# Prove that there is a Galois extension $E$ of $\mathbb{Q}$ such that Gal$(E/\mathbb{Q}) \cong \mathbb{Z}_{18}$

I am trying to find an appropriate Galois extension $E/\mathbb{Q}$ that is isomorphic to $\mathbb{Z}_{18}$ to prove this fact. I am looking for an example or maybe a general proof.

I was trying this:

I want to find and isomorphism $\phi: E/\mathbb{Q} \to \mathbb{Z}_{18}$.

But I don't see how can I choose the elements that I must add to the extension to make it isomorphic. Is there another intuitive way of seeing this?

I want an extension which is Galois,so it must be normal and separable. So we can choose $\mathbb{Q}$ as our ground field and will always be separable since we have zero characteristic field. Now if we just make a map from each element $\alpha_i$ that we add to the ground field to $[i]_{18}$ for $i = 0,...,17$ then we can obtain the required isomorphism.

My problem is how to choose those elements to be added as an extension such that it is a Galois extension. It can be something like $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{18}})$ where each $p_i's$ are prime ?. Thanks for your help.

• It wouldn't look something like Q(sqrt(p1),...,sqrt(p18)). That would be a field extension whose degree may go up to 2^18. – JGA Apr 22 '18 at 22:10

Since $19$ is conveniently prime, the cyclotomic extension $\mathbb{Q}(\zeta_{19})/\mathbb{Q} = \mathbb{Q}[X]/(X^{18} + \cdots + X + 1)$ is cyclic.

$$\varphi(27)=18$$ thus try $$\mathbb{Q}(\zeta_{27}).$$

• Can you elaborate a little bit more ? Thanks. – Richard Clare Apr 23 '18 at 22:40

For convenience, a cyclic extension of degree $n$ will be called a $C_n$-extension, where $C_n$ denotes the cyclic group of order $n$. It is known that $\mathbf Q$ admits a $C_n$-extension for any $n$, see e.g. https://math.stackexchange.com/a/2028709/300700. But let us try more ambitiously to determine all the $C_{18}$ extensions $E/\mathbf Q$. By the Chinese Remainder Theorem, such an extension will be the compositum of a $C_2$-extension and a $C_9$ -extension of $\mathbf Q$ , and the problem boils down to the construction of all the $C_9$-extensions of $\mathbf Q$. We can proceed in two steps :

1) First construct a $C_3$-extension $F/\mathbf Q$. Recalling that $C_3\cong A_3$, classical Galois theory tells us that over a field $K$ of characterictic $\neq 2$, the splitting field $L_f$ of an irreducible polynomial $f\in K[X]$ is a $C_3$-extension iff the discriminant $Disc(f)$ is a square in $K$. This classifies in particular the $C_3$-extensions of $\mathbf Q$. Note that a primitive element of $L_f$ can be explicitly computed using Lagrange resolvents.

2) Introduce then the quadratic field $\mathbf Q(j)$, where $j$ is a primitive $3$-rd root of $1$ , and put $L'=L_f(j)$, which is a $C_3$-extension of $\mathbf Q(j)$. A classical exercise in Kummer theory tells us that $L'/\mathbf Q(j)$ can be embedded into a $C_9$-extension $L''/\mathbf Q(j)$ iff $j$ is a norm from $L'$. More precisely, if $s$ is a generator of $Gal(L'/\mathbf Q(j))$ and $L''=L'(\sqrt[3] a)$, then $L''/\mathbf Q(j)$ is Galois iff $a^{-1}.s(a)= x^3$, hence $N(x)^3=1$; if $N(x)=j$, then $x^3$ is of the form $a^{-1}.s(a)$ by Hilbert's thm. 90, and $L''=L'(\sqrt[3] a)$ will be a $C_9$- extension of $\mathbf Q(j)$. This classifies all $C_9$-extensions of $\mathbf Q(j)$ containing $L'$. The additional necessary and sufficient condition for $L''/\mathbf Q$ to be Galois is that $a^{-1}.t(a)= x^3$ for all $t\in Gal(L'/\mathbf Q)$. If such is the case, because $2$ and $9$ are coprime, $Gal(L''/\mathbf Q)$ (of order $18$) will be a semi-direct product, i.e. there will be a $C_9$- extension $F/\mathbf Q$ s.t. $L''=F(j)$.

These considerations allow us in principle to get hold of all the $C_9$-extensions of $\mathbf Q$. A good practical (but idiotic) exercise would be of course to recover $\mathbf Q(\zeta_{19})$ in this way.