Quadratic Dirichlet Characters

We work out some basic facts about quadratic Dirichlet characters from the perspective of Galois theory.

Recall that the Kronecker–Weber theorem shows that \[\mathbb Q^{\mathrm{ab}}=\bigcup_{N\ge1}\mathbb Q(\zeta_N).\] Thus, there is a canonical isomorphism \[\widehat{\mathbb Z}^\times\to\operatorname{Gal}(\mathbb Q^{\mathrm{ab}}/\mathbb Q)=\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)^{\mathrm{ab}}.\] In the sequel, we will denote this isomorphism by \(\operatorname{Art}_{\mathbb Q}\). This allows us to make the following definition.

Definition 1. Fix a quadratic extension \(K\) of \(\mathbb Q\). Then we define the quadratic character \(\chi_K\) of \(K\) to be the composite \[\widehat{\mathbb Z}^\times\cong\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)^{\mathrm{ab}}\twoheadrightarrow\operatorname{Gal}(K/\mathbb Q)\cong\{\pm1\}\subseteq\mathbb C^\times.\] If \(K=\mathbb Q(\sqrt d)\), we may abbreviate \(\chi_K\) to \(\chi_d\).

Remark 2. It is a bit dishonest to use the Kronecker–Weber theorem to define these characters. Technically, one only needs to know that the quadratic field \(K\) is contained in a cyclotomic field, which can be shown directly. Alternatively, one can simply view the quadratic character \(\chi_K\) as defined on \(\operatorname{Gal}(\mathbb Q^{\mathrm{ab}}/\mathbb Q)\).

We are interested in proving some basic facts on these characters. Here is the kind of calculation we want to do.

Example 3. Fix a quadratic extension \(K\) of \(\mathbb Q\). Then we claim that \[\chi_K(-1)\stackrel?=\begin{cases}+1 & \text{if }K\text{ is real}, \\ -1 & \text{if }K\text{ is imaginary}.\end{cases}\] To start, note \(-1\in\widehat{\mathbb Z}^\times\) goes to the automorphism \(\operatorname{Art}_{\mathbb Q}(-1)\) which sends \(\zeta\) to \(\zeta^{-1}\) for each root of unity \(\zeta\). But \(\overline\zeta=\zeta^{-1}\), so we see that \(\operatorname{Art}_{\mathbb Q}(-1)\) is complex conjugation. It follows that \(\chi_K(-1)=+1\) if and only if complex conjugation acts trivially on \(K\), so the result follows.

If we squint, we can make the above example also work for “finite places.”

Example 4. Fix a quadratic extension \(K\) of \(\mathbb Q\), and let \(p\) be a rational prime unramified in \(K\). Then we claim \[\chi_K(p)\stackrel?=\begin{cases}+1 & \text{if }K\subseteq\mathbb Q_p, \\ -1 & \text{if }K\not\subseteq\mathbb Q_p.\end{cases}\] Indeed, note that \(\operatorname{Art}_{\mathbb Q}(p)\) is acting by the \(p\)th power on a generating set of \(\mathbb Q^{\mathrm{ab}}\), so it is the Frobenius \(\mathrm{Frob}_p\). Accordingly, we embed \(K\hookrightarrow\mathbb Q_p^{\mathrm{unr}}\), and we see that \(\mathrm{Frob}_p\) extends (uniquely) to the usual Frobenius in \(\operatorname{Gal}(\mathbb Q_p^{\mathrm{unr}}/\mathbb Q_p)\), which is a topological generator. We conclude that \(\mathrm{Frob}_p\) acts trivially on \(K\) if and only if \(K\subseteq\mathbb Q_p\).

Remark 5. If we write \(K=\mathbb Q(\sqrt d)\), then the preceding example shows that \[\chi_d(p)\stackrel?=\left(\frac dp\right)\] for any odd prime \(p\) unramified in \(K\). Indeed, \(K\subseteq\mathbb Q_p\) if and only if \(\sqrt d\in\mathbb Q_p\). Because \(p\) does not divide \(d\), this is equivalent to having \(\sqrt d\in\mathbb Z_p\). By Hensel’s lemma (noting that \(p\) is odd for now), this is equivalent to \(\sqrt d\in\mathbb F_p\), which is equivalent to \(\left(\frac dp\right)=+1\).

Remark 6. Here is another way to derive the result of Remark 5: \(\mathrm{Frob}_p\) fixes \(\sqrt d\) if and only if \(p\) splits completely in \(K=\mathbb Q(\sqrt d)\). An understanding of prime-splitting in quadratic fields shows that splitting completely is equivalent to \(\sqrt d\in\mathbb F_p\), which means \(\left(\frac dp\right)=1\).

Remark 7. The same argument as in Remark 5 also works for \(p=2\) when it is unramified, meaning \(d\equiv1\pmod4\). We claim that \[\chi_d(2)=\begin{cases}+1 & \text{if }d\equiv1\pmod 8,\\-1 & \text{if }d\equiv5\pmod 8.\end{cases}.\] To see this, note \(\chi_d(2)=+1\) if and only if \(\mathbb Q(\sqrt d)\) embeds into \(\mathbb Q_2\). This is equivalent to having \(\frac{1+\sqrt d}2\in\mathbb Q_2\), which means that \(x^2-x+\frac{1-d}4\) has a root. This now descends by Hensel’s lemma to having \(x^2-x+\frac{1-d}4\) admit a root over \(\mathbb F_2\), which is equivalent to \(\frac{1-d}4\equiv0\pmod2\), or \(d\equiv1\pmod8\).

We now attempt to compute the conductor of \(\chi_K\). By the nature of the canonical isomorphism \(\widehat{\mathbb Z}^\times\cong\operatorname{Gal}(\mathbb Q^{\mathrm{ab}}/\mathbb Q)\), this means that we need to find the smallest \(N\) for which \(K\subseteq\mathbb Q(\zeta_N)\). This amounts to a classification of the quadratic subfields of cyclotomic fields, for which we pick up the following result.

Proposition 8. 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. Omitted. See this blog post. \(\blacksquare\)

Corollary 9. Let \(d\) be a squarefree integer, and set \(K=\mathbb Q(\sqrt d)\). Then the conductor of \(\chi_K\) is the absolute value of the discriminant of \(K\). In other words, the conductor is \[\begin{cases}\left|d\right| & \text{if }d\equiv1\pmod4, \\ 4\left|d\right| & \text{otherwise}.\end{cases}\]

Proof. Let \(N\) be the conductor, which by definition means that \(\mathbb Q(\zeta_N)\) is the smallest cyclotomic field containing \(K\). By Proposition 8, we see that \(\mathbb Q(\zeta_N)\) containing \(\sqrt d\) means that each prime dividing \(d\) also divides \(N\), so we at least have \(d\mid N\). On the other hand, we see that \(\mathbb Q(\zeta_{4d})\) has square roots of all the prime factors of \(d\) (even \(2\)) as well as \(\sqrt{-1}\), so certainly \(N\le4d\).

We now have the following cases.

• If \(d\) is even, then Proposition 8 tells us that we must also have \(8\mid N\). It follows that \(N\ge4d\), so equality follows.
• If \(d\equiv1\pmod4\), then we see \[d=\prod_{p\mid d}p^*.\] Indeed, this certainly holds up to sign by unique prime factorization, and we have the correct sign by checking\(\pmod4\). It follows that \(N\le d\), so equality follows.
• If \(d\equiv3\pmod4\), then the previous point shows that \(\sqrt{-d}\in\mathbb Q(\zeta_d)\). Thus, because \(\sqrt{-1}\notin\mathbb Q(\zeta_d)\), we are forced to have \(4\mid N\) and so \(N\ge4d\), so equality follows.

The above cases complete the proof. \(\blacksquare\)

Remark 10. It is perhaps worth saying something about why it would be difficult to compute the conductor \(N\) of \(\chi_K\). In brief, a straightforward calculation with the definition of the Frobenius (in Example 4 and Remarks 5 and 6) was able to show that \[\chi_K(p)=\left(\frac dp\right)\] for any odd prime \(p\). (Another way to say this is that this calculation happens entirely on the “Galois side.”) However, establishing a conductor means that \(\chi_K\) should factor through \((\mathbb Z/N\mathbb Z)^\times\), meaning the Jacobi symbol \(\left(\frac dp\right)\) should only depend on \(p\pmod N\). Thus, the existence of a conductor has a similar level of strength as quadratic reciprocity. Indeed, in this blog post, we saw that putting \(K\) inside the correct cyclotomic field allows one to quickly prove quadratic reciprocity.