Let \(K\) be the splitting field of \(x^4 - 2\) over \(\mathbb{Q}\). What is \([K:\mathbb{Q}]\)? Describe the Galois group \(G = \text{Gal}(K/\mathbb{Q})\): either find generators of the group and indicate relations among those generators which determine its multiplication table, or identify a familiar group to which \(G\) is isomorphic. Find all the subgroups of \(G\) and their orders, and indicate the corresponding subfields of \(K\).
Solution
Put \(a=\sqrt[4]{2}>0\). Then \(K=\mathbb Q(a,i)\) has degree \(8\) over \(\mathbb Q\), and its Galois group is the dihedral group of order \(8\). Define its generators by \[ r(a)=ia,\quad r(i)=i,\qquad s(a)=a,\quad s(i)=-i. \] All subgroups and their fixed fields are as follows.
| Subgroup \(H\) | Order | Fixed field \(K^H\) | Degree over \(\mathbb Q\) |
|---|---|---|---|
| \(G\) | \(8\) | \(\mathbb Q\) | \(1\) |
| \(\langle r\rangle\) | \(4\) | \(\mathbb Q(i)\) | \(2\) |
| \(\langle r^2,s\rangle\) | \(4\) | \(\mathbb Q(a^2)\) | \(2\) |
| \(\langle r^2,rs\rangle\) | \(4\) | \(\mathbb Q(ia^2)\) | \(2\) |
| \(\langle r^2\rangle\) | \(2\) | \(\mathbb Q(a^2,i)\) | \(4\) |
| \(\langle s\rangle\) | \(2\) | \(\mathbb Q(a)\) | \(4\) |
| \(\langle r^2s\rangle\) | \(2\) | \(\mathbb Q(ia)\) | \(4\) |
| \(\langle rs\rangle\) | \(2\) | \(\mathbb Q((1+i)a)\) | \(4\) |
| \(\langle r^3s\rangle\) | \(2\) | \(\mathbb Q((1-i)a)\) | \(4\) |
| \(\{1\}\) | \(1\) | \(K\) | \(8\) |
1. The degree and group presentation are as asserted.
The roots of \(x^4-2\) are \(a,ia,-a,-ia\). Their generated field contains \(i=(ia)/a\), so it is \(\mathbb Q(a,i)\). Eisenstein’s criterion at \(2\) gives \([\mathbb Q(a):\mathbb Q]=4\) [@DF04]. Since \(\mathbb Q(a)\) is real and \(i\) is not, \([K:\mathbb Q(a)]=2\). Hence \([K:\mathbb Q]=8\). As a splitting field in characteristic zero, \(K/\mathbb Q\) is Galois, with a group of order \(8\).
The degree \([K:\mathbb Q(i)]=4\) shows that the minimal polynomial of \(a\) over \(\mathbb Q(i)\) is still \(x^4-2\). Sending \(a\) to the root \(ia\) therefore defines a \(\mathbb Q(i)\)-embedding of \(K\) into itself. Its image contains \(i\) and \(ia\), hence also \(a\), so it is an automorphism. This is \(r\). Complex conjugation restricts to the automorphism \(s\) of \(K\).
Their actions on the generators \(a,i\) give \[ r^4=s^2=1,\qquad srs=r^{-1}. \] The eight elements \(r^j s^\varepsilon\), with \(0\leq j<4\) and \(\varepsilon\in\{0,1\}\), are distinct: their actions on \(i\) determine \(\varepsilon\), and their actions on \(a\) then determine \(j\). They are therefore all of \(G\). The displayed relations reduce every word to this form, and multiplication is explicitly \[ (r^j s^\varepsilon)(r^k s^\delta) =r^{j+(-1)^\varepsilon k}s^{\varepsilon+\delta}, \] with exponents reduced modulo \(4\) and \(2\), respectively. Thus these relations give the usual presentation of the dihedral group of order \(8\) and determine its entire multiplication table.
2. The table exhausts the subgroups of \(G\).
Let \(R=\langle r\rangle\), a normal cyclic subgroup of order \(4\) and index \(2\). For any \(H\leq G\), restriction of the quotient map \(G\to G/R\) shows that \([H:H\cap R]\leq2\). The cyclic group \(R\) has precisely the subgroups \(\{1\}\), \(\langle r^2\rangle\), and \(R\).
If \(H\cap R=\{1\}\), then \(H\) is trivial or has order \(2\) and is generated by one of the four elements \(r^j s\) outside \(R\). Each of those elements has order \(2\) by the multiplication formula. This yields the trivial subgroup and the four reflection subgroups.
If \(H\cap R=\langle r^2\rangle\), then either \(H=\langle r^2\rangle\), or \(H=\langle r^2,r^j s\rangle\) has order \(4\). Even \(j\) give \(\langle r^2,s\rangle\), and odd \(j\) give \(\langle r^2,rs\rangle\). These are the two listed subgroups of order \(4\) other than \(R\).
If \(H\cap R=R\), then \(H=R\) or \(H=G\), since \(R\) has index \(2\). The three cases give exactly the ten distinct subgroups in the table.
3. Each field in the table is the fixed field of the indicated subgroup.
The Galois correspondence gives \([K^H:\mathbb Q]=8/|H|\) [@DF04]. It therefore suffices to show that each proposed field is fixed by \(H\) and has this degree.
The generators \(i\), \(a^2=\sqrt2\), and \(ia^2=\sqrt{-2}\) are fixed, respectively, by \(\langle r\rangle\), \(\langle r^2,s\rangle\), and \(\langle r^2,rs\rangle\), as follows directly from the formulas for \(r,s\). Their degrees over \(\mathbb Q\) are \(2\): the first and third are nonreal roots of rational quadratics, and \(a^2\) has minimal polynomial \(x^2-2\).
The subgroup \(\langle r^2\rangle\) fixes both \(a^2\) and \(i\). The field \(\mathbb Q(a^2,i)\) has degree \(4\), since \(\mathbb Q(a^2)\) is real quadratic and adjoining \(i\) has degree \(2\).
For the remaining four subgroups of order \(2\), direct computation gives \[ s(a)=a,\qquad (r^2s)(ia)=ia,\qquad (rs)((1+i)a)=(1+i)a,\qquad (r^3s)((1-i)a)=(1-i)a. \] Let \(b\) be any of these four nonzero fixed generators. Since \(r\) fixes \(i\) and sends \(a\) to \(ia\), the four elements \(b,ib,-b,-ib\) are distinct images of \(b\) under automorphisms of \(K\). All are roots of its minimal polynomial over \(\mathbb Q\), so \([\mathbb Q(b):\mathbb Q]\geq4\). On the other hand, \(b\) belongs to the fixed field of the indicated order-\(2\) subgroup, which has degree \(4\). Thus the degree is exactly \(4\), and \(\mathbb Q(b)\) equals that fixed field.
Finally, the endpoint fields \(K^G=\mathbb Q\) and \(K^{\{1\}}=K\) are part of the Galois correspondence. This proves every row; the subgroup enumeration in step 2 proves that no subfield has been omitted. Field inclusions are obtained by reversing the inclusions among the listed subgroups.