Chapter 3 Abstract algebra (2)

Discrete Mathematics

CHAPTER 3 (part 2)

Let $K = Q (\sqrt{- 3})$ , an imaginary quadratic field.
(a) Does there exists a finite Galois extension L/Q which contains $K$ such that $Gal (L / Q) ≅ S_{3}$ ? (Here $S_{3}$ is the symmetric group in 3 letters $)$
(b) Does there exists a finite Galois extension $L / Q$ which contains $K$ such that $Gal (L / Q) ≅ Z / 4 Z ?$

(a) Determine all automorphisms of the field $Q (\sqrt[3]{2})$ , and of the field $Q (\sqrt[3]{2}, ω)$ , where $ω = e^{2 π i / 3}$ .
(b) Let $K$ be the splitting field over $Q$ of $f (x) = (x^{2} - 2 x - 1) (x^{2} - 2 x - 7)$ . Determine all automorphisms of $K$ .

Let $F$ be a splitting field over $Q$ of the polynomial $x^{8} - 5 \in Q [x]$ . Recall that $F$ is the subfield of $C$ generated by all roots of this polynomial.
(a) Find the degree $[F : Q]$ of the number field $F$ .
(b) Determine the Galois group $Gal (F / Q)$ .

For a positive integer $a$ , consider the polynomial
$f_{a} = x^{6} + 3 a x^{4} + 3 x^{3} + 3 a x^{2} + 1$
Let $F$ be the splitting field of $f_{a}$ . Show that its Galois group is solvable.

A field extension $K / F$ is an algebraic extension if every element of $K$ is algebraic over $F$ . Let $K / F$ and $L / K$ be algebraic field extensions. Prove that $L / F$ is an algebraic extension.

Let $f$ be a polynomial of degree $n$ with coefficients in $F$ and let $K$ be a splitting field for $f$ over $F$ . Prove that $[K : F]$ divides $n!$

A module is called simple if it is not the zero module and if it has no proper submodule.
(a) Prove that any simple $R$ module is isomorphic to an $R$ module of the form $R / M$ where $M$ is a maximal ideal.
(b) Let $φ : S \to S^{'}$ be a homomorphism of simple modules. Prove that $φ$ is either zero, or an isomorphism.
of

Solution

Solution of 1:

a) Yes.
Let $L$ be the splitting field of $f (x) = x^{3} - 2 \in Q [x]$ . Then $L = Q (\sqrt[3]{2}, ω)$ where $ω = \frac{- 1 + \sqrt{- 3}}{2}$ . Since $f (x)$ is separable of degree 3, it follows that $| Gal (L / Q) |^{2} = [L : Q]$ and $Gal (L / Q)$ is a
subgroup of $S_{3}$ , which is the permutation group of the three roots of $f (x) .$ Note that $K = Q (\sqrt{- 3}) = Q (ω) .$ We have $K \subset L$ and that the minimal polynomial $h (x)$ of $\sqrt[3]{2}$ over $K$ , which divides $f (x)$ , is of degree 3. As a result, $[L : Q] = [L : K] [K : Q] = 6$ . Hence $Gal (L / Q) ≅ S_{3}$ .

b) No.
Suppose that such $L$ exists. Then $Gal (L / Q) ≅ Z / 4 Z$ contains a unique element $τ$ of order 2 and hence $τ \in Gal (L / K)$ since $Gal (L / K)$ is a subgroup of $Gal (L / Q)$ of order $2 .$ On the other hand, since $L / Q$ is a finite Galois extension, $L$ is a splitting field of a separable polynomial $f (x) \in Q [x] .$ Let $θ$ be the complex conjugate automorphism of $C$ . Then $θ (f) = f$ and hence $θ_{L} \in Gal (L / Q)$ . Because $\sqrt{- 3} \notin R, θ_{L}$ has order 2 .
Since $Z / 4 Z$ contains a unique order 2 element, $θ_{L} = τ \in$ $Gal (L / Q)$ , a contradiction.

Solution of 3:

(a) We have $F := Q (5^{1 / 8}, ζ)$ , where $ζ = ζ_{8} = e^{π \sqrt{- 1} / 4}$ . Denote by $E := Q (5^{1 / 8})$ and $K := Q (ζ)$
The polynomial $x^{8} - 5 \in Q [x]$ is irreducible by Einsenstein's criterion. So $[E : Q] = 8$ . Similarly $x^{4} + 1$ is irreducible in $Q [x]$ , so $[K : Q] = 4$ . Let $G := Gal (F / Q), N := Gal (F / K)$ , and $H := Gal (F / Q)$ . Clearly
$K / Q$ is cyclic, and its only quadratic subfield is $Q (\sqrt{- 1})$ . So $E \cap K =$ $Q$ , which implies that $[F : Q] = 8 * 4 = 32$ .

(b) We have $N ≅ μ_{8}$ , with elements
$\begin{aligned} σ_{ξ} : 5^{1 / 8} & \mapsto ξ \cdot 5^{1 / 8} & ξ \in μ_{8} \\ ζ & \mapsto ζ \end{aligned}$
$H ≅ (Z / 8 Z)^{\times}$ , with elements
$[a] : 5^{1 / 8} \mapsto 5^{1 / 8} a \in (Z / 8 Z)^{\times}$
$ζ \mapsto ζ^{a}$

The commutation rule is given by $[a] \circ ξ \circ [a^{- 1}]$ , because
$[a] \circ ξ \circ [a^{- 1}] : 5^{1 / 8} \mapsto ξ^{a} \cdot 5^{1 / 8} \forall [a] \in H, \forall ξ \in N$
$ζ \mapsto ζ$
So the Galois group $G = Gal (F / Q)$ is naturally isomorphic to the semi-direct product $(Z / 8 Z)^{\times} ⋊_{φ} μ_{8}$ , when the twisting datum $φ$ is the group homomorphism (actually an isomorphism)
$\begin{aligned} φ : (Z / 8 Z)^{\times} & \mapsto Aut (μ_{8}) \\ a & \mapsto (ξ \mapsto ξ^{a}) ξ \in μ_{8} \end{aligned}$

Solution of :

Let $α \in L;$ we want to show $α$ is algebraic over $F$ . Let $α = \sum_{i = 1}^{m} b_{i} α_{i}$ for $b_{i} \in K, α_{i} \in L$ linearly independent and algebraic over $K .$ Write $α_{i} = \sum_{j = 1}^{n} c_{j} β_{j}$ for $c_{j} \in F, β_{j} \in K$ linearly independent and algebraic over $F$ . Thus, we have $α \in$ $F (β_{1}, \dots, β_{n}) (α_{1}, \dots, α_{m}) .$
Now if $α$ is transcendental over $F$ , then
$[F (β_{1}, \dots, β_{n}) (α_{1}, \dots, α_{m}) : F (β_{1}, \dots, β_{n})] [F (β_{1}, \dots, β_{n}) : F] \geq [F (α) : F] = \infty$
which contradicts it.

Solution of 6 :

We use induction on the degree of $f$ . The claim is trivial if $n = 1$ , so suppose $n > 1$ . Suppose $f$ is reducible, and let $g$ be an irreducible factor of $f$ of degree $m .$ Then, we can choose a subfield $F_{1} \subset K$ such that $g$ splits completely, hence $[F_{1} : F] ∣ m!$ by inductive hypothesis. Similarly, $[K : F_{1}] ∣ (n - m)!$ by inductive hypothesis since $K$ is a splitting field for $f / g$ over $F_{1}$ . Hence by the multiplicative property of the degree , $[K : F] | m! (n - m)! | n!$

Now suppose $f$ is irreducible. Then, $F_{1} := F [x] / (f)$ is a field extension of degree $n$ such that $f$ has a root $α$ . Thus, $[K : F_{1}] ∣ (n - 1)!$ by inductive hypothesis since $K$ is a splitting field for $f / (x - α)$ over $F_{1}$ , and so $[K : F] ∣ n (n - 1)! = n!$ as above.