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I know that if $n$ is prime then $G=\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq \Bbb Z_{n-1}$

But I am unsure what $G$ is when $n$ is not prime. For example when $n=4$:

$\operatorname{Gal}(\mathbb{Q}(\zeta_4)/\mathbb{Q}) \simeq \Bbb Z_4^*$

What is the structure of the group $\Bbb Z_4^*$? I am guessing it is cyclic, but I do not know how to explicitly find its generators.

Also, what is the basis of $\mathbb{Q}(\zeta_4)$ over $\mathbb{Q}$? Is it $\{\zeta, ..., \zeta^4\}$?

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For general $n$, the units of the ring $\Bbb Z/n\Bbb Z$, which I’ll write $(\Bbb Z/n\Bbb Z)^*$, do not form a cyclic group. For, when $n=p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$, then $(\Bbb Z/n\Bbb Z)^*\cong(\Bbb Z/p_1^{e_1}\Bbb Z)^*\oplus(\Bbb Z/p_2^{e_2}\Bbb Z)^*\oplus\cdots\oplus(\Bbb Z/p_m^{e_m}\Bbb Z)^*$, the $p_i$ all being prime.

For odd primes $p$, $(\Bbb Z/p^e\Bbb Z)^*$ is always cyclic, but the structure of $(\Bbb Z/2^e\Bbb Z)^*$ is $C_2\oplus C_{2^{e-2}}$, where by $C_m$ I mean an abstract cyclic group of order $m$. (The explanation of why these groups have this structure is a story for another night.)

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  • $\begingroup$ Thank you. So if I we take $n=18=3^2 \times 2 \implies Gal(\mathbb{Q}(\zeta_{18}/\mathbb{Q}) \simeq \mathbb{Z}/9\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ $\endgroup$ – amiz9 Jun 2 '16 at 13:17
  • $\begingroup$ Take $n=8=2^3 \implies Gal(\mathbb{Q}(\zeta_{8}/\mathbb{Q}) \simeq \mathbb{C_2} \oplus \mathbb{C_2}$? $\endgroup$ – amiz9 Jun 2 '16 at 13:19
  • $\begingroup$ Are these correct? This is a nice classification - can you point me to somewhere where I can read more about the generators of these groups and their structure? Do all cyclotomic extension fields have a similar structure? $\endgroup$ – amiz9 Jun 2 '16 at 13:21
  • $\begingroup$ Don’t forget that (in your notation) $\Bbb Z_2^*$ is trivial, not order two. In fact, $\Bbb Q(\zeta_{18})=\Bbb Q(\zeta_9)$. And yes, $\Bbb Z_8^*\cong C_2\oplus C_2$, generators $3$ and $7$. $\endgroup$ – Lubin Jun 2 '16 at 13:32
  • $\begingroup$ Structure of $\Bbb Q(\zeta_n)$ should be in most books on algebraic number theory. $\endgroup$ – Lubin Jun 2 '16 at 13:35
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I mean, $\zeta_4 = i$ so the extension has degree $2$, so the Galois group is a finite group of order $2$ and there's only one of those, $\Bbb Z/2\Bbb Z$. Like all quadratic extensions, the basis is $\{1, i\}= \{1,\sqrt{-1}\}$. If you want the exact structure, it's mercifully simple: $i\mapsto \overline{i}$ is the automorphism, i.e. complex conjugation is the only one.

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  • $\begingroup$ OK thanks, that all makes sense, except I am unsure why $\zeta_4=i \implies [\mathbb{Q}(\zeta_4) : \mathbb{Q}]=2$.. could you help me understand this please? And is the $i \rightarrow i$ the automorphism because $i \notin \mathbb{Q}$? Thanks $\endgroup$ – amiz9 Jun 2 '16 at 0:16
  • $\begingroup$ Is it because the we must take $i$ to the power of $2$ to return back to our field $\mathbb{Q}$? $\endgroup$ – amiz9 Jun 2 '16 at 0:17
  • $\begingroup$ @amiz9 that's one way to think of it, yeah. $\Bbb Q(i) \cong \Bbb Q[x]/(x^2+1)$. And look at the little bar over the $i$, it means complex conjugate. You can also think of it just as $i\mapsto -i$. $\endgroup$ – Adam Hughes Jun 2 '16 at 0:18
  • $\begingroup$ ok thanks. So in general to find $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, we can calculate the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$. The degree of this polynomial will be the order of our group. Is that right? $\endgroup$ – amiz9 Jun 2 '16 at 0:23
  • $\begingroup$ it will always be $\varphi(n)$ so you don't really need the polynomial. Generally speaking all automorphisms are $\zeta_n\mapsto \zeta_n^k$ for $k\in\Bbb Z/n\Bbb Z^*$. In this case $i\mapsto i^3 = i\mapsto (i)^2\cdot i = -i$ is how it works. $\endgroup$ – Adam Hughes Jun 2 '16 at 0:25
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If $\zeta_n$ is a primitive $n^{th}$ root of unity, then $$Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}_n)^*.$$

What the elements of the group look like:

If $\sigma \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, then there is an $a \in \{1,2,\dots,n\}$ with $a$ relatively prime to $n$ such that $$\sigma(\zeta_n) = (\zeta_n)^a$$

For the structure of $(\mathbb{Z}_n)^*$, see Lubin's answer.

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  • $\begingroup$ Thank you. What can we say about the generators that generate these groups? $\endgroup$ – amiz9 Jun 2 '16 at 13:21

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