# Factoring rational primes over the Eisenstein integers - when can a prime be written as $j^2+3k^2$?

I've been messing around with Eisenstein integers, and comparing them with Gaussian integers. Many things are clear, but I'm struggling with the details underlying which rational primes split, and which are inert. I know that the inert primes are precisely those congruent to $2$, modulo $3$, and the splitting primes are congruent to $1$. (The prime $3$ is the single ramified prime in this ring.)

Now, the corresponding fact for Gaussian primes is that $2$ is ramified, and all of the odd primes either split or are inert, according to whether they are congruent to $1$ or $3$, respectively, modulo $4$. This is something I can prove, and the interesting part of the proof is showing that, if $p=4m+1$, then $p$ can be written as a sum of two squares.

I thought initially that the corresponding fact to establish for Eisenstein integers would be:

1) If a rational prime $p$ can be written as $p=3m+1$, then we can express it in the form $p=j^2+3k^2$. (Here $j$ and $k$ are rational integers.)

Then, I realized that norms don't work that way in Eisenstein, and the fact I really need is:

2) If a rational prime $p$ can be written as $p=3m+1$, then we can express it in the form $p=j^2+jk+k^2$. (with $j$ and $k$ in $\Bbb Z$, as above.)

I am reading that claim (2) "can be shown", but I don't know how to show it. I've tried mimicking the corresponding proof from the Gaussian primes, but there's an intermediate step, where $p=4m+1$ divides a factorial squared, plus $1$. If I could show that $p=3m+1$ divides some square, plus $3$, then I could prove claim (1), I think. For claim (2), I'm not sure how to proceed.

My questions: How to prove claim (2)? Is claim (1) also true? If not what is a counterexample?

There is a demon in charge of obfuscating the mathematics education of humans, sometimes called Apasmara or Maluyakan, and Betsy DeVos is his prophet.

When human mathematicians started discovering facts about quadratic rings a few centuries ago, my colleague demon made a decision made that when $d \equiv 1 \pmod 4$, humans should be guided to always figure in a number $$\theta = \frac{1 + \sqrt d}{2}$$ and use that number for all calculations in $\mathcal O_{\mathbb Q(\sqrt d)}$ whether it makes sense to do so or not. Especially when the latter is the case. Deliciously diabolical!

This often prevents a human math student from realizing that to factor $p$, it might be easier to find $4p = a^2 - db^2$. In the case of $d = -3$ for $\mathbb Z[\omega]$ (where we use $\omega$ instead of $\theta$ just to be arbitrary), this sometimes means figuring out if $a^2 + 3b^2 = 4p$. If $p \equiv 2 \pmod 3$, then $4p \equiv 2 \pmod 3$, but $a^2$ is never $2 \pmod 3$ and $3b^2 \equiv 0 \pmod 3$ always. But if $p \equiv 1 \pmod 3$, then... I better leave it at that, don't want to cross Mulayakan.

So for example $(5 - \sqrt{-3})(5 + \sqrt{-3}) = 28$ and $$\left(\frac{5 - \sqrt{-3}}{2}\right) \left(\frac{5 + \sqrt{-3}}{2}\right) = 7.$$

• This narrative approach would surely be amazingly effective in teaching! :) Also, might be a more convincing explanation that the usual babble about "integral closure"... :) – paul garrett Feb 20 '18 at 23:06

I don't know what proof you're referring to for the case of the Gaussian integers, but in that case the proof I remember uses the Galois norm map from $\mathbb Z[i]$ to $\mathbb Z$, by $N:a+bi\to a^2+b^2$. The idea is that if $p$ is not still a prime in the Gaussian integers, then it factors (because $\mathbb Z[i]$ is Euclidean) $p=\alpha\cdot \beta$ with neither $\alpha$ nor $\beta$ a prime. Taking norms, $p^2=Np=N\alpha\cdot N\beta$. Neither $N\alpha$ nor $N\beta$ can be $\pm 1$, and in fact cannot be negative in any case, so both must be $p$. That is, with $\alpha=a+bi$, $a^2+b^2=1$, as desired. EDIT: forgot to add: for $p$ to remain prime in $\mathbb Z[i]$, it is necessary and sufficient that $\mathbb Z[i]/p$ is a field. Well, $\mathbb Z[i]/p\cong \mathbb Z/p[i]\cong \mathbb F_p[x]/\langle x^2+1\rangle$. For $p=1\mod 4$, the polynomial $x^2+1$ factors over $\mathbb F_p$, so this is not a field... and $p$ is a sum of two squares.

$N(a+b\omega)=a^2+ab+b^2$, with $\omega$ a cube root of unity, functions exactly analogously, because $\mathbb Z[\omega]$ is Euclidean. EDIT: a prime $p$ remains prime in $\mathbb Z[\omega]$ if and only if $\mathbb Z[\omega]/p$ is a field. Since $\mathbb F_p$ contains a cube root of unity if and only if $p=1\mod 3$, we get the analogous result.

The general question about $a^2+b^2N=p$ is of a different nature. David A. Cox wrote a whole book about it: "Primes of the form $a^2+b^2n$".

• It sounds like you're proving, in the Gaussian case, assumes that I know that $p$ is non prime in the Guassians. The proof I'm thinking of is a step on the way to determining the Gaussian primes. In the Eisenstein case, analogously, I'm trying to start at the basics and not assume I already know whether $p$ is an Eisenstein prime or not. – G Tony Jacobs Feb 19 '18 at 19:24
• @GTonyJacobs, ah, yes, I added the relevant discussion in the Gauss case. The Eisenstein case is analogous. – paul garrett Feb 19 '18 at 19:26
• Ok, I see how this is working, although I realize now this is going around the block exactly the opposite way that I was hoping for. I'm not trying to assume any field theory, but rather taking the perspective of working in that direction. Your proof makes sense to me, but I still wonder if there's a lower-tech argument that works..... – G Tony Jacobs Feb 19 '18 at 19:46
• @GTonyJacobs, yes, I'm pretty sure a more-elementary proof exists, akin to the more-elementary proof for Gaussian integers, but the latter I don't remember, since I'm happy with the version I gave above... Sorry it's not what you wanted (but I'd still recommend it as a "memorable" argument). – paul garrett Feb 19 '18 at 20:04
• No apology necessary! Your approach is good, and in a sense, "correct" - certainly worth studying. I'm just a baby number theorist, so those arguments don't occur to me as easily, but eventually they will :) – G Tony Jacobs Feb 19 '18 at 20:11

I just want to fill in a few details glossed over in the previous answers. These details might be very obvious to some, but not at all obvious to others.

First, the symmetry of the Gaussian integers is square, while the symmetry of the Eisenstein integers is hexagonal.

For example, in the case of numbers with norm 13, we see that $(2 + 3i)i = -3 + 2i$, $(-3 + 2i)i = -2 - 3i$, $(-2 - 3i)i = 3 - 2i$ and one more multiplication by $i$ brings us right back to where we started. If you prefer to go clockwise instead, use $-i$ rather than $i$.

A much easier example is 7, with norm 49. We see that $7 \times i = 7i$, $(7i)i = -7$, $-7 \times i = -7i$ and one more multiplication by $i$ brings us right back to where we started.

Moving over to the Eisenstein integers, our "main" unit would be $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2},$$ which has the property that $\omega^3 = 1$.

If a rational prime $p$ can be written as $p = 3m + 1$, then we can express it in the form $p = j^2 + 3k^2$. (Here $j$ and $k$ are rational integers.) Then, I realized that norms don't work that way in Eisenstein ...

They certainly do work that way, but we have to keep in mind the distinction between algebraic numbers that are also algebraic integers and those that are not. To use the example of 7, we see that $7 = 2^2 + 3 \times 1^2$ and indeed $N(2 + \sqrt{-3}) = 7$. We can readily locate $2 + \sqrt{-3}$ on the complex plane.

But when we try $(2 + \sqrt{-3}) \omega$, things get very interesting: $$(2 + \sqrt{-3}) \omega = (2 + \sqrt{-3})\left(-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right) = -\frac{5}{2} + \frac{\sqrt{-3}}{2}.$$

And then $$\left(-\frac{5}{2} + \frac{\sqrt{-3}}{2}\right) \omega = \frac{1}{2} - \frac{3 \sqrt{-3}}{2}.$$

And then $$\left(\frac{1}{2} - \frac{3 \sqrt{-3}}{2}\right) \omega = 2 + \sqrt{-3}.$$ Oops, I thought that was going to take me around the hexagon. But after all, what is a hexagon but two triangles put together?

For norm 13 on the diagram, I'm going to start on $-1 + 2 \sqrt{-3}$ rather than $1 + 2 \sqrt{-3}$. Then we have $$(-1 + 2 \sqrt{-3}) \omega = -\frac{5}{2} - \frac{3 \sqrt{-3}}{2}.$$

Well, maybe norm 13 was not the best choice for the outer example on this diagram. Plus I should have started with a larger diagram before putting on the dots and arrows.

Nevertheless, this shows that while $j + k \sqrt{-3}$ is the best way to locate Eisenstein integers in the complex plane, it is a little bit awkward for calculations. Maybe there is a way to get the best of both notations?

But before we do that, I'm not entirely liking the use of $j$ and $k$, in part because engineers like to use $j = \sqrt{-1}$. Though engineers use $\omega$ for something else, it doesn't bother me to use it here to mean a complex cubic root of 1.

Okay, so then, with $a$ and $b \in \Bbb Z$, we have $N(a + b \sqrt{-3}) = a^2 + 3b^2$ and $$N\left(\frac{a}{2} + \frac{b \sqrt{-3}}{2}\right) = \frac{a^2}{4} + \frac{3b^2}{4}$$ (however, if $a$ and $b$ don't have matching parities, the norm will be rational but not an integer, meaning the number is an algebraic number but not an algebraic integer).

And then, with the proviso that $\alpha$ and $\beta \in \Bbb Z$, we have $N(\alpha + \beta \omega) = \alpha^2 - \alpha \beta + \beta^2$. Furthermore, for $a + b \sqrt{-3} = \alpha + \beta \omega$, $\alpha = a + b$, $\beta = 2b$, and for $$\frac{a}{2} + \frac{b \sqrt{-3}}{2} = \alpha + \beta \omega,$$ $$\alpha = \frac{a + b}{2},$$ $\beta = b$.

To reverse the conversion, $$\alpha + \beta \omega = \frac{a}{2} + \frac{b \sqrt{-3}}{2},$$ $a = 2 \alpha - \beta$, $b = \beta$; simplify fractions as needed.

For example, $$\frac{-5}{2} + \frac{1 \sqrt{-3}}{2} = -2 + \omega,$$ both of which should give 7 when taken through the relevant norm calculations. Then $(-2 + \omega) \omega = -2 \omega + \omega^2$. Keep in mind that $\omega^2 = -1 - \omega$. This means that $-2 \omega + \omega^2 = -2 \omega + (-1 - \omega) = -1 - 3 \omega$.

Note that $(-1)^2 - (-1 \times -3) + (-3)^2 = 1 - 3 + 9 = -2 + 9 = 7$. Since I'm having an extremely difficult time locating $-1 - 3 \omega$ in the complex plane, I perform the conversion $$-1 - 3 \omega = \frac{-2 - (-3)}{2} + \frac{-3 \sqrt{-3}}{2} = \frac{1}{2} - \frac{3 \sqrt{-3}}{2}.$$

Next, $(-1 - 3 \omega) \omega = -\omega - 3 \omega^2 = -\omega - 3(-1 - \omega) = 3 + 2 \omega$. We check that $3^2 - 3 \times 2 + 2^2 = 9 - 6 + 4 = 3 + 4 = 7$. And lastly we check that $$3 + 2 \omega = \frac{4}{2} + \frac{2 \sqrt{-3}}{2} = 2 + \sqrt{-3}$$ and $N(2 + \sqrt{-3}) = 7$.

If a rational prime $p$ can be written as $p = 3m + 1$, then we can express it in the form $p = j^2 + jk + k^2$ (with $j$ and $k$ in $\Bbb Z$, as above).

Hmm, you seem to have a plus where you should have a minus. This detail probably caused half your confusion. Although I probably also made a similar mistake somewhere in the previous paragraphs.

• Well, $\alpha^2 - \alpha \beta + \beta^2 = j^2 + jk + k^2$ if $\alpha = -j$ and $\beta = -k$, or vice-versa. The problem comes in when you try to relate it to $\omega$. – Mr. Brooks Feb 24 '18 at 22:30
• Good point, @Mr.Brooks – Robert Soupe Feb 25 '18 at 3:36