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If $\zeta_n$ is the $n$-th primitive root of unity then $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq Z_n^*$ due to the following map $$\tau(\zeta_n)=\zeta^n$$


I was wondering if we could use this and the Galois Correspondence to find the number of fields between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_n)$. How could we determine this?


I know that if $n$ is prime then $\mathbb{Q}(\zeta_n)$ contains subfields of degree $d$, where $d$ are the divisors of $p-1$.

For example, $Gal(\mathbb{Q}(\zeta_3)/\mathbb{Q}) \simeq Z_3^*$ and since $3-1=2$ has divisors $1$ and $2$, the number of fields between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_3)$ is two... is that right? Do these 'sub'fields actually corresponds to the fields themselves in this case?


What can we say if $n$ is not prime?

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  • $\begingroup$ Not much besides the fact that its Galois Group is abelian and thus it has at least one subgroup of every divisor of its order, which of course is normal and thus every subextension is normal as well. $\endgroup$ – DonAntonio Jun 1 '16 at 21:43
  • $\begingroup$ So say we had $\mathbb{Q}(\zeta_6)$.. can we not determine the number of fields between $\mathbb{Q}(\zeta_6)$ and $\mathbb{Q}$? We only know that there is a degree $1, 2, 3$ and $6$ extension... are subfields of the same degree not isomorphic? Thanks $\endgroup$ – amiz9 Jun 1 '16 at 21:48
  • $\begingroup$ Well, in the particular case of $\;n=6\;$ it is very easy since the extension is of degree $\;3\;$ and thus cyclic. $\endgroup$ – DonAntonio Jun 1 '16 at 21:52
  • $\begingroup$ OK, so $[\mathbb{Q}(\zeta_6) : \mathbb{Q}]=3$. How did you determine this - by finding the minimal polynomial of $\zeta_6$ over $\mathbb{Q}$? And is the only group of order $3$ $\mathbb{Z_3}$? Thanks $\endgroup$ – amiz9 Jun 1 '16 at 21:56
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    $\begingroup$ In my above comment there's a typo: it should be ",,,of degree $\;2\;$ , and this is because $\;\phi(2n)=\phi(n)\;$ for any odd $\;n\;$ , and the degree of the cyclotomic extension is precisely $\;\phi(n)\;$ . In our case, $\;\phi(6)=2\;$ . You can also directly see it: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$ The primitive roots of unit of order $\;6\;$ are from the last quadratic above, as the other quadratic's roots are, of course, the primitive cubic roots. $\endgroup$ – DonAntonio Jun 1 '16 at 22:01
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Recall by the fundamental theorem of Galois theory that the number of fields between $\mathbb Q$ and $\mathbb Q(\zeta_n)$ is the number of subgroups of $\text{Gal}(\mathbb Q(\zeta_n))\cong (\mathbb Z/n\mathbb Z)^\times$. If $n$ is a power of a prime, i.e. $n=p^k$ for some $p$ prime and $k\in\mathbb N$, then $(\mathbb Z/n\mathbb Z)^\times$ is cyclic of order $\varphi(p^k)=p^{k}-p^{k-1}$. By the fundamental theorem of cyclic groups, for all $d|\varphi(p^k)$ there is exactly one subgroup of order $d$. Hence, the number of intermediate fields between $\mathbb Q$ and $\mathbb Q(\zeta_n)$ is the number of divisors of $p^k-p^{k-1}$. (For example, if $n$ is prime, then this is the number of divisors of $n-1$ as you stated.)

If $n$ is not a power of a prime, this question becomes much more difficult (see https://mathoverflow.net/questions/46115/subgroups-of-a-finite-abelian-group). By the way, for any finite abelian group $G$, there exists some cyclotomic field $\mathbb Q(\zeta_n)$ and some intermediate field $\mathbb Q\subset F\subset \mathbb Q(\zeta_n)$ such that $\text{Gal}(F/\mathbb Q)\cong G$. This provides a trivial partial answer to the inverse Galois problem.

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  • $\begingroup$ Thank you this clarifies things a lot. So the number of intermediate fields is equal to the number of intermediate subgroups? And when $n$ is not a power of a prime, say we had $\mathbb{Q}(\zeta_{15}) \simeq \mathbb{Z_{15}}^*$.. we would have subgroups of orders $1, 2, 7, 14$... can we say much more? $\endgroup$ – amiz9 Jun 1 '16 at 22:12
  • $\begingroup$ Yes, the number of intermediate fields equals the number of subgroups of the Galois group. (Remember that a subgroup $H$ of the Galois group corresponds to its fixed field by the Galois correspondence.) In the case $n=15$, you have $\mathbb Q(\zeta_{15})\cong (\mathbb Z/15\mathbb Z)^\times\cong (\mathbb Z/5\mathbb Z)^\times\times (\mathbb Z/3\mathbb Z)^\times\cong \mathbb Z/4\mathbb Z\times\mathbb Z/2\mathbb Z$ (by the Chinese Remainder Theorem). It turns out there are $8$ subgroups of this group, hence 8 intermediate fields (including $\mathbb Q(\zeta_{15})$ and $\mathbb Q$ itself). $\endgroup$ – Michael M Jun 1 '16 at 22:18

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