Past exam of the galois theory course of the University of Oxford / 2023 / 1 / c

Irreducibility of the cyclotomic polynomial gives $[K:\mathbb{Q}]=\phi (15)=8$. The extension is Galois, since $K$ is the splitting field of $T^{15}-1$ in characteristic zero. The injection in part (a) is consequently an isomorphism, and the Chinese remainder theorem givesThe Fundamental theorem of Galois theory identifies proper intermediate fields strictly above $\mathbb{Q}$ with proper nontrivial subgroups. Such subgroups have order two or four. The two-torsion subgroup is $C_2\times C_2$, with three nonidentity elements and hence three order-two subgroups. There are also three order-four subgroups: they are the kernels of the three nonzero group homomorphisms $C_2\times C_4\to C_2$. Distinct such maps have distinct kernels because the target has only one nonidentity element. Therefore

Put $\omega ={\zeta }^5$, $\tau ={\zeta }^3$, $a=1+2\omega$, and $b=1+2(\tau +{\tau }^{-1})$. The primitive cubic and quintic root identities give $a^2=-3$, $b^2=5$. These elements therefore provide the following six generators:

The three quadratic fields are distinct because $-3,5,-15$ are distinct rational square classes. To justify the fifth entry, the two quadratic fields $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$ are distinct, so their compositum has degree four. Moreover $a+b\ne 0$ andThus their sum alone generates the compositum.

The field $\mathbb{Q}(\tau )$ has degree four by the quintic cyclotomic polynomial. Its Galois group is $C_4$ and its unique quadratic subfield is $\mathbb{Q}(b)$, so it cannot contain $a$. It is therefore different from $\mathbb{Q}(a,b)$. The final field has degree four by part (b) and is real; the preceding two quartic fields contain nonreal $\tau$ and $a$, respectively. Thus all three quartics are distinct, and the count in part (i) proves the list exhaustive. This gives the subfields of the fifteenth cyclotomic field without relying on a choice of square-root signs.