Of interest to me is the following question (it would be nice to find out what is known in its direction):

Given a Galois number field $K/\mathbb{Q}$ and a completely and principally split prime number $$p = \prod_{\sigma \in \mathrm{Gal}(K/\mathbb{Q})}{\sigma(q)},$$

where $q \in \mathcal{O}_K$ is a prime element, "how probable", in a loose sense, is it that the images of the numbers $\sigma(q)$, $(\sigma \neq \mathrm{id})$ in the residue field $\mathcal{O}_K/(q)$ $(\cong \mathbb{F}_p)$ have multiplicative orders of one's choice? For instance, given some choice of distinct elements ${\sigma_1, \sigma_2, \ldots, \sigma_r \in \mathrm{Gal}(K/\mathbb{Q})}$, can one always find "many" $p$ such that the numbers $\sigma_1(q), \ldots, \sigma_r(q) \in \mathrm{Gal}(K/\mathbb{Q})$ are quadratic residues in the field $\mathcal{O}_K/(q)$, and the remaining numbers $\sigma(q),\; \sigma \neq \sigma_j\; (j = 1, \ldots, r),\; \sigma \neq \mathrm{id}$ are quadratic nonresidues?

  • $\begingroup$ So, to state a concrete version, if we look at primes that are $p \equiv 1 \bmod 8$ and we write $p = a^2+b^2$, we would like to know whether $\left( \frac{a}{p} \right)$ is $1$ and $-1$ equally often. (For primes that are $5 \bmod 8$, switching $a$ and $b$ switches the residue class, so we don't get a good question.) It seems hard to imagine the answer could be no, but I also see no attack. $\endgroup$ May 8, 2014 at 18:24
  • $\begingroup$ Whoa, I take it back! I just checked the first 93 primes which are $1 \bmod 8$, and $a$ is always square! (To fill in a few details above, $\left( \frac{a-bi}{a+bi} \right) = \left( \frac{2a}{a+bi} \right) = \left( \frac{2a}{p} \right)$, but $\left( \frac{2}{p} \right) \equiv 1$ since $p \equiv 1 \bmod 8$.) $\endgroup$ May 8, 2014 at 18:31

2 Answers 2


It depends on how large the image $U$ of the global units $\mathcal{O}_K^\times$ in the group $$\prod_\sigma (\mathcal{O}_K/(\sigma(q)))^\times$$ happens to be. It's almost never the full group (because the unit rank is always smaller than the field degree), and in general it will be very far from it.

In the rare case when $U$ is all of that, you can multiply $q$ by a unit to move its images in the other factors (i.e. $\sigma \neq \mathrm{id}$) anywhere you want without changing the prime ideal it generates. But usually $U$ will be much smaller, even when $\mathcal{O}_K^\times$ happens to surject onto each factor $(\mathcal{O}_K/(\sigma(q)))^\times,$ and certainly when it doesn't. For an example, take $K=\mathbb{Q}(\zeta_4)$ and $p=17$: you only have the fourth roots of unity available to adjust $q$. The image of $q=4+\zeta_4$ in the other residue class field $\mathcal{O}_K/(4-\zeta_4)$ is $8$, and you can move it to any of $-2, -8, 2$; but these are all quadratic residues.

All this is not yet a complete answer - it also depends on how, as $p$ varies, the subgroups of $\mathbb{F}_p^\times$ fit into the additive structure of $\mathbb{F}_p$. But in general I expect you'll find that few $p$ will work.

Edit: I was being sloppy. In what ways we can move the images of $q$ in the other residue class fields actually depends on the image $U'$ of $\mathcal{O}_K^\times$ in the group $$\prod_{\sigma\neq\mathrm{id}} (\mathcal{O}_K/(\sigma(q)))^\times$$ (one factor less), and this isn't automatically so small in the totally real case where we have $\mathrm{deg}(K/\mathbb{Q})-1$ independent fundamental units, as many as there are factors in the latter product. I can't complete an argument offhand but now I'm inclined to think that infinitely many $p$ might be viable in this case.

Edit 2014-05-08: Here is a partial positive result. Assume that

  1. $K$ has class number $1$,
  2. $K$ has infinitely many units,
  3. $K$ is abelian over $\mathbb{Q}$,
  4. and a suitable set of generalized Riemann hypotheses.

Then there are infinitely many $q$ (and thus $p$) for which you can fully control what happens in at least one factor $\mathcal{O}_K/(\sigma(q))$ ($\sigma\neq\mathrm{id}$).

P. J. Weinberger proved in 1973 (On Euclidean Rings of Algebraic Integers) that $\mathcal{O}_K$ is Euclidean in this case, not necessarily for the norm. (This doesn't depend on $K$ being abelian.) A key step in the proof is the existence of sufficiently many prime ideals $\mathfrak{q}$ such that a given fundamental unit $\varepsilon\in\mathcal{O}_K^\times$ maps to a primitive root mod $\mathfrak{q}$. (Being fundamental means $\varepsilon$ is not a proper power of a unit and, in particular, not a root of unity.) Somewhat more precisely, his Theorem 4 yields infinitely many such $\mathfrak{q}$ in a given (ray) ideal class; combining this with our assumption 3 we can get infinitely many such $\mathfrak{q}=(q)$ generated by an element congruent to $1$ modulo the conductor and thus fully split. (This may be overkill.)

Now if the image of $\varepsilon$ generates all of $(\mathcal{O}_K/(q))^\times$, then so does $\sigma(\varepsilon)$ in $(\mathcal{O}_K/(\sigma(q)))^\times$ for each $\sigma$, since the image is the same residue class in $\mathbb{Z}/p\mathbb{Z}$. Take your favorite $\sigma\neq\mathrm{id}$ and multiply $q$ by a suitable power of $\sigma(\varepsilon)$ to move the image of this product mod $(\sigma(q))$ to any desired prime residue class, QED.

But it is not clear whether you can then continue to do this in additional residue class fields without destroying what you already have. The first obstacle is that the $\sigma(\varepsilon)$ may not be multiplicately independent. (They certainly won't be unless $K$ is totally real. Even then, $\varepsilon$ might have norm $\pm1$ in a subfield strictly between $K$ and $\mathbb{Q}$.) The second and more serious obstacle is that any further tweaking of $q$ must be done by units with image $1$ in each of the residue class fields we've already dealt with, and we may run out of units mapping to primitive roots before we've dealt with the whole lot.


In this answer, I will analyze the case of the Gaussian integers, and explain the peculiar phenomena pointed out in my comment above. So we are considering whether or not $a+bi$ is a square modulo $a-bi$, where $a^2+b^2$ is a prime $p$.

As GNiklasch points out in his answer, if $i$ is not a square modulo $a-bi$, then one of $(a+bi)$ and $i(a+bi)$ will be square modulo $a-bi$ and there other will not, so there is no interesting question here. The element $i$ will be square modulo $a+bi$ if and only if there is a primitive $8$-th root of unity in $\mathbb{Z}[i]/(a+bi) \cong \mathbb{F}_p$, so if and only if $p \equiv 1 \bmod 8$. So the interesting problem is when $p \equiv 1 \bmod 8$.

It turns out that, in this case, $a-bi$ is always a square modulo $a+bi$. To prove this, we will use quartic reciprocity. If $\pi$ is a Gaussian prime not equal to $1+i$, and $z$ is a Gaussian integer relatively prime to $\pi$, then $\left[ \frac{z}{\pi} \right]$ is defined to be the unique power of $i$ so that $z^{(N(\pi)-1)/4} \equiv \left[ \frac{z}{\pi} \right] \bmod \pi$.

We need two key properties: $$\left[ \frac{\overline{\alpha}}{\overline{\beta}} \right] = \overline{ \left[ \frac{\alpha}{\beta} \right] }$$ which is straightforward from the definition and $$\left[ \frac{\alpha}{\beta} \right] \left[ \frac{\beta}{\alpha} \right]^{-1} = (-1)^{\frac{N(\alpha)-1}{4} \cdot \frac{N(\beta)-1}{4}}$$ which is quartic reciprocity.

Putting these together, if $a^2+b^2 \equiv 1 \bmod 8$, then $$\left[ \frac{a+bi}{a-bi} \right] = \overline{ \left[ \frac{a-bi}{a+bi} \right]} \ \mbox{and} \ \left[ \frac{a+bi}{a-bi} \right] = \left[ \frac{a-bi}{a+bi} \right].$$

So $\left[ \frac{a+bi}{a-bi} \right] = \pm 1$. This means that the quadratic character $\left( \frac{a+bi}{a-bi} \right)=\left[ \frac{a+bi}{a-bi} \right]^2$ is $1$.

We can generalize this to prove something like: If $K$ is a field containing the $m$-th roots of unity and Galois over $\mathbb{Q}$ and $\pi$ is an element generating a prime ideal and congruent to $1$ modulo a sufficiently high power of $m$, then for any $\sigma \in G$ we have $$\left( \frac{\sigma(\pi)}{\pi} \right)_m = \sigma \ \left( \frac{\sigma^{-1}(\pi)}{\pi} \right)_m.$$ Here $\left( \frac{\ }{\ } \right)_m$ is the power residue symbol.

I did an experiment to see what would happen in $\mathbb{Q}(\sqrt{-2})$. This is a PID, the primes that split are $p \equiv 1$ or $3 \bmod 8$ and the unit $-1$ is a square when $p \equiv 1 \bmod 8$. Out of the first $241$ primes that are $1 \bmod 8$, there are $117$ where $a+b \sqrt{-2}$ is a square modulo $a-b \sqrt{-2}$ and $124$ where it is a nonsquare, with no obvious pattern to which primes are which. So, at least some of the time, it looks like we do have equidistribution. I have no idea how to prove it though; it doesn't resemble any equidistribution theorem or conjecture I know of.

  • $\begingroup$ I could add that my own small computation said that, for instance, in the (cubic) splitting field of $x^3 - 3x + 1$, all four possible combinations of quadratic residues/nonresidues occur. $\endgroup$
    – Albertas
    May 9, 2014 at 20:41
  • $\begingroup$ With equal densities? $\endgroup$ May 10, 2014 at 0:45
  • 1
    $\begingroup$ A root $\alpha$ of $x^3 - 3x + 1$ gives a power basis as a $\mathbb{Z}$-module for the corresp. ring of integers. One can then write any integral element as $a + b\alpha + c(\alpha^2 - 2)$. If, given a prime $p$ that splits (that is, = 1 or 8 mod 9), one chooses a factor $q$ with smallest value of $|a|$, then, for $p < 10000$, residues of $\sigma(q), \sigma^2(q)$ modulo $q$ seem to be squares/nonsquares equally likely: (square, square) - 102 times, (square, nonsquare) - 99 times, (nonsquare, square) - 101 times, (nonsquare, nonsquare) - 101 times. $\endgroup$
    – Albertas
    Jun 5, 2014 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.