Fields and Galois theory

Every cyclotomic field \(L=\mathbb Q(\zeta_m)\) is the splitting field of \(x^m-1\), which is separable in characteristic zero, so \(L/\mathbb Q\) is finite Galois. An automorphism sends \(\zeta_m\) to \(\zeta_m^a\) for some unit \(a\) modulo \(m\). This embeds its Galois group into the abelian group \((\mathbb Z/m\mathbb Z)^\times\): the image of the generator determines the automorphism, and composition multiplies the exponents. Thus its Galois group is abelian. Every subgroup is normal, so every intermediate field is Galois over \(\mathbb Q\) by the Galois correspondence [@DF04].

But \(E=\mathbb Q(\sqrt[3]{2})\) is not normal over \(\mathbb Q\). Indeed, \(x^3-2\) is irreducible by Eisenstein at \(2\) [@DF04], and \(E\subseteq\mathbb R\) contains its real root but not its two nonreal roots. Hence \(E\) cannot be an intermediate field of a cyclotomic extension.

Write \(q=|\mathbb F|\), choose a root \(a\) of \(f\) in an algebraic closure, and put \(E=\mathbb F(a)\). Irreducibility gives \([E:\mathbb F]=n\), so \(E\) has \(q^n\) elements. Its multiplicative group has order \(q^n-1\), hence every \(b\in E\) satisfies \(b^{q^n}=b\), including \(b=0\). Therefore \[ x^{q^n}-x=\prod_{b\in E}(x-b). \] The equality follows because both sides are monic of degree \(q^n\) and have exactly these distinct roots.

The minimal polynomial of \(a\) over \(\mathbb F\) divides \(x^{q^n}-x\), since the latter vanishes at \(a\). It is a nonzero scalar multiple of \(f\). Thus \(f\) splits in \(E\). The field generated by all its roots is contained in \(E\) and contains \(\mathbb F(a)=E\), so equals \(E\) and has degree \(n\).

For any homomorphism \(\varphi:S_4\to A_3\), the image of a transposition has order dividing two and also dividing \(|A_3|=3\). Its order is therefore one. Every permutation is a product of transpositions: each cycle \((a_1\,a_2\,\ldots\,a_k)\) equals \((a_1\,a_k)\cdots(a_1\,a_2)\), and disjoint cycles give every permutation. Hence \(\varphi\) is trivial and cannot surject onto the three-element group \(A_3\).

Let \(X\) consist of the three unordered pair partitions \[ A=\{\{1,2\},\{3,4\}\},\quad B=\{\{1,3\},\{2,4\}\},\quad C=\{\{1,4\},\{2,3\}\}. \] Permuting the four labels sends a pair partition to another pair partition and respects composition. This defines a homomorphism \(\psi:S_4\to\operatorname{Sym}(X)\cong S_3\). The permutation \((1\,2)\) fixes \(A\) and interchanges \(B,C\), whereas \((1\,2\,3)\) sends \(A\) to \(C\), \(C\) to \(B\), and \(B\) to \(A\). The image therefore contains a transposition and a three-cycle. A subgroup containing them has order divisible by both two and three; its order divides six, so it is all of \(S_3\) by Lagrange’s theorem [@DF04]. Thus \(\psi\) is surjective.

Use the same action on \(X\). The powers of \((1\,2\,3)\) send \(A\) to each of \(A,B,C\), as explicitly computed in step <1>4. The orbit of \(A\) is all of \(X\), proving transitivity on this three-element set.

an \(F\)-isomorphism \(L \to M\) extends to an automorphism of \(\overline F\) fixing \(F\), so \(M = \sigma(L)\) for some \(\sigma\).

any degree-\(2\) extension in characteristic \(0\) is Galois (it is the splitting field of a quadratic).

\(K/F\) is Galois, so it is stable under every \(F\)-automorphism of \(\overline F\).

\(\sigma|_K\) is an automorphism of \(K\), and \(\sigma(L)\) is a \(K\)-conjugate of \(L\).

a Galois extension is stable under all \(K\)-automorphisms of \(\overline F\).

if \(\sigma|_K = \tau|_K\), then \(\sigma^{-1}\tau\) fixes \(K\), so \(\sigma^{-1}\tau(L) = L\), i.e. \(\sigma(L) = \tau(L)\).

\(|\operatorname{Aut}_F(K)| = [K:F] = 2\).

\(\sigma|_K \in \operatorname{Aut}_F(K)\), which has two elements.

By step <1>4, \(\sigma(L)\) is determined by \(\sigma|_K\), and by step <1>5 there are at most two such restrictions.

Step <1>6 is the claim.

Normality and separability of the Galois extension imply that \(m\) splits in \(K\) with \(d\) distinct roots [@DF04]. Every \(\sigma\in G\) fixes its coefficients, so sends each root to a root. These permutations respect composition and give the stated homomorphism.

The fixed field \(K^G\) is \(F\) [@DF04]. Since \(\alpha\notin F\), some \(\sigma\in G\) satisfies \(\sigma(\alpha)\ne\alpha\). Thus the root action is not trivial and \(\ker\rho\ne G\).

For \(n\geq5\), the group \(A_n\) is simple [@DF04]. Hence \(G\cong A_n\) has no normal subgroups other than \(1\) and \(G\). The kernel of a homomorphism is normal, and step <1>1 excludes the latter possibility. Therefore \(\ker\rho=1\).

Injectivity gives \(|A_n|=n!/2\leq |S_d|=d!\). If \(d<n\), then \(d\leq n-1\), so \[ d!\leq(n-1)!<\frac n2(n-1)!=\frac{n!}{2}, \] where strictness follows from \(n\geq5>2\). This contradicts \(n!/2\leq d!\). Thus \(d\geq n\), as required.

The polynomial splits in the algebraically closed field \(K\), and its derivative is \(-1\), so its \(p^m\) roots are distinct. In characteristic \(p\), the binomial theorem and iteration give \((u+v)^{p^m}=u^{p^m}+v^{p^m}\). The fixed elements of the map \(u\mapsto u^{p^m}\) are therefore closed under addition and subtraction, and also under multiplication and nonzero inversion. They contain \(0,1\), so they form a field with exactly \(p^m\) elements. We denote this subfield of \(K\) by \(\mathbb F_{p^m}\). Its degree over \(\mathbb F_p\) is \(m\), by counting elements in a finite-dimensional vector space over \(\mathbb F_p\).

Write \(q=p^n\). Since \(f(\alpha)=0\), we have \(\alpha^q=\alpha-1\). For \(a\in\mathbb F_q\), step <1>1 gives \(a^q=a\). Therefore \[ f(\alpha+a)=\alpha^q+a^q-\alpha-a+1=0. \] Put \(\beta=\alpha^p-\alpha+1\). Taking \(q\)th powers gives \[ \begin{aligned} \beta^q &=(\alpha^q)^p-\alpha^q+1\\ &=(\alpha-1)^p-(\alpha-1)+1\\ &=\alpha^p-\alpha+1=\beta. \end{aligned} \] Hence \(\beta\in\mathbb F_q\), again by step <1>1.

Starting with \(\alpha^{p^n}=\alpha-1\), induction on \(j\geq1\) gives \[ \alpha^{p^{jn}}=\alpha-j, \] where the integer \(j\) is viewed in the prime field. Indeed, raising the equality to the \(p^n\)th power subtracts one more, because every element of the prime field is fixed by Frobenius. At \(j=p\) we obtain \(\alpha^{p^{np}}=\alpha\), proving the claim.

Let \(\alpha\) be any root. With \(n=1\), step <1>3 puts \(\mathbb F_p(\alpha)\) inside \(\mathbb F_{p^p}\). Its degree \(d\) over \(\mathbb F_p\) divides \(p\) by the tower law, so \(d=1\) or \(p\). But \(a^p-a+1=1\) for every \(a\in\mathbb F_p\), so \(\alpha\notin\mathbb F_p\) and \(d\ne1\). Thus \(d=p\). The monic minimal polynomial of \(\alpha\) divides \(X^p-X+1\) and has the same degree, so it equals that polynomial.

It takes value \(1\) at both \(0\) and \(1\), so it has no linear factor. The only monic irreducible quadratic over \(\mathbb F_2\) is \(q(X)=X^2+X+1\): a monic quadratic \(X^2+aX+b\) without a root at zero must have \(b=1\), and absence of a root at one then forces \(a=1\).

Modulo \(q\), the residue class \(x\) satisfies \(x^2=x+1\) and \(x^4=(x+1)^2=x^2+1=x\). Consequently \(x^4+x+1=1\), so \(q\) does not divide the polynomial. Any reducible quartic without a linear factor is a product of two irreducible quadratics. The only possible quadratic factor has just been excluded, so the quartic is irreducible.

If \(f\) is irreducible and \(\alpha\) is a root, then \([\mathbb F_p(\alpha):\mathbb F_p]=\deg f=p^n\). Step <1>3 and the tower law imply \[ p^n\mid np,\qquad\text{hence }p^{n-1}\mid n. \] For \(n\geq3\), the inequality \(2^{n-1}>n\) holds: it is true at \(n=3\), and doubling the left side preserves strict inequality at the next integer since \(2n\geq n+1\). Then \(p^{n-1}\geq2^{n-1}>n\), contradicting the divisibility. Thus \(n=1\) or \(2\). In the latter case the divisibility gives \(p\mid2\), hence \(p=2\). Steps <1>4 and <1>5 establish irreducibility in both surviving cases, completing every part.

Choose an isomorphism with the symmetries of a square, taking \(r\) to a quarter-turn and \(s\) to a reflection. The displayed elements and relations describe these eight symmetries. Since \(K/F\) is Galois, its fixed field \(K^G\) is \(F\). The fixed-field theorem for a finite group of field automorphisms gives \([K:F]=|G|=8\), so the finite Galois correspondence applies with the asserted degree and normality properties [@DF04].

Let \(R=\langle r\rangle\), a cyclic subgroup of order four. Its only subgroups are \(\{1\}\), \(\langle r^2\rangle\), and \(R\): containing \(r\) or \(r^3\) forces all of \(R\), and any other subgroup is contained in \(\{1,r^2\}\).

For any \(H\leq G\), put \(T=H\cap R\). If \(H\subseteq R\), this already lists all possibilities. Otherwise choose a reflection \(t=r^j s\in H\). Every reflection has order two, by \(sr^j=r^{-j}s\). The product of two reflections is a rotation. Hence each element of \(H\) outside \(R\) lies in the coset \(tT\), and conversely \(tT\subseteq H\). Therefore \[ H=T\mathbin{\sqcup}tT. \]

If \(T=\{1\}\), this gives \(H=\{1,r^j s\}\), one of four distinct order-two subgroups. If \(T=\langle r^2\rangle\), it gives \(H=\{1,r^2,r^j s,r^{j+2}s\}\). This set is a subgroup: \(r^2\) commutes with the order-two element \(r^j s\), so their four products form a group. The set depends precisely on the parity of \(j\), giving the two displayed subgroups of order four. Finally, \(T=R\) gives \(H=G\). Together with the subgroups contained in \(R\), these cases exhaust every \(H\leq G\). Removing \(\{1\}\) and \(G\) leaves exactly the eight subgroups stated above, without repetitions.

The subgroup orders in step <1>2 and the degree formula in step <1>1 give exactly the stated fields and degrees. Different subgroups have different fixed fields, so none of these entries is duplicated. The subgroup \(G\) corresponds to \(F\) and the subgroup \(\{1\}\) to \(K\).

All three subgroups of order four have index two and are normal: both the left and right cosets of any element outside an index-two subgroup are its complement. The subgroup \(\langle r^2\rangle\) is central, since \(r\) commutes with \(r^2\) and \(sr^2s=r^{-2}=r^2\). It too is normal. No reflection subgroup is normal, because \[ r(r^j s)r^{-1}=r^{j+2}s\ne r^j s. \] Conjugation takes its unique nonidentity element outside that subgroup. The normality criterion in step <1>1 therefore gives precisely the last column of the table.

The roots of \(x^p-2\) are \(\alpha^q(\zeta^q)^j\), \(0\leq j<p\); the roots of \(x^q-2\) are \(\alpha^p(\zeta^p)^k\), \(0\leq k<q\). All lie in \(\mathbb Q(\zeta,\alpha)\). Conversely, the splitting field contains \(\alpha^q\) and \(\alpha^p\), and their nonzero-root ratios give \(\zeta^q\) and \(\zeta^p\). Choose integers \(a,b\) with \(pa+qb=1\). Then \[ \alpha=(\alpha^p)^a(\alpha^q)^b, \qquad \zeta=(\zeta^p)^a(\zeta^q)^b \] also belong to the splitting field. Negative exponents are allowed because all four elements are nonzero. This proves equality of the two fields.

The polynomial \(x^n-2\) is Eisenstein at two, so \([\mathbb Q(\alpha):\mathbb Q]=n\) [@DF04]. The cyclotomic degree formula gives \([L:\mathbb Q]=\varphi(pq)=(p-1)(q-1)=m\) [@DF04]. Neither \(p\) nor \(q\) divides \(m\): for \(p\) this uses \(p\nmid q-1\), and for \(q\) it follows from \(p-1<q\) and \(q\nmid q-1\). Thus \(\gcd(n,m)=1\).

The degree \([E:\mathbb Q]\) is divisible by both \(n\) and \(m\), since \(E\) contains both subfields. It is consequently divisible by their product \(nm\). On the other hand, \(E=L(\alpha)\) and \(\alpha\) satisfies \(x^n-2\) over \(L\), so \([E:L]\leq n\) and \([E:\mathbb Q]\leq nm\). Both bounds force equality. In particular \(x^n-2\) remains irreducible over \(L\), and \([E:L]=n\).

Since \(x^n-2\) is irreducible over \(L\), sending \(\alpha\) to its root \(\zeta\alpha\) defines an \(L\)-embedding of \(E\) into itself. Its image contains \(\alpha=\zeta^{-1}(\zeta\alpha)\), so it is the automorphism \(\tau\), of order exactly \(n\).

For each unit \(b\) modulo \(n\), the cyclotomic automorphism \(\gamma_b\) of \(L\) sends \(\zeta\) to \(\zeta^b\) [@DF04]. Every element of \(E\) has a unique expression \(\sum_{j=0}^{n-1}c_j\alpha^j\), \(c_j\in L\). Define \(\sigma_b\) by applying \(\gamma_b\) to each coefficient. This preserves addition. It preserves multiplication because products are reduced using \(\alpha^n=2\), whose right side is fixed by \(\gamma_b\). Its inverse is \(\sigma_{b^{-1}}\). Thus these are field automorphisms with the asserted values, and \(\sigma_b\sigma_c=\sigma_{bc}\).

The conjugation identity follows by evaluating both sides on \(\alpha\) and \(\zeta\), which generate \(E\). The maps \(\tau^a\sigma_b\) are all distinct: their values on \(\zeta\) determine \(b\), and their values on \(\alpha\) then determine \(a\). The splitting field is Galois in characteristic zero and has degree \(nm\), so these \(nm\) automorphisms exhaust its Galois group [@DF04].

The composition rule in step <1>3 makes \(H_0\) a subgroup isomorphic to \((\mathbb Z/n\mathbb Z)^\times\), of order \(m\). Its conjugate \(H_1\) is a subgroup of the same order. The displayed description of \(H_1\) follows from \(\sigma_b\tau^{-1}=\tau^{-b}\sigma_b\). Every element of \(H_0\) fixes \(\alpha\), while every element of \(H_1\) fixes \(\tau(\alpha)=\zeta\alpha\). But \(\sigma_{-1}\in H_0\) sends \(\zeta\alpha\) to \(\zeta^{-1}\alpha\ne\zeta\alpha\), since \(\zeta\) has odd order \(n>2\). Hence \(\sigma_{-1}\notin H_1\), proving that these two subgroups are distinct.

Since \(|G|=pqm\) and \(p\nmid qm\), a Sylow \(p\)-subgroup has order \(p\). The element \(\tau^q\) has order \(p\), so its subgroup \(P\) is Sylow. Conjugation by \(\tau\) fixes \(\tau^q\), and conjugation by \(\sigma_b\) sends it to \(\tau^{bq}\), another generator of \(P\) because \(p\nmid b\). The elements \(\tau\) and all \(\sigma_b\) generate \(G\), so \(P\) is normal in \(G\). Sylow conjugacy then makes it the unique Sylow \(p\)-subgroup [@DF04].

Put \(M=\mathbb Q(\zeta,\alpha^p)\), noting that \(\alpha^p=\sqrt[q]{2}\). The subgroup \(P\) fixes both generators, because \((\tau^q)(\alpha^p)=\zeta^{pq}\alpha^p=\alpha^p\). Thus \(M\subseteq E^P\). Also \(E=M(\alpha)\) and \(\alpha\) satisfies \(T^p-\alpha^p\in M[T]\), so \([E:M]\leq p\). The \(p\) elements \(\tau^{qj}(\alpha)=\zeta^{qj}\alpha\), \(0\leq j<p\), are distinct, and each is a root of the minimal polynomial of \(\alpha\) over \(E^P\), because these automorphisms fix its coefficients. Hence \([E:E^P]\geq p\). The tower law yields \[ p\leq[E:E^P]\leq[E:M]\leq p, \] so equality holds throughout and \([E^P:M]=1\). Therefore \(E^P=M\), as claimed.

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].

Since \([\mathbb Q(i):\mathbb Q]=2\), step <1>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 \(8\).

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.

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 <1>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.

The roots \(a,\zeta a,\zeta^2a\) lie in \(\mathbb Q(a,\zeta)\). Conversely, the field generated by them contains \(a\) and \(\zeta=(\zeta a)/a\), so it is precisely \(K\). Eisenstein’s criterion at \(2\) makes \(x^3+2\) irreducible over \(\mathbb Q\), giving \([\mathbb Q(a):\mathbb Q]=3\) [@DF04]. The field \(\mathbb Q(a)\) is real, whereas \(\zeta\) is nonreal and satisfies \(\zeta^2+\zeta+1=0\). Thus adjoining \(\zeta\) has degree \(2\), and the tower law gives \([K:\mathbb Q]=6\). As a splitting field in characteristic zero, this is a Galois extension [@DF04].

The polynomial \(x^2+x+1\) is the minimal polynomial of \(\zeta\) over \(\mathbb Q\), since its roots are nonreal. The tower law therefore gives \([K:\mathbb Q(\zeta)]=3\). Hence \(x^3+2\) remains the minimal polynomial of \(a\) over \(\mathbb Q(\zeta)\). Sending \(a\) to \(\zeta a\) defines an embedding of \(K\) into itself fixing \(\zeta\). Its image contains \(a\) as well, so it is the automorphism \(r\). Complex conjugation restricts to \(s\).

Their actions on the root indices are \[ r:j\longmapsto j+1,\qquad s:j\longmapsto-j. \] Thus \(r\) is the cycle \((0\,1\,2)\) and \(s\) the transposition \((1\,2)\). Their six products \(r^us^v\), with \(0\leq u<3\) and \(v\in\{0,1\}\), act by \(j\mapsto u+(-1)^vj\) and are distinct permutations. They exhaust the Galois group, whose order is \(6\). Equivalently, these are all permutations of the three roots, since the roots generate \(K\) and the action is faithful. The relations are \(r^3=s^2=1\) and \(srs=r^{-1}\).

An automorphism \(r^us^v\) fixes \(\zeta\) exactly when \(v=0\). Consequently \[ \operatorname{Gal}(K/\mathbb Q(\zeta))=\langle r\rangle \cong C_3, \] acting on \(a_0,a_1,a_2\) by cyclic rotations. The extension \(\mathbb Q(\zeta)/\mathbb Q\) is the splitting field of \(x^2+x+1\) and has degree \(2\). Its nonidentity automorphism \(\bar s\) interchanges the roots \(\zeta,\zeta^2\). Thus its Galois group is the full two-letter permutation group \(S_2\cong C_2\).

Restriction to \(\mathbb Q(\zeta)\) sends \(r\) to the identity and \(s\) to \(\bar s\), so is onto and has kernel \(\langle r\rangle\). This gives \[ 1\longrightarrow\operatorname{Gal}(K/\mathbb Q(\zeta)) \longrightarrow\operatorname{Gal}(K/\mathbb Q) \xrightarrow{\mathrm{res}} \operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q) \longrightarrow1. \] The map sending \(\bar s\) to \(s\) is a homomorphic section, because both have order \(2\). Thus \(G\cong C_3\rtimes C_2\), where the nonidentity element of \(C_2\) acts on \(C_3\) by inversion, as the relation \(srs=r^{-1}\) explicitly shows. This is not a direct product, since \(r\ne r^{-1}\).

The polynomial \(T^3-2\) is Eisenstein at two and therefore irreducible over \(\mathbb Q\) [@DF04]. Thus \([E:F]=3>1\). The field \(E\) is real, whereas \(\zeta\) is nonreal, so \(E\subset K\) is also strict. The roots \(a,\zeta a,\zeta^2a\) generate \(K\), because their ratio recovers \(\zeta\). Hence \(K\) is a splitting field in characteristic zero and \(K/F\) is Galois [@DF04]. But \(E\) contains only the real root of the irreducible polynomial \(T^3-2\), so \(E/F\) is not normal and is not Galois.

For \(a\in K\), let \(p_a(T)\in F[T]\) and \(q_a(T)\in E[T]\) be its monic minimal polynomials over \(F\) and \(E\). The extension \(K/F\) is algebraic, so both exist. Since \(p_a(a)=0\), division by the minimal polynomial over \(E\) gives \(q_a\mid p_a\) in \(E[T]\). Normality and separability of \(K/F\) imply that \(p_a\) splits into distinct linear factors in \(K[T]\). Its divisor \(q_a\) therefore does the same. Thus every element of \(K\) is separable over \(E\), and every irreducible polynomial over \(E\) with a root in \(K\) splits over \(K\). This proves that \(K/E\) is algebraic, normal, and separable, hence Galois.

The polynomial is nonzero at both \(0\) and \(1\). A reducible cubic over a field has a linear factor, so this polynomial is irreducible over \(\mathbb F_2\). The quotient is consequently a field [@DF04]. Polynomial division gives the basis \(1,[T],[T^2]\) over \(\mathbb F_2\), so it has exactly \(2^3=8\) elements.

Let \(L\) be the splitting field of \(T^5+T-1\) over \(\mathbb Q\). This is a finite Galois extension, since the base field has characteristic zero [@DF04]. Its automorphisms act faithfully on the distinct roots, because those roots generate \(L\). There are at most five such roots, so this action embeds \(\operatorname{Gal}(L/\mathbb Q)\) into \(S_5\), adding fixed letters if necessary. Lagrange’s theorem gives \([L:\mathbb Q]=|\operatorname{Gal}(L/\mathbb Q)|\mid120\). For an intermediate field \(E\), the tower law makes \([E:\mathbb Q]\) a divisor of \([L:\mathbb Q]\) and hence of \(120\). Since \(7\nmid120\), degree seven is impossible.

The polynomial \(T^2-t\) has no root in \(F\): an equality \((A(t)/B(t))^2=t\), with nonzero polynomials \(A,B\), would give \(2\deg A=1+2\deg B\), impossible by parity. It is therefore irreducible, giving \([K:F]=2\). In \(K[T]\) it equals \((T-u)^2\), so the minimal polynomial of \(u\) has a repeated root. Hence \(K/F\) is not separable.

Eisenstein’s criterion at two proves irreducibility over \(\mathbb Q\) [@DF04]. Let \(b=\sqrt[5]{2}\) and choose a primitive fifth root of unity \(\xi\). The roots of \(f\) are precisely \(\xi^j b\), \(0\leq j<5\). They all lie in the radical tower \[ \mathbb Q\subset\mathbb Q(\xi) \subset\mathbb Q(\xi,b): \qquad \xi^5=1\in\mathbb Q,\quad b^5=2\in\mathbb Q(\xi). \] Each step adjoins an element having a positive integral power in the preceding field. Thus every root of the irreducible fifth-degree polynomial lies in a radical extension, which is exactly solvability by radicals.