Let $K$ be the quadratic field $\mathbb{Q}(\sqrt d)$ where $d\in\mathbb{N}_{\geq1}$ is square-free, and let $\mathbb{Z}_K$ be the ring of integers of $K$. Define the fundamental unit of $\mathbb{Z}_K$ as the minimal invertible element of $\mathbb{Z}_K$ larger than $1$.

We can necessarily write this fundamental unit as $a+b\sqrt d$ where $a,b\in\mathbb{Q}^+$. Then, since it is invertible, its field norm $a^2-db^2$ is $\pm1$. Here is what the sign of this norm ends up being for the first bunch of values of $d$.

$2$ $-$
$3$ $+$
$5$ $-$
$6$ $+$
$7$ $+$
$10$ $-$
$11$ $+$
$13$ $-$
$14$ $+$
$15$ $+$
$17$ $-$
$19$ $+$
$21$ $+$
$22$ $+$
$23$ $+$
$26$ $-$
$29$ $-$
$30$ $+$

Given $d$, is there a way to tell what this sign will be, than computing the fundamental unit and its product with its conjugation? I'm also curious what more advanced concepts this sign may be related to.

  • $\begingroup$ Possibly related to the pell-equation and when $a^2-db^2=-1$ has a solution. But I wonder whether a norm actually can be negative. $\endgroup$
    – Peter
    Commented Aug 24, 2017 at 9:38
  • 1
    $\begingroup$ @Peter yes, definitely related. The norm of $a+b\sqrt d$ as in the question is $a^2-db^2$. One just has to be careful about what values $a$ and $b$ can take. They can always lie in $\mathbb{Z}$, but if $d\equiv_41$ then there is also the possibility that both lie in $\mathbb{Z}+\{\frac{1}{2}\}$. Moreover, there is a solution to $a^2-db^2=-1$ in this way if and only if the corresponding fundamental unit has negative norm, because it cyclically generates the whole unit group. $\endgroup$
    – j0equ1nn
    Commented Aug 24, 2017 at 22:08
  • $\begingroup$ ...up to sign. That is $\mathbb{Z}_K=\{\pm(a+b\sqrt d)^\ell\mid\ell\in\mathbb{Z}\}$. $\endgroup$
    – j0equ1nn
    Commented Aug 28, 2017 at 1:24
  • 1
    $\begingroup$ The group of units and the fundamental unit are involved in the functional equation of the Dedekind zeta function and the class number formula. Also here they suggest the case $N(\varepsilon) =-1$ is not totally understood. $\endgroup$
    – reuns
    Commented Aug 28, 2017 at 2:13

2 Answers 2


Introductory Algebraic Number Theory by Alaca & Williams deals with this in a fair level of detail in Chapter 11.

  • Theorem 11.5.4: given prime $d \equiv 1 \pmod 4$, the fundamental unit has norm $-1$. e.g., $d = 5$, $N(\phi) = -1$.
  • Theorem 11.5.5: if $d$ is divisible by some prime $p \equiv 3 \pmod 4$, the fundamental unit has norm 1. e.g., $N(2 + \sqrt 3) = N(5 + 2 \sqrt 6) = 1$.
  • Theorem 11.5.6: if $d = 2p$ and $p \equiv 5 \pmod 8$, the fundamental unit has norm $-1$. e.g., $N(3 + \sqrt{10}) = N(5 + \sqrt{26}) = -1$.
  • Theorem 11.5.7: if $d = pq$ with distinct primes $p \equiv q \equiv 1 \pmod 4$ and $$\left(\frac{p}{q}\right) = \left(\frac{q}{p}\right) = -1$$ (I know that by quadratic reciprocity that's a bit redundant) then the fundamental unit has norm $-1$. e.g., $N(8 + \sqrt{65}) = -1$.

These are from my notes, not directly from the book; hopefully I haven't made any mistakes of transcription. I don't remember if they say anything about the $d$ not covered by these four theorems. Proofs are included for all four of them, I think one or two of them is given more than one proof.

  • 3
    $\begingroup$ I've obtained a copy of that book, thanks! It does give a very accessible treatment of the topic. $\endgroup$
    – j0equ1nn
    Commented Aug 28, 2017 at 1:22

There are some partial results; if $d\equiv3\pmod 4$ then the fundamental unit has norm $+1$. The same is true if $d$ has a positive factor congruent to $3$ modulo $4$. The proof is just working modulo $4$.

On the other hand if $d=p\equiv1\pmod 4$ is a prime, then the fundamental unit has norm $-1$. To prove this one can show that if $\varepsilon>0$ is a unit of norm $1$ in $K=\Bbb Q(\sqrt d)$ then $\sqrt{\varepsilon}\in K$ also.

Things get more complicated when there are more prime factors. If $d=pq$ where $p$ and $q$ are primes $\equiv1\pmod 4$ then one can have both $+1$ and $-1$ as the norm of the fundamental unit.

  • $\begingroup$ Could you say confidently that this is the extent of the known results (not counting experimental probabilistic evidence)? $\endgroup$
    – j0equ1nn
    Commented Aug 24, 2017 at 22:18

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