Quadratic Subfields of Cyclotomic Fields

We use Gauss sums to work out the quadratic subfields of cyclotomic fields.

Here is the main character for today.

Definition 1. Fix a Dirichlet character \(\chi\pmod p\) for a prime \(p\). Then we define the Gauss sum \[\tau(\chi)=\sum_{a\in\mathbb F_p^\times}\chi(a)\zeta_N^a.\]

Proposition 2. Fix a prime \(p\) and a nontrivial Dirichlet character \(\chi\pmod p\). Then \(\left|\tau(\chi)\right|^2=N\).

Proof. This is a direct calculation. Observe \[\tau(\chi)\overline{\tau(\chi)}=\sum_{a,b\in\mathbb F_p^\times}\chi(a)\overline\chi(b)\zeta_p^{a-b}.\] Note \(\chi(a)\overline\chi(b)=\chi(a/b)\), so we may re-index the sum as \[\sum_{a,b\in\mathbb F_p^\times}\chi(ab)\overline\chi(b)\zeta_p^{ab-b}=\sum_{a,b\in\mathbb F_p^\times}\chi(a)\zeta_p^{(a-1)b}.\] We would like to isolate the sum on \(b\) so that we get a sum of roots of unity, but this does not exactly work yet because \(b\) only varies over \(\mathbb F_p^\times\) instead of all \(\mathbb F_p\). Well, adding in \(b=0\) does nothing because \(\sum_a\chi(a)=0\) anyway (recall \(\chi\) is nontrivial!), so our sum is actually \[\sum_{a\in\mathbb F_p^\times}\Bigg(\chi(a)\sum_{b\in\mathbb Z/N\mathbb Z}\zeta_p^{(a-1)b}\Bigg).\] Now, if \(a\ne1\), then the inner sum vanishes. Thus, we are left with \(\chi(1)\cdot N=N\). \(\blacksquare\)

Corollary 3. For each odd prime \(p\), we have \[\sqrt{\left(\frac{-1}p\right)p}\in\mathbb Q(\zeta_p).\]

Proof. Consider the nontrivial character \(\chi=\left(\frac{\cdot}p\right)\) on \(\mathbb F_p^\times\). Then we know that \(\tau(\chi)\overline{\tau(\chi)}=p\) by Proposition 7. However, \(\chi\) outputs to real numbers, so \[\overline{\tau(\chi)}=\sum_{a\in\mathbb F_p^\times}\chi(a)\zeta_p^{-a}\] equals \(\chi(-1)\tau(\chi)\) after re-indexing the sum. Thus, \(\tau(\chi)^2=\left(\frac{-1}p\right)p\). \(\blacksquare\)

Notation 4. For an odd prime \(p\), we define \(p^*=\left(\frac{-1}p\right)p\).

Remark 5. To extend to \(2\), we note that \(\zeta_8+\zeta_8^{-1}=\sqrt2\).

We can now classify quadratic sub-extensions of \(\mathbb Q(\zeta_N)\).

Proposition 6. Fix a positive integer \(N\). Let \(S\) denote the set consisting of \(p^*\) for each odd prime divisor \(p\) of \(N\), along with \(-1\) if \(4\mid N\) and \(2\) if \(8\mid N\). For a squarefree integer \(d\), we have \(\sqrt d\in\mathbb Q(\zeta_N)\) if and only if \(d\) is a product of elements in \(S\).

Proof. In one direction, note that having a square root in \(\mathbb Q(\zeta_N)\) is preserved by taking products, so it is enough to show that each element of \(S\) has a square root in \(\mathbb Q(\zeta_N)\). For odd primes \(p\), this is Corollary 3. For \(2\), this is Remark 5. Lastly, for \(-1\), this is just noting that \(\zeta_4=i\).

It remains to show that \(\mathbb Q(\zeta_N)\) admits no more square roots, which will do by a counting argument. We will use Galois theory to turn this into a group theory problem. Note that \(\mathbb Q(\sqrt d)\) is a distinct quadratic extension of \(\mathbb Q\) for each \(d\), so we have thus far exhibited \(2^{\#S}-1\) quadratic subextensions of \(\mathbb Q(\zeta_N)\). We will show that this is all of them. Well, such extensions are in bijection with index-\(2\) subgroups of the Galois group, which are in bijection with nontrivial homomorphisms \[\operatorname{Gal}(\mathbb Q(\zeta_N)/\mathbb Q)\to\mathbb Z/2\mathbb Z.\] Adding \(1\) for the trivial homomorphism and recalling that \(\operatorname{Gal}(\mathbb Q(\zeta_N)/\mathbb Q)\cong(\mathbb Z/N\mathbb Z)^\times\), we see that we need to show \[\#\operatorname{Hom}((\mathbb Z/N\mathbb Z)^\times,\mathbb Z/2\mathbb Z)=2^{\#S}.\] This is now a group theory problem. By the Chinese remainder theorem, we may reduce this to counting \[\operatorname{Hom}((\mathbb Z/N\mathbb Z)^\times,\mathbb Z/2\mathbb Z)=\prod_{\text{prime }p}\operatorname{Hom}\left(\Big(\mathbb Z/p^{\nu_p(N)}\mathbb Z\Big)^\times,\mathbb Z/2\mathbb Z\right).\] Now, for each odd prime \(p\) dividing \(N\), we know \((\mathbb Z/p^{\nu_p(N)}\mathbb Z)^\times\) is cyclic of even size, so we get a factor of \(2\). For \(p=2\), we note that \[(\mathbb Z/2^\nu\mathbb Z)^\times=\begin{cases} 0 & \text{if }4\nmid N, \\(\mathbb Z/2\mathbb Z)\times(\mathbb Z/2^{\nu-2}\mathbb Z) & \text{if }4\mid N. \end{cases}\] Thus, we get no additional factors of \(2\) if \(N\) is odd or not divisible by \(4\), and we get one additional factor of \(2\) if \(N\) is divisible by \(4\), and we get another additional factor of \(2\) if \(N\) is divisible by \(8\). These factors of \(2\) exactly describe the size of \(S\), so we are done. \(\blacksquare\)

Remark 7. It is in fact possible to directly write down a Gauss sum which squares to \(\pm d\) for any odd \(d\), but it requires using more general Dirichlet characters than we have bothered with.

Because it is not very hard, let’s explain how we can use these ideas to prove quadratic reciprocity.

Theorem 8 (Quadratic reciprocity). Fix distinct odd primes \(p\) and \(q\). Then \[\left(\frac pq\right)=\left(\frac{q^*}p\right)\]

Proof. We will show that either side of the given equation equals \(1\) if and only if \(\mathrm{Frob}_p\) fixes \(\sqrt{q^*}\).

• On one hand, \(\mathrm{Frob}_p\) fixing \(\sqrt{q^*}\) is equivalent to having \(\mathrm{Frob}_p\) fix the quadratic Gauss sum \[\tau=\sum_{a\in\mathbb F_q^\times}\left(\frac aq\right)\zeta_q^a.\] However, \(\mathrm{Frob}_p\colon\zeta_q\mapsto\zeta_q^p\), so a quick reindexing shows that \(\mathrm{Frob}_p(\tau)=\left(\frac pq\right)\tau\).

  Alternatively, one can use Galois theory for this point: \(\mathrm{Frob}_p\) needs to be in the unique index-\(2\) subgroup of \((\mathbb Z/q\mathbb Z)^\times\), which consist of the squares.

• On the other hand, \(\mathrm{Frob}_p\) fixes \(\sqrt{q^*}\) if and only if \(\mathbb Q(\sqrt{q^*})\) is fixed by \(\mathrm{Frob}_p\) once embedded in \(\mathbb Q_p^{\mathrm{unr}}\). Because \(\mathrm{Frob}_p\) is a topological generator of \(\operatorname{Gal}(\mathbb Q_p^{\mathrm{unr}}/\mathbb Q_p)\), this is equivalent to \(\mathbb Q(\sqrt{q^*})\subseteq\mathbb Q_p\), which is equivalent to having \(\sqrt{q^*}\in\mathbb Q_p\). By Hensel’s lemma, this is equivalent to \(\left(\frac{q^*}p\right)=1\).

Combining the above two points completes the proof. \(\blacksquare\)