Chapter 3 Abstract algebra (2)

Solution of 1:

a) Yes.
Let $L$ be the splitting field of $f(x)=x^{3}-2 \in \mathbb{Q}[x]$. Then $L=\mathbb{Q}(\sqrt[3]{2}, \omega)$ where $\omega=\frac{-1+\sqrt{-3}}{2}$. Since $f(x)$ is separable of degree 3, it follows that $|\operatorname{Gal}(L / \mathbb{Q})|^{2}=[L: \mathbb{Q}]$ and $\operatorname{Gal}(L / \mathbb{Q})$ is a
subgroup of $S_{3}$, which is the permutation group of the three roots of $f(x) .$ Note that $K=\mathbb{Q}(\sqrt{-3})=\mathbb{Q}(\omega) .$ 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, $[L: \mathbb{Q}]=[L: K][K: \mathbb{Q}]=6$. Hence $\operatorname{Gal}(L / \mathbb{Q}) \cong S_{3}$.

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

Solution of 3:

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

(b) We have $N \cong \mu_{8}$, with elements
$$ \begin{aligned} \sigma_{\xi}: 5^{1 / 8} & \mapsto \xi \cdot 5^{1 / 8} & & \xi \in \mu_{8} \\ \zeta & \mapsto \zeta & & \end{aligned} $$
$H \cong(\mathbb{Z} / 8 \mathbb{Z})^{\times}$, with elements
$$ [a]: 5^{1 / 8} \mapsto 5^{1 / 8} \quad a \in(\mathbb{Z} / 8 \mathbb{Z})^{\times} $$
$$ \zeta \mapsto \zeta^{a} $$

The commutation rule is given by $[a] \circ \xi \circ\left[a^{-1}\right]$, because
$$ [a] \circ \xi \circ\left[a^{-1}\right]: 5^{1 / 8} \mapsto \xi^{a} \cdot 5^{1 / 8} \quad \forall[a] \in H, \forall \xi \in N $$
$\zeta \mapsto \zeta$
So the Galois group $G=\operatorname{Gal}(F / \mathbb{Q})$ is naturally isomorphic to the semi-direct product $(\mathbb{Z} / 8 \mathbb{Z})^{\times} \rtimes_{\varphi} \mu_{8}$, when the twisting datum $\varphi$ is the group homomorphism (actually an isomorphism)
$$ \begin{aligned} \varphi:(\mathbb{Z} / 8 \mathbb{Z})^{\times} & \mapsto \operatorname{Aut}\left(\mu_{8}\right) \\ a & \mapsto\left(\xi \mapsto \xi^{a}\right) \quad \xi \in \mu_{8} \end{aligned} $$

Solution of :

Let $\alpha \in L ;$ we want to show $\alpha$ is algebraic over $F$. Let $\alpha=\sum_{i=1}^{m} b_{i} \alpha_{i}$ for $b_{i} \in K, \alpha_{i} \in L$ linearly independent and algebraic over $K .$ Write $\alpha_{i}=\sum_{j=1}^{n} c_{j} \beta_{j}$ for $c_{j} \in F, \beta_{j} \in K$ linearly independent and algebraic over $F$. Thus, we have $\alpha \in$ $F\left(\beta_{1}, \ldots, \beta_{n}\right)\left(\alpha_{1}, \ldots, \alpha_{m}\right) .$
Now if $\alpha$ is transcendental over $F$, then
$$ \left[F\left(\beta_{1}, \ldots, \beta_{n}\right)\left(\alpha_{1}, \ldots, \alpha_{m}\right): F\left(\beta_{1}, \ldots, \beta_{n}\right)\right]\left[F\left(\beta_{1}, \ldots, \beta_{n}\right): F\right] \geq[F(\alpha): F]=\infty $$
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 $\left[F_{1}: F\right] \mid m !$ by inductive hypothesis. Similarly, $\left[K: F_{1}\right] \mid(n-m) !$ 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 $\alpha$. Thus, $\left[K: F_{1}\right] \mid(n-1) !$ by inductive hypothesis since $K$ is a splitting field for $f /(x-\alpha)$ over $F_{1}$, and so $[K: F] \mid n(n-1) !=n !$ as above.