Find a normal extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}_2\times \mathbb{Z}_2$

Question

Find a subfield $E$ of the cyclotomic field $\mathbb{Q}(\zeta_{15})$ with $Gal(E/\mathbb{Q}) = \mathbb{Z}_2\times \mathbb{Z}_2$ and find a primitive element in $E$.

Let $G = Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$. A similar question was answered here and following that process, I am wondering if we want $E$ s.t. $[E:\mathbb{Q}] = 4$? because if $F = \mathbb{Q}(\zeta_{15})$ ($\zeta_{15}$ is a primitive root), then $[F:\mathbb{Q}] = 8 = [F:E][E:\mathbb{Q}]$

If that is true, $[F:E] = 2\implies$ we are looking for a subgroup $H\leq G$ s.t. $H\cong C_2$ and does $\zeta_{15}+\zeta_{15}^{-1}$ become a primitive element of E? I don't understand how different answers would be for $Gal(E/\mathbb{Q}) = \mathbb{Z}_2\times \mathbb{Z}_2$ vs when $Gal(E/\mathbb{Q}) = \mathbb{Z}_4$. Thanks.

Answer 1

Let $w=\exp\left({\large{{\frac{2i\pi}{15}}}}\right)$, and let $K=\mathbb{Q}(w)$.

Let $a=w^5$, and let $b=w^3$.

Then $a$ is a primitive cube root of $1$, with minimal polynomial $$f(t)=t^2+t+1$$ hence $i\sqrt{3}\in {\mathbb{Q}}(a)$.

Explicitly we have $i\sqrt{3}=2a+1$.

Also, $b$ is a primitive $5$-th root of $1$, with minimal polynomial $$g(t)=t^4+t^3+t^2+t+1$$ Since the discriminant of $g$ is $125$, which must be a square in ${\mathbb{Q}}(b)$, it follows that $\sqrt{5}\in {\mathbb{Q}}(b)$.

Explicitly we have $$\sqrt{5}=-{\small{\frac{1}{5}}}(b-b^2)(b-b^3)(b-b^4)(b^2-b^3)(b^2-b^4)(b^3-b^4)$$

Let $E={\mathbb{Q}}(i\sqrt{3},\sqrt{5})$.

Then $\mathbb{Q}\subset E\subset K$, and it's easily verified that

• $E$ is a normal extension of $\mathbb{Q}$.$\\[6pt]$
• $\text{Gal}(E/\mathbb{Q})=\mathbb{Z}_2 \times \mathbb{Z}_2$.$\\[6pt]$
• $i\sqrt{3}+\sqrt{5}$ is a primitive element of $E/\mathbb{Q}$.

Answer 2

For two numbers $a\ne b$ both not multiples of squares (i.e. no prime repeats in their factorization) the field generated by their square roots together is such a field you are looking for.

Your job is to find such integers $a,b$ with $\sqrt a, \sqrt b \in \mathbf{Q}[\zeta_{15}]$.

Hint: From $\omega=\zeta_{15}^5$, a cube root of unity, show that $\sqrt {-3}$ is available in $ \mathbf{Q}[\zeta_{15}]$.

Answer 3

For an odd prime $p$, it is classically known that the unique quadratic field contained in $\mathbf Q(\zeta_p)$ is $\mathbf Q (\sqrt {p^*})$, where $p^*=(-1)^{\frac {p-1}2}p$. This is usually shown by computing the discriminant of $\mathbf Q(\zeta_p)$, and constitutes the first step in one of the numerous proofs of the quadratic reciprocity law. To answer your question, it suffices to take a prime $p\equiv 1$ mod $4$ and another prime $q\equiv -1$ mod $4$, and the cyclotomic field $\mathbf Q(\zeta_{pq})$ will contain the biquadratic field $\mathbf Q(\sqrt p,\sqrt {-q})$.