- For \(u=\sqrt{2+\sqrt2}\), determine the Galois group of \(\QQ(u)/\QQ\) and all intermediate fields.
Solution
Set \[ L=\QQ(u), \qquad u=\sqrt{2+\sqrt2}. \]
1. The minimal polynomial of \(u\) over \(\QQ\) is\[ f(x)=x^4-4x^2+2, \]and \([L:\QQ]=4\).
From \[ u^2=2+\sqrt2 \] we obtain \[ (u^2-2)^2=2, \] so \(f(u)=0\). Also \[ \sqrt2=u^2-2\in L, \] so \(\QQ(\sqrt2)\subseteq L\).
Let \(E=\QQ(\sqrt2)\). If \(u\in E\), then \(2+\sqrt2\) would be a square in \(E\). But \[ N_{E/\QQ}(2+\sqrt2)=(2+\sqrt2)(2-\sqrt2)=2, \] which is not a square in \(\QQ\), whereas the norm of a square in \(E\) is a square in \(\QQ\). Hence \(u\notin E\), so \[ [L:E]=2. \] Since \([E:\QQ]=2\), one gets \([L:\QQ]=4\). Therefore the degree-\(4\) polynomial \(f\) is the minimal polynomial of \(u\).
2. The polynomial \(f\) splits in \(L\).
Put \[ v=\sqrt{2-\sqrt2}. \] Then \[ uv=\sqrt{(2+\sqrt2)(2-\sqrt2)}=\sqrt2, \] so \[ v=\frac{\sqrt2}{u}\in L. \] The four roots of \(f\) are \[ \pm u,\qquad \pm v, \] all of which lie in \(L\). Thus \(L/\QQ\) is the splitting field of the separable polynomial \(f\), hence is Galois.
3. The Galois group is cyclic of order \(4\).
Define \(\sigma\in\Gal(L/\QQ)\) by \[ \sigma(u)=v. \] Since \[ \sqrt2=u^2-2, \] we have \[ \sigma(\sqrt2)=v^2-2=-\sqrt2. \] Using \(v=\sqrt2/u\), \[ \sigma(v)=\frac{\sigma(\sqrt2)}{\sigma(u)} =\frac{-\sqrt2}{v} =-u. \] Therefore \[ u\xmapsto{\sigma}v\xmapsto{\sigma}-u\xmapsto{\sigma}-v\xmapsto{\sigma}u. \] Thus \(\sigma\) has order \(4\). Since \(|\Gal(L/\QQ)|=[L:\QQ]=4\), it follows that \[ \Gal(L/\QQ)=\langle\sigma\rangle\cong C_4. \]
4. The only proper nontrivial intermediate field is\[ \QQ(\sqrt2). \]
A cyclic group of order \(4\) has a unique subgroup of order \(2\), namely \(\langle\sigma^2\rangle\). By the Galois correspondence, \(L/\QQ\) therefore has a unique intermediate field of degree \(2\) over \(\QQ\).
Now \(\sigma^2(u)=-u\), so \[ \sigma^2(\sqrt2)=\sigma^2(u^2-2)=u^2-2=\sqrt2. \] Hence \(\QQ(\sqrt2)\) is fixed by \(\langle\sigma^2\rangle\). Since it has degree \(2\) over \(\QQ\), it is exactly the corresponding fixed field.
5. Hence the full intermediate-field lattice is\[ \QQ\subsetneq\QQ(\sqrt2)\subsetneq L=\QQ(\sqrt{2+\sqrt2}). \]
This is 4 together with the bottom and top fields.