This question asks you to describe the splitting field \(K\) of the polynomial \(x^4 - 5\) over \(\mathbb{Q}\).
What is \([K:\mathbb{Q}]\)?
What is the Galois group \(G\) of \(K\) over \(\mathbb{Q}\)?
Find all the subgroups of \(G\) and describe their fixed fields.
Which of the fixed fields in part (c) is Galois over \(\mathbb{Q}\)?
Solution
Put \(a=\sqrt[4]{5}>0\). The splitting field is \(K=\mathbb Q(a,i)\), of degree \(8\), and its Galois group is dihedral of order \(8\). Use generators \[ r(a)=ia,\quad r(i)=i,\qquad s(a)=a,\quad s(i)=-i. \] The subgroups of \(G\) and their fixed fields are:
| Subgroup \(H\) | Order | Fixed field \(K^H\) | Degree over \(\mathbb Q\) | Galois over \(\mathbb Q\)? |
|---|---|---|---|---|
| \(G\) | \(8\) | \(\mathbb Q\) | \(1\) | Yes |
| \(\langle r\rangle\) | \(4\) | \(\mathbb Q(i)\) | \(2\) | Yes |
| \(\langle r^2,s\rangle\) | \(4\) | \(\mathbb Q(a^2)\) | \(2\) | Yes |
| \(\langle r^2,rs\rangle\) | \(4\) | \(\mathbb Q(ia^2)\) | \(2\) | Yes |
| \(\langle r^2\rangle\) | \(2\) | \(\mathbb Q(a^2,i)\) | \(4\) | Yes |
| \(\langle s\rangle\) | \(2\) | \(\mathbb Q(a)\) | \(4\) | No |
| \(\langle r^2s\rangle\) | \(2\) | \(\mathbb Q(ia)\) | \(4\) | No |
| \(\langle rs\rangle\) | \(2\) | \(\mathbb Q((1+i)a)\) | \(4\) | No |
| \(\langle r^3s\rangle\) | \(2\) | \(\mathbb Q((1-i)a)\) | \(4\) | No |
| \(\{1\}\) | \(1\) | \(K\) | \(8\) | Yes |
1. The field degree and group presentation are as stated.
The four roots are \(a,ia,-a,-ia\). They generate \(\mathbb Q(a,i)\), since their ratio \((ia)/a\) is \(i\). Eisenstein’s criterion at \(5\) makes \(x^4-5\) irreducible over \(\mathbb Q\), so \(\mathbb Q(a)\) has degree \(4\) [@DF04]. It is a real field and does not contain \(i\), whose polynomial \(x^2+1\) has degree \(2\). Hence \([K:\mathbb Q]=4\cdot2=8\). As a splitting field in characteristic zero, this is a Galois extension.
Since \([K:\mathbb Q(i)]=4\), the minimal polynomial of \(a\) over \(\mathbb Q(i)\) is still \(x^4-5\). Sending \(a\) to its root \(ia\) 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 constructs \(r\). Complex conjugation constructs \(s\). Evaluation on \(a,i\) gives \[ r^4=s^2=1,\qquad srs=r^{-1}. \] The eight elements \(r^j s^e\), for \(0\leq j<4\) and \(e\in\{0,1\}\), are distinct: their actions on \(i\) determine \(e\), and their actions on \(a\) determine \(j\). They exhaust the order-\(8\) Galois group. The relations reduce any word to this form, with multiplication \[ (r^j s^e)(r^k s^f)=r^{j+(-1)^e k}s^{e+f}, \] reducing exponents modulo \(4\) and \(2\). Thus they present the dihedral group of order \(8\).
2. The table contains every subgroup, exactly once.
Let \(R=\langle r\rangle\), a normal subgroup of index \(2\). For \(H\leq G\), the quotient map to \(G/R\) has kernel \(H\cap R\) on \(H\), so \([H:H\cap R]\leq2\). The only subgroups of the cyclic group \(R\) are \(1\), \(\langle r^2\rangle\), and \(R\).
If \(H\cap R=1\), then \(H=1\) or \(H\) has order \(2\) and is generated by one of \(s,rs,r^2s,r^3s\). Each squares to \(1\) by the multiplication formula, and the four elements are distinct.
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 \(\langle r^2,rs\rangle\).
If \(H\cap R=R\), then \(H=R\) or \(G\). These three disjoint cases give precisely the ten subgroups in the table.
3. Every field in the table is the indicated fixed field.
The Galois correspondence gives \([K^H:\mathbb Q]=8/|H|\) [@DF04]. It suffices to show that the proposed generators are fixed by \(H\) and generate a field of this degree.
The three quadratic generators \(i\), \(a^2=\sqrt5\), and \(ia^2=\sqrt{-5}\) are fixed respectively by \(\langle r\rangle\), \(\langle r^2,s\rangle\), and \(\langle r^2,rs\rangle\). Their degrees are \(2\): \(i\) and \(ia^2\) are nonreal roots of rational quadratics, while \(x^2-5\) is irreducible by Eisenstein. The generator \(r^2\) fixes both \(a^2\) and \(i\). The field \(\mathbb Q(a^2,i)\) has degree \(4\), since adjoining \(i\) to the real quadratic field \(\mathbb Q(\sqrt5)\) has degree \(2\).
For the remaining order-two subgroups, direct substitution 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 \(b\) denote any of these four nonzero fixed generators. Since \(r\) fixes \(i\) and sends \(a\) to \(ia\), the four distinct elements \(b,ib,-b,-ib\) are images of \(b\) under automorphisms of \(K\). They are roots of its minimal polynomial over \(\mathbb Q\), so \([\mathbb Q(b):\mathbb Q]\geq4\). On the other hand, \(\mathbb Q(b)\) lies in the fixed field of the indicated subgroup of order \(2\), which has degree \(4\). Equality follows, proving every remaining quartic row.
Finally \(K^G=\mathbb Q\) and \(K^{\{1\}}=K\). By step 2 and the Galois correspondence, the table lists every intermediate field of \(K/\mathbb Q\); inclusions between fields reverse the inclusions of their subgroups.
4. Exactly the six fields marked Yes are Galois over \(\mathbb Q\).
An intermediate field \(K^H\) is Galois over \(\mathbb Q\) exactly when \(H\lhd G\) [@DF04]. The trivial and whole subgroups are normal, as are all three subgroups of order \(4\), since they have index \(2\). The subgroup \(\langle r^2\rangle\) is central: \(r\) commutes with \(r^2\) and \(sr^2s=r^{-2}=r^2\).
None of the four reflection subgroups is normal, since \[ r(r^j s)r^{-1}=r^{j+2}s\ne r^j s \] and a subgroup of order \(2\) has only one nonidentity element. Thus the Galois intermediate fields are precisely \[ \mathbb Q,\quad \mathbb Q(i),\quad \mathbb Q(\sqrt5),\quad \mathbb Q(\sqrt{-5}),\quad \mathbb Q(\sqrt5,i),\quad K. \]