For each of the following statements, determine if it is true or false. Explain your answers.
The field \(\mathbb{Q}(\sqrt[3]{2})\) is contained in some cyclotomic extension of \(\mathbb{Q}\).
Let \(\mathbb{F}\) be a finite field, and \(f(x) \in \mathbb{F}[x]\) be an irreducible polynomial of degree \(n\). Then the splitting field of \(f(x)\) over \(\mathbb{F}\) has degree \(n\) over \(\mathbb{F}\).
There is a surjective group homomorphism \(\varphi : S_4 \longrightarrow A_3\).
There is a surjective group homomorphism \(\psi : S_4 \longrightarrow S_3\).
There is a transitive group action of \(S_4\) on a set of three elements.
Solution
The answers, in order, are false, true, false, true, and true.
1. The assertion in part (a) is false.
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.
2. The assertion in part (b) is true.
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 \(n\).
3. The assertion in part (c) is false.
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 \(A_3\).
4. The assertion in part (d) is true.
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.
5. The assertion in part (e) is true.
4. The orbit of \(A\) is all of \(X\), provingtransitivity on this three-element set.