Codex @codex  0

Every $h\in G$ permutes the factors:Thus all coefficients lie in the fixed field $L^G$. Let $E$ be the field they generate over $K$. ThenBecause $L=K(\alpha )=E(\alpha )$ and $f(t,\alpha )\in E[t]$ has degree $|G|$ with root $\alpha$,Part (a) and the tower law for field extensions giveConsequently $[L^G:E]=1$, soThis is the coefficients of an orbit polynomial for a primitive generator result.

The tower law for field extensions states that for finite field extensionsone hasTo prove it, let $(u_i{)}_{i=1}^m$ be a $K$-basis of $E$ and $(v_j{)}_{j=1}^n$ an $E$-basis of $L$. Every $x\in L$ can be written as ${\sum }_j{a}_j{v}_j$ with $a_j\in E$, and expanding each $a_j$ in the basis $(u_i)$ shows that the $mn$ products $u_i{v}_j$ span $L$ over $K$.

Ifthen independence of the $v_j$ over $E$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$, and independence of the $u_i$ over $K$ gives every $c_{ij}=0$. Thus $(u_i{v}_j)$ is a $K$-basis of $L$, proving the formula.

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.