Exercise 7.3.5

Proof. Consider the extension $F\subset ℂ(t^4)\subset L=ℂ(t)$, where $t$ is a variable.

(a)

$t^4\in F$, thus $f=x^4-t^4\in F[x]$, and $f=(x-t)(x+t)(x-\mathit{it})(x+\mathit{it})$.

$f$ splits completely on $L$, and the roots of $f$ in $L$ are $t,\mathit{it},-t,-\mathit{it}$. The splitting field of $f$ over $ℂ(t^4)$ is so $ℂ(t^4)(t,\mathit{it},-t,-\mathit{it})=ℂ(t^4,t)=ℂ(t)$, since $t^4\in ℂ(t)$.

$ℂ(t)$ being the splitting field of the separable polynomial $f$ over $ℂ(t^4)$, $ℂ(t^4)\subset ℂ(t)$ is a Galois extension.

(b)

$t\notin ℂ(t^4)$, otherwise $t=u(t^4)∕v(t^4),u,v\in F[x],v\ne 0$, where $t$ is transcendental over $ℂ$, and the identity $u(t^4)-\mathit{tv}(t^4)=0$ is impossible, since all the monomials in $u(t^4)$ have even degree, and all the monomial in $\mathit{tv}(t^4)\ne 0$ have odd degree. Consequently the other roots of $f$ in $L$, which are $-t,\mathit{it},-\mathit{it}$, are not in $ℂ(t^4)$.

If $f$ was reducible over $F$, $f$ would be the product of two polynomials $p,q\in F[x]$ of degree 2, each gathering two factors of the form $x-i^{k}t$:

$$p=(x-i^{k}t)(x-i^{l}t)\in F[x],0\le k,l\le 3.$$

But then the coefficient of degree 0 in $x$, which is $i^{k+l}{t}^2$ is in $F$, therefore $t^2\in F$ :

$$t^2=\frac{u(t^4)}{v(t^4)},u,v\in F[x],v\ne 0,u\wedge v=1.$$

Then $s=t^2$ is transcendental over $ℂ$ (otherwise $t$ would be algebraic over $ℂ$) and satisfies

$$s=\frac{u(s^2)}{v(s^2)}.$$

The identity $u(s^2)-\mathit{sv}(s^2)=0$, with $s$ transcendental, implies $u(x^2)=\mathit{xv}(x^2)$ where $x$ is a variable, so is impossible, since all the monomials in $u(x^2)$ have even degree, and all the monomial in $\mathit{xv}(x^2)\ne 0$ have odd degree.

$f=x^4-t^4$ is so irreducible over $ℂ(t^4)$.

(c)

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=4$, and $f$ is monic irreducible, so is the minimal polynomial of $t$ over $F$. Therefore $$|\mathrm{Gal}(L∕F)|=[L:F]=[ℂ(t^4,t):ℂ(t^4)]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=4.$$

Let $\sigma \in G=\mathrm{Gal}(L∕F)$. As $t$ is a root of $f\in F[x]$, $\sigma (t)$ is a root of $f$, thus $\sigma (t)\in \{t,\mathit{it},i^{2}t,i^{3}t\}$. Moreover $L=ℂ(t)=ℂ(t^4)(t)$, so $\sigma$ is uniquely determined by the image of $t$. As $|G|=4$, these four possibilities occur and correspond to an element of $G$: if $0\le k\le 3$, there exists one and only one ${\sigma }_k\in G$ such that

$${\sigma }_k(t)=i^{k}t.$$

Let $\sigma ={\sigma }_1:t\mapsto \mathit{it}$. Then ${\sigma }^k(t)=i^{k}t={\sigma }_k(t)$, so ${\sigma }^k={\sigma }_k$, and $G=⟨\sigma ⟩$ is cyclic.

$$G=\{e,\sigma ,{\sigma }^2,{\sigma }^3\}\simeq \mathbb{Z}∕4\mathbb{Z}.$$

(d)

The only non trivial subgroup of $G$ is $H=⟨{\sigma }^2⟩=\{e,{\sigma }^2\}$,where $G$ is an Abelian group, so $H$ is normal in $G$. Let $L_H$ its fixed field. By the Fundamental Theorem of Galois Theory, there exists thus a unique intermediate field distinct of $ℂ(t^4)$ and $ℂ(t)$, which is thus $ℂ(t^2)$ : $$L_H=ℂ(t^2).$$

The Galois correspondence is between the two chains:

$$\begin{array}{ccc}ℂ(t^4)\hfill & \subset ℂ(t^2)\hfill & \subset ℂ(t),\hfill \\ G=⟨\sigma ⟩\hfill & \supset ⟨{\sigma }^2⟩\hfill & \supset \{e\}.\hfill \end{array}$$