Cyclotomic extension of $\mathbb{Q}$

Question

Let $L$ be an extension of $\mathbb{Q}$ with $[L:\mathbb{Q}]=10$. Let $L^*$ be the multiplicative group $(L\setminus\{0\}, \cdot)$ and $T(L^*)$ the torsion group of $L^*$. Prove that there are exactly four possibilities for the order of $T(L^*)$, namely $|T(L^*)|\in\{2, 4, 6, 22\}$. For each case, exhibit the extension $L$.

I thought about the extension $L=\mathbb{Q}(e^{2\pi i/11})$, which satisfies $[L:\mathbb{Q}]=10$. In this case, $T(L^*)=\{e^{2\pi ki/11}\mid k \in\{0,...,10\}\}$, which are all the primitive $11^{th}$-roots of unity, so $|T(L^*)|=11\notin\{2, 4, 6, 22\}$. Am I missing something?

Answer 1

Let $\zeta_n$ be a primitive $n^{th}$ root of unity generating the torsion subgroup of $L^*$. Then as $F=\Bbb Q(\zeta_n)\subseteq L$ and we know that $n$ must be even for this to be so, as if $n$ is odd, $-\zeta_n$ has twice the order, so it cannot be a generator. Write $n=2k$. Then $2k$ is the order of the torsion subgroup, and we require $\phi(2k)|10$. Now for each prime, $p|k$ we know that $\phi(2k)\ge p-1$, so the only primes which can divide $k$ are $2,3,5,7,11$ since $\phi(2k)|10$. We conclude $k=2^\alpha 3^\beta 5^\gamma 7^\delta 11^\epsilon$.

If $\epsilon >1$ then all the others are zero and we have $\epsilon=1$ by our inequalities, and $L=\Bbb Q(\zeta_{11}$. Next if $\delta, \gamma>0$ then $6|10$ or $4|10$ which is crazy so both of those are zero. If $\beta>1$ then $3|10$ and finally if $\alpha >1$ then $4|10$. So our options are

$$\begin{cases} 0\le\alpha \le 1 \\ 0\le\beta\le 1 \\ \gamma = \delta = 0 \\ 0\le \epsilon \le 1\end{cases}$$

Which can happen? We can rule out $\alpha = \beta = 1 $ since this would give $\phi(12) = 4 | 10$, again impossible. We already covered $\epsilon >0$, so all other cases are with $\alpha, \beta$ possibly non-zero. So overall all our cases are:

$$\begin{cases} \alpha = \beta = 0\iff \text{Tor }L^* = \{\pm 1\} \\ \alpha = 1,\beta = 0 \iff \text{Tor }L^*= \{\pm i, \pm 1\} \\ \alpha = 0, \beta = 1 \iff \text{Tor }L^* = \langle \zeta_6\rangle \\ \epsilon = 1 \iff \text{Tor }L^* = \langle\zeta_{22}\rangle \end{cases}$$

And these have all the right orders to make up your candidate list. Now to find the $L$. You've already gotten the last case taken care of, what of the others?

Note that the maximal, real subfield of $\Bbb Q(\zeta_{25})$ has degree ${\phi(25)/2} = 10$. Since this is totally real, the number of roots of unity is $2$, and so if $\tau$ is complex conjugation, $L=\Bbb Q(\zeta_{25})^\tau$, the fixed field of complex conjugation is such an $L$. For orders $4$ and $6$, we need an extension of $\Bbb Q(i)$ or $\Bbb Q(\zeta_3)$ of degree $5$. Can you see how to proceed in these cases based on how we've done things so far?