Galois Theory HT25, Flashcards for computing the Galois group

Each $\mathbb Q$-automorphism $\sigma$ of $\mathbb Q(\alpha _ 1,\dots,\alpha _ n)$ must send every root $\alpha _ i$ of $f$ to another root, because

Hence

is a permutation, giving an injective homomorphism
$\rho\colon\operatorname{Gal}(f)\hookrightarrow S _ n$. @important~

Even when $f$ is irreducible, its roots may satisfy algebraic relations that any $\mathbb Q$-automorphism must preserve.
For $f(t)=t^3-3t+1$ the roots are

and they obey
$\alpha _ 2=\alpha _ 1^2-2$ and $\alpha _ 3=\alpha _ 2^2-2$.
A permutation such as $(13)$ would break these identities, so it is forbidden; one checks that the only allowed permutations form $A _ 3$, not $S _ 3$. @important~

Let $\varphi\colon F\to\tilde F$ be a field isomorphism and let

If $\varphi\bigl(m _ {F,\alpha}\bigr)(\tilde\alpha)=0$, then $\varphi$ extends uniquely to an isomorphism

Overall idea:
$\operatorname{Gal}(K/F)$ acts on the $n$ roots; irreducibility ensures any one root generates $K$ over $F$, and separability gives enough automorphisms to move any root to any other.

• Because $f$ is irreducible and separable, for any two roots $\alpha _ i,\alpha _ j$ there exists an $F$-embedding sending $\alpha _ i$ to $\alpha _ j$ (apply the extending-identity lemma).
• Such embeddings lie in $\operatorname{Gal}(K/F)$, so the action on the root set is transitive. @important~

Since $K/F$ is Galois and generated by $\alpha$,

Thus the Galois group’s cardinality equals the degree of the minimal polynomial of $\alpha$.

• $n=2$: $C _ 2$
• $n=3$: $S _ 3$, $C _ 3$
• $n=4$: $S _ 4$, $A _ 4$, $D _ 8$, $V _ 4$, $C _ 4$
• $n=5$: $S _ 5$, $A _ 5$, $F _ {20}\cong\mathbb Z _ 5\rtimes\mathbb Z _ 4$, $D _ {10}$, $C _ 5$

• $S _ n$: one transposition and one $n$-cycle suffice.
• $C _ n$: any single $n$-cycle (a generator) suffices.

If $f\in\mathbb Q[t]$ has prime degree $p$ and exactly $p-2$ real roots, then $\operatorname{Gal} _ {\mathbb Q}(f)\cong S _ p$. @important~

Let $f=g _ 1\cdots g _ k$ over $F$, with each $g _ i$ irreducible and $K _ i$ its splitting field.
Set $L:=K _ 1\cdots K _ k$.
If the extensions $K _ i/F$ are pairwise linearly disjoint—equivalently $K _ i\cap K _ j=F$—then

For $f(x)=(x^3-2)(x^3-1)$ over $\mathbb Q$ we have
$\operatorname{Gal} _ \mathbb Q(x^3-2)\cong S _ 3$,
$\operatorname{Gal} _ \mathbb Q(x^3-1)\cong C _ 3$,
yet the splitting field is $\mathbb Q(\sqrt[3]{2},\omega)$ and

not $S _ 3\times C _ 3$, because both factors share the root of unity $\omega$ so their splitting fields overlap.

• Degree $\le3$: check for rational roots.
• Eisenstein’s criterion (possibly after a substitution).
• Reduction modulo a prime: if the reduction is irreducible, so is $f$.
• Galois-action criterion: $f$ is irreducible iff its Galois group acts transitively on its roots.

Always $\lvert\operatorname{Aut} _ F(K)\rvert\le [K:F]$; equality holds iff the extension is Galois.

They correspond bijectively to the roots of the minimal polynomial $m _ {F,\alpha}$ lying in $K$.
Thus the number of $F$-homomorphisms equals the number of such roots.

Compute

No quadratic factor lies in $\mathbb Q[x]$, so $m _ {\alpha,\mathbb Q}(x)=x^4-4x^2-1$.

Overall idea:
Show the splitting field has degree $8$ and identify two automorphisms generating $D _ 4$.

Let $\alpha=\sqrt{2+\sqrt5}$, $\beta=\sqrt{2-\sqrt5}$.
The splitting field is $S=\mathbb Q(\alpha,\beta)$, and

Automorphisms:
$\sigma\colon\beta\mapsto-\beta, \alpha\mapsto\alpha$ (order 2)
$\tau\colon\alpha\mapsto\beta, \beta\mapsto-\alpha$ (order 4)

They satisfy the relations of the dihedral group $D _ 4$, and $ \vert \langle\sigma,\tau\rangle \vert =8$, so
$\operatorname{Gal}(S/\mathbb Q)\cong D _ 4$.

Overall idea:
Verify the extension has degree 4 and find an automorphism of order 4.

Let $\alpha=\sqrt{d+\sqrt b}$, $\beta=\sqrt{d-\sqrt b}$.
The polynomial $f(x)=(x^2-d)^2-b$ is separable and has roots $\pm\alpha,\pm\beta$, all lying in $K:=\mathbb Q(\alpha)$, hence $K/\mathbb Q$ is Galois.
Using the tower law,

Define $\sigma(\alpha)=\beta$. A short computation shows $\sigma^2(\alpha)=-\alpha$, so $\sigma$ has order 4 and generates the whole group:
$\operatorname{Gal}(K/\mathbb Q)=\langle\sigma\rangle\cong C _ 4$.

The Frobenius automorphism $\phi:x\mapsto x^p$ has order $n$ (since $\phi^n=\operatorname{id}$ and no smaller power fixes all elements).
All $\mathbb F _ p$-automorphisms are powers of $\phi$, so
$\operatorname{Gal}(\mathbb F _ {p^n}/\mathbb F _ p)=\langle\phi\rangle\cong C _ n$. @important~

Via $\sigma(\zeta _ n)=\zeta _ n^{ a}$ one obtains

with equality when $\Phi _ {n,F}$ is irreducible (e.g. $F=\mathbb Q$).

One has an injection

If $K/F$ is Galois and contains $\zeta _ n$, the map is an isomorphism.

Overall idea:
Use translations by elements of $\mathbb F _ p$ to produce $p$ distinct automorphisms.

$f$ is separable and irreducible; its splitting field is $S=F(\alpha)$ for any root $\alpha$.
For $k\in\mathbb F _ p$ define $\sigma _ k(\alpha)=\alpha+k$.
Because $(\alpha+k)^p-(\alpha+k)=a$, $\sigma _ k$ permutes the roots, giving $p$ automorphisms with
$\sigma _ k\sigma _ {k'}=\sigma _ {k+k'}$.
Thus $\operatorname{Gal}(f)\cong C _ p$.

Overall idea:
The $25$ roots form an additive coset, giving two independent $5$-torsion translations.

Let $\alpha$ be one root. All roots are $\alpha+\lambda$ with $\lambda\in\mathbb F _ {25}$.
Translations by $1$ and by a non-trivial element $u\in\mathbb F _ {25}\!\setminus\!\mathbb F _ 5$ give commuting automorphisms of order $5$:

Since $\langle\sigma,\tau\rangle$ has order $25$ and the extension degree is $25$,
$\operatorname{Gal}(f)\cong C _ 5\times C _ 5$.

The $124$ non-zero elements of $\mathbb F _ {125}$ are the units; multiplying a fixed root $\alpha$ by any unit $v$ gives another root.
Choose a generator $g$ of $\mathbb F _ {125}^\times$ and define $\sigma(\alpha)=g\alpha$.
Then $\sigma$ has order $124$ and the extension degree is $124$, hence
$\operatorname{Gal}= \langle\sigma\rangle\cong C _ {124}$.

Overall idea:
The roots form an additive coset of $\mathbb F _ {13}$ inside the degree-13 extension; a single translation generates all automorphisms.

$f$ is separable and irreducible (Eisenstein with prime $13$).
Let $\alpha$ be a root; then $\alpha+k$ is again a root for every $k\in\mathbb F _ {13}$.
Translation $\sigma(\alpha)=\alpha+1$ has order $13$, the extension degree equals $13$, and therefore
$\operatorname{Gal}(f)\cong C _ {13}$.

For $f=t^3+pt+q$ irreducible with discriminant $\Delta$:

• $\Delta$ square in $F$ $\Longrightarrow$ $\operatorname{Gal}(f)\cong A _ 3$
• $\Delta$ non-square $\Longrightarrow$ $\operatorname{Gal}(f)\cong S _ 3$ @important~

Overall idea:
The extension is the compositum of two quadratic fields, giving a Klein four group.

Degree: $[K:\mathbb Q]=4$.
Automorphisms:

They generate $V _ 4\cong C _ 2\times C _ 2$.
Fixed fields:

Overall idea:
The splitting field of $x^3-2$ has group $S _ 3$; list its subgroups and fixed fields.

Generators:
$\sigma(\sqrt[3]{2})=\omega\sqrt[3]{2}, \sigma(\omega)=\omega$ (3-cycle)
$\tau(\sqrt[3]{2})=\sqrt[3]{2}, \tau(\omega)=\omega^2$ (transposition on $\omega$).

Subgroups ↔ subfields: