Weak cubic reciprocity in Diamond and Shurman

Question

In Diamond and Shurman, I am trying to understand the weak form of cubic reciprocity in section 4.11 - how it comes from the more standard version. In the following $A$ is the Eisenstein integers, and $\mu_3=e^{\frac{2\pi i}{3}}$.
The book says:

Let $d\in\mathbb{Z}$ be cubefree and $N=3\prod_{p|d}p$. Then there exists a character
$$\chi:A/NA)^*\to \{1,\mu_3,\mu_3^2\}$$ such that the multiplicative extension of $\chi$ to all of $A$ is trivial on $A^*$ and on primes $p \nmid N$, while on elements $\pi$ of $A$ such that $\pi\overline{\pi}$ is a prime $p\nmid N$ it is trivial if and only if $d$ is a cube modulo $p$

The standard version of cubic reciprocity also has a character the cubic residue character, however

1.The cubic residue character (the extended version) is defined for pairs $(\frac{\alpha}{\beta})$ where $\alpha,\beta\in A$ with $\beta$ coprime to $3$, whereas $\chi$ doesn't have pairs, it is defined on $(A/NA)^*$.
2.$\chi$ is supposed to be trivial on the units of $A$, and I believe the cubic reciprocity one is not.
3.$\chi$ is also (if I read correctly) trivial on rational primes $p\equiv 2\pmod{3}$, and I don't think the standard one is.

It seems the character he wants must be based on the standard one, but I don't see how.

Answer 1

Let me start with the analogous question for quadratic reciprocity which may be more familiar.

The analogue of ''weak quadratic reciprocity'' is: Given a squarefree $d$, let $N=4|d|$. Then there exists a quadratic Dirichlet character $\chi$ on $(\Bbb Z/N\Bbb Z)^\times$ such that for any prime $p\nmid N$, $\chi(p)=1$ if and only if $N$ is a square mod $p$.

The character in the second part of your question is the Jacobi symbol: $\left(\frac db\right) = \prod_i\left(\frac d{p_i}\right)^{k_i}$ where $b = \prod_i p_i^{k_i}$ and $2\nmid b$.

Now for fixed $b$, $\left(\frac-b\right)$ is trivially a character mod $b$. The content of the weak quadratic reciprocity law above is that conversely for fixed $d$, $\left(\frac d-\right)$ is a Dirichlet character (*) (but in general mod $4|d|$, not mod $|d|$). This follows from the usual quadratic reciprocity law: For simplicity take $d=q$ to be an odd prime. Then for odd $n$, define $\chi(n) = (-1)^{(n-1)(p-1)/4}\left(\frac np\right)$. This is easily seen to define a Dirichlet character mod $N=4p$, and by quadratic reciprocity $\left(\frac pq\right)=\chi(q)$ for an odd prime $q\ne p$. Also note that $\chi$ is trivial on $\Bbb Z^\times$.

Now you are in the exact same situation, except we are dealing with cubic reciprocity. Again in the second part you have the cubic analogue of the Jacobi symbol (*): $\left(\frac \alpha\beta\right)_3 = \prod_i\left(\frac\alpha{\pi_i}\right)_3^{k_i}$. Now weak cubic reciprocity is saying that $\left(\frac{d}-\right)_3$ is a character mod $N$, and again this follows from the usual cubic reciprocity law.

(*) (there is a small problem in my answer: not every non-zero integer is of the form $\prod_i p_i^{k_i}$ with primes $p_i$, similarly not every non-zero element in $A$ is of the form $\prod_i \pi_i^{k_i}$ with primes $\pi_i$, in both cases the problem is units. The best way to fix this is to consider ideals instead. Then the character $\chi$ is really defined on the ray class group modulo $(N)$ which can be naturally identified with the quotient of $(A/NA)^\times$ by the image of $A^\times$. This also explains how it fits into class field theory: $\chi$ is the Artin symbol for the extension $K(\sqrt[3]d)/K$ with $K=\Bbb Q(\sqrt{-3}) =\Bbb Q(\omega)$)

Answer 2

I figured out an answer of the type I wanted. I'll call the character $\chi_N$ instead of $\chi$ to indicate its dependence on $N$. The requirements for $\chi_N$ are:
(a) It must be trivial on units
(b) It must be trivial on rational primes $p$ not dividing $N$
(c) If $\pi\in A$ is such that $\pi\overline{\pi}$ is a rational prime $p$ not dividing $N$, then $\chi(\pi)=1\iff d$ is a cube modulo $p$.

A basic fact from Wikipedia is that rational primes which are coprime to $3$ fall into two categories:

• Those equivalent to $1$ modulo 3. These split into $2$ primes $\pi$ and $\overline{\pi}$ as described in (c) above.
• Those equivalent to $2$ modulo $3$. These remain prime in $A$.

It is easy to see that the elements $p$ in (c) are the split rational primes, i.e. $\equiv 1\pmod{3}$, and so for proving requirement (b) we only need to consider $p\equiv 2\pmod{3}$.
In the Wikipedia article he defines a primary as an element coprime to $3$ and equivalent to $\pm 1\pmod{3}$. Under this definition, he then states that for any element $\lambda\in A$ coprime to $3$, exactly one of $\lambda,\mu_3\lambda\text{, and }\mu_3^2\lambda$ is primary. The definition has the weakness that the unique factorization that he states isn't quite correct because the negative of a primary is primary. Others define a primary as only being equivalent to $2\pmod{3}$, and state that any $\lambda$ coprime to $3$ has exactly one associate which is primary. Under this definition the unique factorization works. But I will use the Wikipedia definition. It is a minor difference, but I wanted to note it.

As suggested, use the character $\chi_N(r)=(\frac{N}{r})$, where $(\frac{N}{r})$ is the cubic character extended to non-primes. It is initially defined on all $R_N:=\{r\in A\ |\ r\text{ is coprime to } N\}$. Any such $r$ reduces an element of $A_N:= (A/NA))^*=R_N/NR_N$. But we need to show it is well-defined on $A_N$; I will show this below, but for now I will just use the cubic character itself.

First note that (a) is a general property of the cubic character. If the rational prime $p$ is $\equiv 2\pmod{3}$, then $3\nmid p-1$, so on $(\mathbb{Z}/p\mathbb{Z})^*$ cubing has trivial kernel, so is surjective. Therefore $d$ is a cube, say $N=N_1^3\pmod{p}$, and so $(\frac{N}{p})=(\frac{N_1}{p})^3=1$. So we only need to prove (c)

The next step is to break up $N$ into $3Part\times N_1$ where $N_1$ is coprime to $3$ and $3Part$ is $3$ or $9$. Correspondingly $\chi_N$ breaks into $\chi_{3Part}\chi_{N_1}$..
It is enough to show that if $r_1\equiv r_2\pmod{N}$, both coprime to $N$. then $\chi_{3Part}(r_1)=\chi_{3Part}(r_2)$ and $\chi_{N1}(r_1)=\chi_{N1}(r_2)$. By multiplying $r_1$ by a unit $u$, we can make it primary. So we can replace $r_1$ and $r_2$ by $r_1u$ and $r_2u$, giving the same value for $\chi_3$ and $\chi_1$. But now $r_1$ being primary and $r_1\equiv r_2\pmod{N}\implies r_1\equiv r_2\pmod{3}\implies r_2$ is primary.

The supplementary law for $3$ shows $\chi_{3Part}(r_1)=1=\chi_{3Part}(r_2)$. Also, since $r_1$ and $r_2$ are both primary, cubic reciprocity says that $\chi_{N_1}(r_1)=(\frac{r_1}{N_1})$ and $\chi_{N_1}(r_2)=(\frac{r_2}{N_1})$. But since $r_1\equiv r_2\pmod{N}$, $r_1\equiv r_2\pmod{N_1}$, and therefore by the equal property of the cubic character for numerators modulo the denominator, $(\frac{r_1}{N_1})=(\frac{r_2}{N_1})$, and by the same law again $\chi_{N_1}(r_1)=\chi_{N_1}(r_2)$. So $\chi$ is well-defined on $(A/NA)^*$.

Consider the requirement (c) by considering a rational prime $p\equiv 1 \pmod{3}$ with $N(\pi)=\pi\overline{\pi}=p$.
Lemma. $\pi\mid N^k-1\iff N\text{ is a cube }\pmod{\pi}\iff N\text{ is a cube }\pmod{p}$.
Proof. $A/pA$ has $p^2$ elements, and it breaks up into the direct sum of $A/\pi A$ and $A/\overline{\pi}A$, so these each have $p$ elements. Therefore the field inclusion $\mathbb{Z}/p\mathbb{Z}\to A/\pi A$ is an isomorphism, with cubes mapping to cubes. Q.E.D. This really works because $N$ is an ordinary integer.

The lemma shows that $$(\frac{N}{\pi})=1\iff N\text{ is a cube }\pmod{p}$$ whereas for part c) we need
$$(\frac{N}{\pi})=1\iff d\text{ is a cube }\pmod{p}$$

This is quite similar, but there are two differences (replacement of prime powers) between $d$ and $N$:
(1) Primes $q\ne 3$ dividing $d$ may occur as a square in $d$ but not in $N$.
(2) The power of $3$ in the definition of $N$ may be different.

Fix any prime $q$ dividing $N$ (so it is prime to $p$.).
Lemma. $q$ is a cube $\pmod{p}\iff q^2$ is.
$\implies$ Obvious. $\impliedby$ If $q^2\equiv x^3\pmod{p}$, then $$q^4\equiv x^3\pmod{p}\implies q\equiv (q^{-1}x)^3\pmod{p}$$ Since $A_N$ breaks up as a product over all primes dividing $N$ (including $3$), and since an element of $A_N$ is a cube if and only if each component is, we just need to consider cases 1) and 2)
For case 1), the lemma implies the replacement of $q^2$ in $d$ with $q$ in $N$ doesn't matter. For case 2), the possibilities are

• $3$ doesn't divide $d$, in which case the 3Part acts trivially and so can be ignored
• $3$ divides $d$ but $9$ doesn't, in which case $3Part$ is $3^2$, and the lemma suffices
• $9$ divides $d$, in which case the 3 part of $N$ is $9$, the same as the previous item

I hope this is complete and typed in correctly..