t=x - {b\over 3a}\), yielding \begin{align*} f(t) = t³ + pt + q \implies \Delta = -4p³-27q² .\end{align*}
-
General:
\begin{align*}
f(x) = ax^3 + bx^2 + cx + d \implies \Delta = b²c² - 4ac³ - 4b³d - 27a²d² + 18abc
.\end{align*}
Some useful facts:
- \(\Delta = 0 \iff f\) has a repeated root.
- \(G\hookrightarrow A_n \iff \Delta\) is a perfect square in \(k\).
Worked Examples
If \(\deg f = 5\) with exactly 3 real roots and one non-real complex conjugate pair, then \(G(f) = S_5\). \(G\) contains a transposition, namely complex conjugation on the conjugate pair. This already implies \(G\neq A_5\), since a transposition is an odd number of even cycles.
The claim is that \(G\) contains an element of order 5, i.e. a 5-cycle, which is enough to generate \(S_5\). This follows because
- Galois acts transitively, so there is a length 5 orbit.
- By Orbit-Stabilizer, 5 divides \({\sharp}G\).
- By Sylow, there is an element of order 5.
So \(G = S_5\).
Quadratics
Every degree 2 extension \(L/k\) is Galois, except possibly in characteristic 2:
-
If \(\alpha\in L \setminus k\) then \(\min_{\alpha}(x) \in L[x]\) must split in \(L[x]\), so \(L\) is automatically a splitting field.
- Why? \(\alpha\in L \implies \min_{\alpha}(x) = (x- \alpha)g(x)\) which forces \(\deg(g) = 1\).
- If \(\operatorname{ch}(k) \neq 2\), then \(L\) is separable since \begin{align*} \min_{ \alpha}(x)' = 2x + \cdots \not\equiv 0 ,\end{align*}
One can complete the square for quadratics: \begin{align*} f(x)=x^{2}+\alpha x+\beta=\left(x-\frac{\alpha}{2}\right)^{2}+\beta-\frac{\alpha^{2}}{4} \text { . } .\end{align*}
Thus it suffices to consider quadratics of the form \(x^2+a\).
-
\(G(x^2-m) = C_2\) for \(m\) not a perfect square.
- \(x^2-m = (x+\sqrt{m})(x-\sqrt{m})\), so the splitting field is \({\mathbf{Q}}(\sqrt{m})\) of degree 2.
- Since \(G\leq S_2\) and has order 2, \(G= S_2 \cong C_2\).
- Concretely, take \(m=2\), then \(G = \left\{{\operatorname{id}, \tau}\right\}\) where \(\tau: \sqrt{2} \mapsto -\sqrt{2}\), and correspondingly \(a+b\sqrt{2} \mapsto a-b\sqrt{2}\).
-
\(G((x^2-2)(x^2-3)) = C_2\times C_2\).
- Since \(G\) must permute irreducible factors, labeling the roots \(r_1, r_2 = \pm \sqrt{2}\) and \(r_3, r_4 = \pm \sqrt{3}\), we have \begin{align*} G \subseteq \left\langle{(1, 2), (3, 4)}\right\rangle = \left\{{\operatorname{id}, (1,2), (3,4), (1,2)(3,4)}\right\} \cong C_2\times C_2 .\end{align*}
- \({\sharp}G = 4\), taking the tower \({\mathbf{Q}}(\sqrt 2, \sqrt 3) / {\mathbf{Q}}(\sqrt 2)/ {\mathbf{Q}}\) and noting \(\sqrt 3 \not\in {\mathbf{Q}}(\sqrt 2)\) which makes each step degree 2. So this forces \(G \cong C_2\times C_2\).
Cubics
Tricks/reminders:
- Try the rational roots test to check irreducibility, since reducible implies there’s a linear factor.
- Nice situation: one real and two complex roots. Try Calculus and MVT to reason about real roots. This immediately yields \(S_3\).
- Otherwise, 3 real roots (or no easy way to check root types). Try discriminant classification: put in the form \(t^3+pt+q\), potentially using \(t=x-b/3a\), then \(\Delta = -4p^3-27q^2\).
If \(n \coloneqq\deg f\) is odd, then \(f\) has at least one real root. If \(f\) has two non-real roots, then \(G\) contains a transposition. If \(n\) is prime, then \(G\) contains an \(n{\hbox{-}}\)cycle (by transitivity and Cauchy), forcing \(G = S_n\).
Let \(f(x) = x^3-2\).
Then \(f'(x) = 3x^2\), so \(f\) is monotone increasing. By the MVT, checking \(f(-2)<0\) and \(f(2)>0\), \(f\) has a single real root in \([-2, 2]\). The other two must be a complex conjugate pair.
Alternatively, just factor the darn thing: \(f(x) = (x-\omega)(x-\zeta_3\omega)(x-\zeta_3^2\omega)\) where \(\omega \coloneqq 2^{1\over 3}\) and \(\zeta_3^3=1\).
So \(G(f)\) contains a transposition, and since \(\deg f = 3\) is prime, \(G\) contains a 3-cycle and \(G(f) = S_3\).
Let \(f(x) = x^3-4x+5\).
Then \(f'(x) = 3x^2-4 = 0\) when \(x = \pm \sqrt{4/3}\). Checking \(f''(x) = 6x\) yields a max at \(-\sqrt{4/3}\) and a min at \(\sqrt{4/3}\). Checking \begin{align*} f(4/3) = (4/3)^3-4(4/3)+5 = (64/27)-(16/3)+5 = 55/27 > 0 ,\end{align*} so knowing the general shape of a cubic, there is exactly one real root, somewhere in \((-\infty ,-\sqrt{4/3})\). So \(G(f) = S_3\).
Away from \(\operatorname{ch}k = 2\), Galois groups of cubics are entirely determined by discriminants:
There are only two possibilities: \(S_3\) or \(A_3 \cong C_3\).
- If \(\sqrt{\Delta}\in k\), then \(G\cong A_3\).
- Otherwise, \(G\cong S_3\).
-
\(G(x^3+x+1) = S_3\):
- Irreducible because it has no rational roots (by the rational roots test)
- \(f'(x) = 3x^2+1>0\) so \(f\) increases everywhere and can only have one real root \(r\), so \({\mathbf{Q}}(r)/{\mathbf{Q}}= \deg f = 3\).
- The other roots are a non-real conjugate pair \(w, \overline{w}\), so \({\mathbf{Q}}(w, r)/{\mathbf{Q}}(r) = \deg f(x)/(x-r) = 2\).
- So \([\operatorname{SF}(f): {\mathbf{Q}}] = 6\), and the only transitive subgroup of order 6 in \(S_3\) is \(S_3\) itself.
-
\(G(x^3-x+1) = S_3\):
- This is already a depressed cubic, so use \(\Delta = -4 (-1)^3 - 27(1) = -23\).
- \(\Delta = -23 \not\in {\mathbf{Q}}^2\), so \(G\not\leq A_4\) which forces \(G = S_4\).
-
\(G(x^3-3x+1) = A_3\):
- This is already a depressed cubic, so \(\Delta = -4(-3)^3 + 27(1) = -3(27) = 81\).
- \(\Delta = 81 \in {\mathbf{Q}}^2\), so \(G\leq A_3\).
Quartics
If \begin{align*} f(x)=x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0} \end{align*} then define the resolvent of \(f\) by \begin{align*} R_{4}(t)=t^{3}-a_{2} t^{2}+\left(a_{1} a_{3}-4 a_{0}\right) t+4 a_{0} a_{2}-a_{1}^{2}-a_{0} a_{3}{ }^{2} .\end{align*}
Alternatively, it can be defined in terms of the roots \(r_i\): \begin{align*} \left(x-\left(r_{1} r_{2}+r_{3} r_{4}\right)\right)\left(x-\left(r_{1} r_{3}+r_{2} r_{4}\right)\right)\left(x-\left(r_{1} r_{4}+r_{2} r_{3}\right)\right) .\end{align*}
For depressed quartics, \begin{align*} f(X)=X^{4}+c X+d \Longrightarrow R_{3}(X)=X^{3}-4 d X-c^{2} .\end{align*}
The Galois groups of irreducible quartics can be determined using discriminants, resolvents, and checking irreducibility:
-
If \(\sqrt{\Delta}\in {\mathbf{Q}}\), then \(G = A_4, C_2^2\).
- If resolvent is irreducible: \(A_4\). Otherwise \(C_2^2\).
-
If \(\sqrt{\Delta}\not\in {\mathbf{Q}}\) then \(G = S_4, D_4, C_4\).
- Is resolvent is irreducible: \(S_4\). Otherwise, \(D_4\) or \(C_4\), argue by cycle types (or if \(f\) is irreducible in \({\mathbf{Q}}(\sqrt{\Delta})\), \(D_4\)).
-
Summary:
A flow chart summarizing the full process:
See Hungerford 273 for classification.
-
\(G(x^4-x-1) = S_4\):
- Check \(f\) is irreducible in \({ \mathbf{F} }_2[x]\).
- \(R_3(t) = t^4+4t-1\)
-
\(G(x^4+8x+12) = A_4\):
- The resolvent cubic is \(x^3-48x+64\), which has no rational roots.
- Now check \begin{align*} \Delta = (-27)(8^4) + (256)(12^3)=(81)(2^{12}) \in {\mathbf{Q}}^2 ,\end{align*} so \(G=A_4\).
-
\(G(x^4+3x+3) = D_4\):
- The resolvent cubic is \(g(x) = x^3-12x+9=(x-3)(x^2+3x-3)\) and \(\Delta = 3^3 5^2 7\), so \(G = C_4, D_4\).
- Check \(D \coloneqq\Delta_g = 21\).
-
Check if \(g\) is irreducible in \({\mathbf{Q}}(\sqrt{21})\): suppose \(x^{4}+3 x+3=\left(x^{2}+a x+b\right)\left(x^{2}-a x+c\right)\), then \(-a^{2}+b+c=0, a(c-b)=0, b c=3\)
- From \(a(c-b)=0\), if \(a=0\) then \(b=-c\) and \(c^2=3\), but \(\sqrt{-3}\not\in {\mathbf{Q}}(D)\). Otherwise \(c=b\) and \(c^2 = 3\), but \(\sqrt{3}\not\in{\mathbf{Q}}(D)\).
- So \(G= D_4\).
Cyclotomic Fields
\({ \mathsf{Gal}}({\mathbf{Q}}(\zeta_n)/{\mathbf{Q}}) \cong ({\mathbf{Z}}/n)^{\times}\) and is generated by maps of the form \(\zeta_n \mapsto \zeta_n^j\) where \((j, n) = 1\). I.e., the following map is an isomorphism: \begin{align*} ({\mathbf{Z}}/n)^{\times}&\to { \mathsf{Gal}}({\mathbf{Q}}(\zeta_n), {\mathbf{Q}}) \\ [r] &\mapsto (\phi_r: \zeta_n \mapsto \zeta_n^r ) \end{align*}
The splitting field of \(x^p-1\) is \({\mathbf{Q}}(\zeta_p)\), and the splitting field of \(x^p+1\) is \({\mathbf{Q}}(\zeta_{2p})\).
-
\(x^p - a\) factors as \(\prod_{k=0}^{p-1} (x- \zeta_p^k \omega)\) where \(\omega \coloneqq a^{1\over p}\), so this splits in \({\mathbf{Q}}(\zeta_p, \omega)\) which has degree
\begin{align*}
\phi(p) \cdot \deg \min_\omega(x) = (p-1)p
.\end{align*}
- This yields two cyclic subgroups \(C_{p-1}, C_p\) where \(C_p{~\trianglelefteq~}G\), and thus some semidirect product \(C_{p-1} \curvearrowright C_{p}\).
- \(x^p+a\) factors as \(\prod_{k=0}^{p-1}(x - \zeta_p^k \omega )\) for \(\omega \coloneqq(-a)^{1\over p}\).
Also use that splitting fields over \({\mathbf{Q}}\) are always normal, so it suffices to check that \(f\) is separable and irreducible to show extensions are Galois.
Degree 3:
-
\(G(x^3-2): S_3\)
- The roots are \(\zeta_3^k \omega\) for \(0\leq k \leq 3\), \(\omega \coloneqq 2^{1\over 3}\).
- The splitting field is \({\mathbf{Q}}(\omega, \zeta_3)\) which has degree \(3\phi(3) =6\).
- The possibilities are \(G = A_3\cong C_3, S_3\), and order 6 forces \(G=S_3\).
-
Useful alternative:
- Note that there is exactly one real root and one conjugate pair, so \(G\) contains a transposition \((23)\).
- There is a 3-cycle \((123)\) given by fixing \(\omega\) and sending \(\zeta_3 \mapsto \zeta_3\omega\), and this is enough to generate \(D_3 \cong S_3\).
Degree 4:
-
\(G(x^4-1) = C_2\):
- The roots are \(\zeta_4^k\) for \(0\leq k \leq 3\).
- The splitting field is \({\mathbf{Q}}(\zeta_4) = {\mathbf{Q}}(i)\) which has degree \(\phi(4) = 2\).
- But this is not a reducible polynomial! Use that that Galois is defined as \(\mathop{\mathrm{Aut}}(\operatorname{SF}(f) / {\mathbf{Q}})\) and quadratic extensions are Galois.
-
\(G(x^4-2) = D_4\):
-
The roots are \(\zeta_4^k \omega\) for \(0\leq k \leq 3\), where \(\zeta_4 = i, \omega = 2^{1\over 4}\).
- Explicitly, \(r_i \in \left\{{\omega, i\omega, -\omega, -i\omega}\right\}\)
- The splitting field is \({\mathbf{Q}}(\omega, \zeta_4)\), which has degree \(4\phi(4) = 8\) since \(\min_{\zeta_4} = x^2+1\), which is still irreducible over \({\mathbf{Q}}(\omega) \subseteq {\mathbf{R}}\).
- \(D_4 \leq S_4\) is the only transitive subgroup of order 8.
-
Useful note on bounding the size:
- Any \(\sigma \in G\) must preserves roots of \(x^4-2\) but also \(x^2+1\). So there are at most 4 possibilities for \(\sigma(\omega)\), and at most 2 for \(\sigma(\zeta_4)\), so \({\sharp}G \leq 8\) and \(G\neq S_4\).
-
Explicitly, there is a 4-cycle \(\sigma = (1,2,3,4)\) generated by \(\omega \mapsto \zeta_4 \omega\) and a 2-cycle \((2,4)\) given by complex conjugation, and this generates \(D_4\) since \(\gcd(4-2, 4) \neq 1\).
- Why this is a 4-cycle: check \(\sigma(i)=i\), and: \begin{align*} \sigma(r_1) &= r_2 = ir_1 \\ \sigma(r_2) &= \sigma(ir_1) = ir_2 = r_3 \\ \sigma(r_3) &= \sigma(ir_2) = ir_3 = i(ir_2) = -r_2 = r_4 .\end{align*}
-
The roots are \(\zeta_4^k \omega\) for \(0\leq k \leq 3\), where \(\zeta_4 = i, \omega = 2^{1\over 4}\).
General cases:
-
\(G(x^p-1) = C_p^{\times}\):
- ?
-
\(G(x^4+1) = C_2^2\):
- This is irreducible because it is irreducible (by having no roots) mod \(3\).
- The roots are \(\zeta_8^k\) for \(k=1,3,5,7\) coprime to 8, since this is \(\Phi_8(x)\).
- The splitting field is \({\mathbf{Q}}(\zeta_8) = {\mathbf{Q}}(i, \sqrt 2)\), noting that \(\zeta_8 = e^{2\pi i \over 8} = e^{\pi i \over 4} = \cos(\pi/4) + i\sin(\pi/4) = (1/2)(\sqrt 2 + i \sqrt{2})\) so we have containment and both are degree \(\phi(8)=4\) extensions.
- This restricts to \(C_4, C_2^2\).
- Reduce \(\operatorname{mod}5\) to get \((x^2+2)(x^2+2)\) of cycle type \((2, 2)\), forcing \(C_2^2\).
-
\(G(x^4+2) = D_4\):
- The roots are \(\zeta_8^k \omega\) for \(\omega = 2^{1\over 4}, k = 1,3,5,7\) coprime to 8.
- The splitting field is \({\mathbf{Q}}(\zeta_8, \omega) = {\mathbf{Q}}(\zeta_4, \omega)\).
-
\(G(x^4+3) = D_4\):
- The roots are \(\zeta_8^k, \omega\) for \(\omega = 3^{1\over 4}, k=1,3,5,7\) coprime to 8.
- The splitting field is \({\mathbf{Q}}(\omega, \zeta_8)\) of degree \(\phi(8) = 8\)
Finite Fields
\({ \mathsf{Gal}}({ \mathbf{F} }_{p^n}/{ \mathbf{F} }_p) \cong {\mathbf{Z}}/ \left\langle{ n }\right\rangle\), a cyclic group generated by powers of the Frobenius automorphism: \begin{align*} \varphi_p: { \mathbf{F} }_{p^n} &\to { \mathbf{F} }_{p^n} \\ x &\mapsto x^p \end{align*}
See D&F p.566 example 7.
Lattices
\(n=2\):
- \(D_2 \cong 1\)
- \(A_2 \cong 1\)
- \(S_2 \cong C_2\)
\(n=3\):
- \(D_3 \cong S_3\)
- \(A_3 \cong C_3\).
-
-
\(n=4\):
\(n=5\):
Misc:
