In my notation, $\mathbb{Z}_n$ = the integers modulo $n$.
Find a Galois extension whose Galois group is $\mathbb{Z}_4 \times \mathbb{Z}_4$. (The additive group).
I thought about using cyclotomic polynomials, since I know the Galois group of the $k$-th cyclotomic polynomial (the polynomial whose roots are the primitive $k$-th roots of unity) is $\mathbb{Z}_k^\star$ (the units of the multiplicative group $\mathbb{Z}_k$). The order of this group is $\phi(k)$. Here $\phi$ is Euler's totient function.
I also know these $\mathbb{Z}_k^\star$ groups are isomorphic to a cross product of copies of (possibly equal or different) additive groups of $\mathbb{Z}_{m_{i}}$'s.
I know there is a theorem that states that any $\mathbb{Z}_{k}^\star$ for $k = p^a$ or $k=2p^a$ for odd primes $p$ are cyclic.
I first started searching the (non-cyclic) $\mathbb{Z}_{k}^\star$ for $k\in\mathbb{N}$ s.t. $\phi(k)=16$. Neither of them happened to contain two disjoint copies of $\mathbb{Z}_4$. (I checked by running code on RStudio).
I then started searching for $\phi(k) = 32$, with the idea of extracting two copies of $\mathbb{Z}_4$ as a subgroup, since $16|32$.
I found that $\mathbb{Z}_{80}^\star$ happend to contain two disjoint copies of $\mathbb{Z}_4$, in fact, $\mathbb{Z}_{80}^\star \cong \mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_4$.
If I were to quotient it by $\mathbb{Z}_2$, I would get an isomorphism with $\mathbb{Z}_4 \times \mathbb{Z}_4$.
Therefore, the extension $\mathbb{Q}(\zeta_{80})$, where $\zeta_{80}$ is a primitive $80$-th root of unity seems like a good candidate to extract a subgroup from.
Note $[\mathbb{Q}(\zeta_{80}):\mathbb{Q}] = 32$ since $\phi(80) = 32$.
By the Fundamental Theorem of Galois Theory, I need to search for extension fields $L$ such that $\mathbb{Q} \leq L \leq \mathbb{Q}(\zeta_{80})$ and $[L:\mathbb{Q}] = 16$.
I know the subgroup generated by complex conjugation is isomorphic to $\mathbb{Z}_2$, and the complex conjugation must belong to $Gal(\Phi_{80})$ (where $\Phi_{80}$ is the $80$-th cyclotomic polynomial) which contains $\zeta_{80}$ as a root by definition. This since $\zeta_{80}$ is itself not real, and non-real roots come in conjugate pairs (making the complex conjugation automorphism swap them).
I then thought quotienting by complex conjugation wouldn't work since any non-trivial intermediate extension of $\mathbb{Q}$ must contain non-real elements.
On the other hand, the subgroups of $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_4$ of order $16$ are:
- $\{0\} \times \mathbb{Z}_4 \times \mathbb{Z}_4$
- $\mathbb{Z}_2 \times 2 \mathbb{Z}_4 \times \mathbb{Z}_4$
- $\mathbb{Z}_2 \times \mathbb{Z}_4 \times 2 \mathbb{Z}_4$,
the two latter being isomorphic.
My intuition tells me that taking $\mathbb{Q}(\zeta_{80}^2)$ would cut the degree in half, yielding $[\mathbb{Q}(\zeta_{80}^2):\mathbb{Q}] = 16$, however as I discussed the impossibility of quotienting by conjugation, this extension's Galois group doesn't seem to belong to the correct class of isomorphism.
Is this the right path?
I got this problem in my Abstract Algebra II final a while ago, this one seemed to be the hardest one and don't know anyone who solved it back then, so I'm asking here because I'm curious and want to know the answer.
Best Answer
If $L/K$ and $F/K$ are both Galois extensions then $LF/K$ is Galois and we have an embedding ${\rm Gal}(LF/K) \hookrightarrow {\rm Gal}(L/K) \times {\rm Gal}(F/K)$ by $\sigma \mapsto (\sigma|_L, \sigma|_F)$. The image is $\{(\tau,\tau') : \tau = \tau' {\rm on } \ L \cap F\}$, so if $L \cap F = K$ then the image is everything: ${\rm Gal}(LF/K) \cong {\rm Gal}(L/K) \times {\rm Gal}(F/K)$. So you just need to find two Galois extensions of a field $K$ with Galois group cyclic of order $4$ and the fields have intersection $K$.
Example 1. For prime $p$, ${\rm Gal}(\mathbf Q(\zeta_p)/\mathbf Q) \cong (\mathbf Z/p\mathbf Z)^\times$, which is cyclic of order $p-1$. If $p \equiv 1 \bmod 4$ then there is a unique intermediate field $L$ such that $[L:\mathbf Q] = 4$, so ${\rm Gal}(L/\mathbf Q)$ is cyclic of order $4$ (all quotient groups of a cyclic group are cyclic). For distinct primes $p$ and $q$, $\mathbf Q(\zeta_p) \cap \mathbf Q(\zeta_q) = \mathbf Q$, so any subfields of $\mathbf Q(\zeta_p)$ and $\mathbf Q(\zeta_q)$ have intersection $\mathbf Q$. Let $L = \mathbf Q(\zeta_5)$ and $F$ be the subfield of $\mathbf Q(\zeta_{13})$ where $[F:\mathbf Q] = 4$. Explicitly, $F = \mathbf Q(\gamma)$ where $\gamma = \zeta_{13} + \zeta_{13}^3 + \zeta_{13}^9$. Then $LF = \mathbf Q(\zeta_5,\gamma)$ is a Galois extension of $\mathbf Q$ with Galois group that's a product of two cyclic groups of order $4$.
Alternatively, we could use Kummer theory. If $i \in K$ (meaning $K$ contains a primitive $4$th root of unity) then $K(\sqrt[4]{a})/K$ is Galois with Galois group cyclic of order $4$ as long as the extension has degree $4$. And if you have another such extension $K(\sqrt[4]{b})/K$ that is "disjoint" from the first one, meaning their intersection is $K$, you'll have $K(\sqrt[4]{a},\sqrt[4]{b})/K$ being a Galois extension with the desired Galois group.
Example 2. Let $K = \mathbf C(x)$ be a rational function field over $\mathbf C$. Then $K(\sqrt[4]{x},\sqrt[4]{x+1})/K$ is an example of the desired type, although proving $K(\sqrt[4]{x}) \cap K(\sqrt[4]{x+1}) = K$ might not be something you can easily do using only the math that is typically taught in a first course on Galois theory.