Let \(K\) be the splitting field of \(x^4 - 6\) over \(\mathbb{Q}\).
What is the dimension of \(K\) over \(\mathbb{Q}\)?
What is the Galois group of \(K\) over \(\mathbb{Q}\)? You must not only identify the group, but describe how it acts on the roots of \(x^4 - 6\).
How many subfields does \(K\) have of each possible dimension over \(\mathbb{Q}\)?
How many of those subfields in each dimension are Galois over \(\mathbb{Q}\)?
Solution
Set \(a=\sqrt[4]{6}>0\), and label the roots by \(a_j=i^ja\) for \(j\in\mathbb Z/4\mathbb Z\).
1. The field is \(K=\mathbb Q(a,i)\) and\([K:\mathbb Q]=8\).
All four roots lie in \(\mathbb Q(a,i)\). Conversely, the field generated by the roots contains \(a\) and \(i=(ia)/a\). The polynomial \(x^4-6\) is Eisenstein at \(2\), so \([\mathbb Q(a):\mathbb Q]=4\) [@DF04]. This field lies in \(\mathbb R\), whereas \(i\) does not. The minimal polynomial of \(i\) over it is therefore \(x^2+1\), and the tower law gives \([K:\mathbb Q]=2\cdot4=8\). As a splitting field in characteristic zero, \(K/\mathbb Q\) is Galois [@DF04].
2. The Galois group \(G\) is the dihedral group of order \(8\),with generators\[ r(a)=ia,\quad r(i)=i,\qquad s(a)=a,\quad s(i)=-i. \]On the root indices, these act as\(r:j\mapsto j+1\) and \(s:j\mapsto-j\).
Since \([\mathbb Q(i):\mathbb Q]=2\), step 1 gives \([K:\mathbb Q(i)]=4\). Thus \(x^4-6\) is still the minimal polynomial of \(a\) over \(\mathbb Q(i)\). Substitution of its root \(ia\) for \(a\) therefore defines a \(\mathbb Q(i)\)-embedding of \(K\) into itself. Its image contains \(i\) and \(a=-i(ia)\), so it is the automorphism \(r\). Complex conjugation gives the automorphism \(s\).
Their actions on \(a_j\) give the stated permutations. In particular, \(r=(0\,1\,2\,3)\) and \(s=(1\,3)\). The eight products \(r^us^v\), with \(0\leq u<4\) and \(v\in\{0,1\}\), act by \[ j\longmapsto u+(-1)^vj\pmod4. \] They are distinct: evaluating at \(0\) determines \(u\), and evaluating at \(1\) distinguishes the two values of \(v\). The Galois group has order \(8\), so these are all its elements. They satisfy \(r^4=s^2=1\) and \(srs=r^{-1}\) and give precisely the four rotations and four reflections of the square on the root indices. This identifies \(G\) with the dihedral group of order \(8\).
3. The subgroups of \(G\), listed without repetitions, are\[ \{1\},\quad \langle r^2\rangle,\quad \langle r^j s\rangle\ (j=0,1,2,3),\quad \langle r\rangle,\quad \langle r^2,s\rangle,\quad \langle r^2,rs\rangle,\quad G. \]
Write \(R=\langle r\rangle\), the cyclic rotation subgroup of order \(4\). Its only subgroups are \(\{1\}\), \(\langle r^2\rangle\), and \(R\): a subgroup containing \(r\) or \(r^3\) is \(R\), and any other subgroup is contained in \(\{1,r^2\}\).
Let \(H\leq G\), and put \(T=H\cap R\). If \(H\subseteq R\), it is one of those three subgroups. Otherwise choose \(r^j s\in H\). Every element of \(H\) outside \(R\) lies in the coset \((r^j s)T\), since the quotient of two reflections is a rotation in \(H\). Hence \[ H=T\mathbin{\sqcup}(r^j s)T. \] If \(T=\{1\}\), this gives the four order-\(2\) subgroups \(\langle r^j s\rangle\). Each is a subgroup because \((r^j s)^2=1\), and the four are distinct because their nonidentity elements are distinct.
If \(T=\langle r^2\rangle\), this gives \(\{1,r^2,r^j s,r^{j+2}s\}\), which depends exactly on the parity of \(j\). These are the two subgroups \(\langle r^2,s\rangle\) and \(\langle r^2,rs\rangle\); the element \(r^2\) commutes with each reflection, so both displayed sets are indeed subgroups of order \(4\). Finally, \(T=R\) gives all of \(G\). These cases exhaust the possible intersections \(T\), proving completeness of the list.
4. The subfield counts, including \(\mathbb Q\) and \(K\),are as follows.
| \([L:\mathbb Q]\) | Number of subfields \(L\) | Number Galois over \(\mathbb Q\) |
|---|---|---|
| \(1\) | \(1\) | \(1\) |
| \(2\) | \(3\) | \(3\) |
| \(4\) | \(5\) | \(1\) |
| \(8\) | \(1\) | \(1\) |
The Galois correspondence sends a subgroup \(H\) to its fixed field \(K^H\), with \([K^H:\mathbb Q]=8/|H|\). Moreover, \(K^H/\mathbb Q\) is Galois exactly when \(H\) is normal in \(G\) [@DF04]. Step 3 lists one subgroup of order \(1\), five of order \(2\), three of order \(4\), and one of order \(8\). This gives the middle column; no other degree is possible by the tower law.
The subgroups \(\{1\}\) and \(G\) are normal. All three order-\(4\) subgroups have index \(2\) and are normal: both the left and right coset of an element outside such a subgroup are its complement. The subgroup \(\langle r^2\rangle\) is central, since \(r^2\) commutes with \(r\) and \(sr^2s=r^{-2}=r^2\).
None of the four reflection subgroups is normal, since \[ r(r^j s)r^{-1}=r^{j+2}s\ne r^j s. \] Conjugation therefore takes its unique nonidentity element outside that subgroup. These are all the order-\(2\) subgroups apart from \(\langle r^2\rangle\), so exactly one quartic subfield is Galois. This proves the last column.