Show that \(\sqrt{2+\sqrt{2}}\) is a root of \(p(x) = x^4 - 4x^2 + 2 \in \mathbb{Q}[x]\).
Prove that \(\mathbb{Q}(\sqrt{2 + \sqrt{2}})\) is a Galois extension of \(\mathbb{Q}\) and find its Galois group.
Let \(f(x) = x^3 - 5\). Determine the splitting field \(K\) of \(f(x)\) over \(\mathbb{Q}\) and the Galois group of \(f(x)\). Give an example of a proper sub-extension \(\mathbb{Q} \subset L \subset K\), such that \(L/\mathbb{Q}\) is Galois.
The source prints \(p(x)=x^2-4x^2+2\). This is a misprint for \(x^4-4x^2+2\): at \(x=\sqrt{2+\sqrt2}\) the printed expression evaluates to \(-4-3\sqrt2\ne0\).
Hint
The number \(\sqrt{2-\sqrt2}\) is another root of \(p(x)\).
Solution
Put \(a=\sqrt{2+\sqrt2}>0\) and \(b=\sqrt{2-\sqrt2}>0\).
1. The number \(a\) has minimal polynomial\(p(x)=x^4-4x^2+2\) over \(\mathbb Q\).
Since \(a^2-2=\sqrt2\), squaring gives \((a^2-2)^2=2\), or \(a^4-4a^2+2=0\). The polynomial is Eisenstein at two, so it is irreducible over \(\mathbb Q\) [@DF04]. Thus it is the minimal polynomial and \([\mathbb Q(a):\mathbb Q]=4\).
2. The extension \(\mathbb Q(a)/\mathbb Q\) isGalois with cyclic group of order four.
The four roots of \(p\) are \(a,-a,b,-b\): their squares are the two roots \(2+\sqrt2,2-\sqrt2\) of \(y^2-4y+2\). These four real roots are distinct. Moreover \(ab=\sqrt2=a^2-2\), so \[ b=\frac{a^2-2}{a}\in\mathbb Q(a). \] Thus this field contains every root and is the splitting field of \(p\), a separable polynomial. It is Galois [@DF04].
Sending \(a\) to the root \(b\) of its minimal polynomial defines a \(\mathbb Q\)-embedding \(\sigma\) of \(\mathbb Q(a)\) into itself. Its image is a four-dimensional rational subspace, so it is onto and is an automorphism. Applying it to \(\sqrt2=a^2-2\) gives \(\sigma(\sqrt2)=b^2-2=-\sqrt2\). Consequently \(\sigma(b)=\sigma(\sqrt2/a)=-\sqrt2/b=-a\). The successive images are \[ a\longmapsto b\longmapsto-a\longmapsto-b\longmapsto a. \] Thus \(\sigma\) has order four and its powers exhaust the Galois group of order four. The group is \(C_4\), not the Klein four-group.
3. For \(f(x)=x^3-5\), the splitting field and Galoisgroup are\[ K=\mathbb Q(c,\zeta),\qquad \operatorname{Gal}(K/\mathbb Q)\cong S_3, \quad c=\sqrt[3]{5}>0,\quad \zeta=e^{2\pi i/3}. \]
The roots are \(c,\zeta c,\zeta^2c\). Their field contains \(c\) and their ratio \(\zeta\), so it is exactly \(K\). Eisenstein at five makes \(x^3-5\) irreducible [@DF04]. Thus \(\mathbb Q(c)\) has degree three. It is real, whereas \(\zeta\) is nonreal and satisfies \(x^2+x+1=0\). Adjoining \(\zeta\) therefore has degree two, giving \([K:\mathbb Q]=6\).
As a splitting field in characteristic zero, \(K\) is Galois. Its automorphisms permute the three roots faithfully because the roots generate \(K\). Its order six equals \(|S_3|\), so it acts as every permutation. More explicitly, a three-cycle is \(r(c)=\zeta c\), \(r(\zeta)=\zeta\); it exists because \([K:\mathbb Q(\zeta)]=3\) makes \(x^3-5\) still minimal over that field. Complex conjugation \(s\) fixes \(c\) and interchanges the other two roots. These give \(r^3=s^2=1\) and \(srs=r^{-1}\) and generate the group.
4. A proper Galois subextension in part (3) is\(L=\mathbb Q(\zeta)\).
The field \(L\) is the splitting field of \(x^2+x+1\), with two distinct roots \(\zeta,\zeta^2\), so it is Galois over \(\mathbb Q\). Its degree is two because its roots are nonreal and hence not rational. Since \([K:\mathbb Q]=6\), both inclusions \(\mathbb Q\subsetneq L\subsetneq K\) are strict.