Let $\zeta$ be a primitive cubic root of unity, i.e., $\zeta$ is a complex number such that $\zeta^{3}=1$ and $\zeta\neq1$.

Let us consider the Galois extension $\mathbb{Q}(\zeta)/\mathbb{Q}$ (the Eisenstein field) and the norm map given by \begin{equation} N(a+b\zeta)=a^{2}-ab+b^{2}, \end{equation} where $a$ and $b$ are rational numbers.

Since the norm defines a homomorphism on the multiplicative groups $\mathbb{Q}(\zeta)^{\times}\to\mathbb{Q}^{\times}$, we know already that the image will be a multiplicative subgroup of $\mathbb{Q}^{\times}$. Can we describe this group nicely?

I've tried putting some values and trying to figure out the form of this group inside $\mathbb{Q}^{\times}$, but I can't really see a pattern going on. Taking primes $p\equiv 2\pmod3$, we can see that this norm map is not surjective, but I don't know the group $\mathbb{Q}^{\times}$ really well to know its subgroup structure.

I'd guess that Global Class Field Theory gives an answer to this question, but I don't know much about it.

Thanks for any help or references.

  • $\begingroup$ what exactly is a norm map? Is it "just a homomorphism" from the extension field to the base field? I've never seen this before. Thanks! $\endgroup$ May 23, 2016 at 16:40
  • $\begingroup$ If $L$ is a finite extension of $\mathbb{Q}$, you can the norm of an element $x\in L^{\times}$ as $N_{L}(x)=\prod \sigma(x)$, where $\sigma$ varies through the field automorphisms $L\to L$. If the extension is "well-behaved" in a certain sense (Galois extensions), then there are $[L\colon K]$ such automorphisms. In $\mathbb{Q}(\zeta)=\{a+b\zeta\}$ the automorphisms are given by $\varphi_{1}\colon \zeta\to\zeta$ and $\varphi_{2}\colon \zeta\to\zeta^{2}$. So, we have $N(a+b\zeta)=(a+b\zeta)(a+b\zeta^{2})$, which gives the question homomorphism when you use $\zeta^{2}=-1-\zeta$ $\endgroup$
    – Shoutre
    May 23, 2016 at 16:42
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    $\begingroup$ Since this is a quadratic form, the elements of the group are completely determined by Hasse-Minkowski principle: as long as there are $p$-adic solutions to $x^2-xy+y^2 = a$ for each $p$ (and a real solution), then there is a rational one. In this case I believe it amounts to "$a > 0$ and no prime of the form $3k+2$ occurs to an odd power in $a$". $\endgroup$
    – Erick Wong
    May 23, 2016 at 16:45
  • $\begingroup$ Thanks for your comment. I was looking for some argument (ideally) that would cover also higher degree fields, where the norms are no longer quadratic forms, but I guess this is a difficult problem in general. Thanks, though. $\endgroup$
    – Shoutre
    May 23, 2016 at 16:49
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    $\begingroup$ @Shoutre Just consider the homogeneous equation $f(x,y) - az^2 = 0$ where $z$ is a third variable. $\endgroup$
    – Erick Wong
    May 24, 2016 at 15:16

2 Answers 2


The idea is to reduce the problem to integer arithmetic. Since $\mathbb Q(\zeta)$ is the field of fractions of $\mathbb Z[\zeta]$, we can write $$x+y\zeta = \frac{a+b\zeta}{c+d\zeta}$$ where $x,y\in\mathbb Q$ and $a,b,c,d\in \mathbb Z$, and where the numerator and the denominator on the right hand side are coprime. Here we are using the fact that $\mathbb Q(\zeta)$ has class number $1$.

In particular, $\frac mn\in\mathbb Q^\times$ (written in lowest terms) is a norm from $\mathbb Q(\zeta)^\times$ if and only if $m,n$ are both norms from $\mathbb Z[\zeta]$.

We can calculate using quadratic reciprocity that a prime $p$ is a norm from $\mathbb Z[\zeta]$ if and only if $p=3$ or $p\equiv 1 \pmod 3$. It follows that $$n = p_1^{e_1}\cdots p_k^{e_k}$$ is a norm from $\mathbb Z[\zeta]$ if and only if $e_i$ is even whenever $p_i\equiv 2\pmod 3$. (Note that all norms are positive in this case.)

We deduce that the image $N(\mathbb Q(\zeta)^\times)$ is the subgroup of $\mathbb Q^\times$ generated by $$\{p:p\equiv 1\pmod 3\}\cup\{3\}\cup\{p^2 : p\equiv 2\pmod 3\}.$$

To illustrate the difficulty of generalising this when $K$ does not have class number $1$, neither $2$ nor $3$ are norms from $\mathbb Z[\sqrt{-5}]$, but $$\frac 32 = \frac 64 =N\left(\frac{1+\sqrt{-5}}2\right) .$$

We can get around this by considering factorisation of ideals. Let $K$ be a number field, and let $x\in K^\times$. Write $$(x) = \frac{\mathfrak a}{\mathfrak b},$$ where $\mathfrak{a,b}$ are coprime ideals of $\mathcal O_K$. Since $\frac{\mathfrak a}{\mathfrak b}$ is principal, both $\mathfrak a$ and $\mathfrak b$ lie in the same class of the ideal class group. In particular, we can find some integral ideal $\mathfrak c$ such that $\mathfrak{ac}$ and $\mathfrak {bc}$ are both principal.

Hence, $$|N(x)| = \frac{N(\mathfrak a)}{N(\mathfrak b)}=\frac{N(\mathfrak {ac})}{N(\mathfrak {bc})}=\frac{|N(a)|}{|N(b)|},$$ for some $a,b\in\mathcal O_K$.

It follows that the image $|N(K^\times)|\subset \mathbb Q_{>0}^\times$ is generated by the elements of $\mathbb Z$ which are norms from $\mathcal O_K$. There should be some way to sort out the issue with signs.


In dealing with rational values of rational binary quadratic forms, I think that it’s much simpler to use Hasse’s norm principle for cyclic extensions in place of the Minkowski-Hasse principle for quadratic forms suggested by @Erik Wong (but of course both principles come from CFT). Everything is known « in a nice way » according to the following process :

Diagonalizing the quadratic form if necessary, we are led to solve the equation $a = x^2+ dy^2$ in $\mathbf Q$, where $d$ is a given square free rational number, or equivalently to solve the normic equation $a = N(\alpha)$, where $N$ denotes the norm map from $K = \mathbf Q(\sqrt –d) $ to $\mathbf Q$. Hasse’s principle asserts that $a$ is a global norm in $K/\mathbf Q$ iff $a$ is a norm in all the local extensions $K_v/\mathbf Q_{p}$, where $\mathbf Q_{p}$ is the field of $p$-adic numbers (resp. $\mathbf R$) if $p$ is a finite (resp. the infinite) prime, and $K_v$ is the completion of $K$ at a prime $v$ above $p$. The local condition at $v$ is equivalent to $(a, d)_p$ = 1, where $(a, d)_p$ is the Hilbert symbol at $p$ (which generalizes the Legendre symbol), and this latter criterion is effective in the sense that we have explicit formulae for the Hilbert symbol, and moreover we need to compute only a finite number of them. All this is explained in great detail e.g. in chapters II and III of Serre’s « A Course in Arithmetic ». Note that we don’t need to suppose that the class number of K is 1 as in the case treated by @Mathmo123.

Going a step further, we can ask for integral values of integral binary quadratic forms, as e.g. in the questions posted by @Henry and @Kieren MacMillan. This is genuine arithmetic, which needs CFT over the quadratic field $K$, and no longer above $\mathbf Q$ alone.. See e.g. D. Cox’s book « Primes of the form $x^2 + ny^2 »$.


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