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Using Chebotarev's density theorem, asymptotically, what is the probability that a rational prime remains prime in $\mathbb Z[i, \sqrt{-3}]$?

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  • $\begingroup$ I had this question for a while. Then it occurred to me the answer below. But I am not very sure my proof is correct or if there’s a more direct answer using in a better fashion the Cheboratev density theorem $\endgroup$
    – Ponce
    Jan 19, 2019 at 3:19
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    $\begingroup$ One method is to say one of $x^2+1,x^2+3, x^2-3$ is not irreducible $\bmod p$. The other method is $(p)$ is prime in $\mathbb{Z}[i,\sqrt{-3}]$ means $\mathbb{Z}[i,\sqrt{-3}]/(p)$ is a field with $p^4$ elements so $Gal(\mathbb{Z}[i,\sqrt{-3}]/(p)/ \mathbb{Z}/(p))$ is cyclic of order $4$. But since we are in Galois extension $Gal(\mathbb{Z}[i,\sqrt{-3}]/ \mathbb{Z})\to Gal(\mathbb{Z}[i,\sqrt{-3}]/(p)/ \mathbb{Z}/(p))$ is surjective and since the former doesn't have a cyclic subgroup of order $4$, that $p$ is prime never happens. $\endgroup$
    – reuns
    Jan 19, 2019 at 3:40

4 Answers 4

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If you're referring to $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$, the ring of algebraic integers of $\mathbb{Q}(\zeta_{12})$ (see its page in the LMFDB), then the answer is that no prime from $\mathbb{Z}$ remains prime in $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$.

The intermediate fields are $\mathbb{Q}(i)$, $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$. For a prime from $\mathbb{Z}$ to be prime in both $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$, it must be $p \equiv 5 \bmod 12$. But then this means that $p \equiv 1 \bmod 4$ and you know what that means for $\mathbb{Z}[i]$...

P.S. A Naiade discovered the density theorem twelve centuries before Chebotarev.

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    $\begingroup$ The density of prime ideals before Galois invented number fields, before the logarithm was discovered ? And you don't need quadratic reciprocity to say one of $(\frac{-3}{p}),(\frac{3}{p}),(\frac{-1}{p})$ is not $-1$ $\endgroup$
    – reuns
    Jan 20, 2019 at 2:28
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    $\begingroup$ @reuns I think he meant "naiad." I still find the postscript preposterous, though for different reasons. And then there's the issue of whether or not the OP meant to include numbers like $$\frac{\sqrt 3}{2} + \frac{i}{2}.$$ $\endgroup$ Jan 20, 2019 at 2:55
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    $\begingroup$ @RobertSoupe What would be your argument for the ring of integers (which can be shown to be $\mathbb{Z}[\zeta_{12}]$) ? $(x^2+1)(x^2+3)$ is separable $\bmod p$ for $p \ne 2,3$ so $O_{\mathbb{Z}[i,\sqrt{-3}]} \subset \mathbb{Z}[i,\sqrt{-3},2^{-1},3^{-1}]$ and $O_{\mathbb{Z}[i,\sqrt{-3}]}$ will still be ramified at $2,3$ (thus $(2),(3)$ are not prime) because it contains $O_{\mathbb{Z}[i]},O_{\mathbb{Z}[\sqrt{-3}]}$. $\endgroup$
    – reuns
    Jan 20, 2019 at 3:12
  • $\begingroup$ Where I can read the theory behind this identity $\mathbb{Z}[\zeta_{12}]$ = $\mathbb{Z}[i,\sqrt {-3}]$ $\endgroup$
    – Ponce
    Jan 20, 2019 at 4:04
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    $\begingroup$ @Kronecker I don't know of any books or webpages, but asking Wolfram Alpha is i + sqrt(-3) an algebraic integer? might give you some interesting information to cross-check with the LMFDB. $\endgroup$ Jan 20, 2019 at 4:30
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The simplest answer, which is already implicit in several of the other answers is as follows: Let $p \geq 5$ be prime. Then at least one of $-1$, $-3$ and $3$ are square modulo $p$.

If $a$ is a square root of $-1$ modulo $p$ then I claim that $\langle p \rangle = \langle p, a - i \rangle \langle p, a + i \rangle$. To see this, note that $\langle p, a - i \rangle \langle p, a + i \rangle = \langle p^2, p(a - i), p(a+i), a^2 + 1 \langle$. Each term on the right is clearly divisible by $p$, so $\langle p \rangle \supseteq \langle p, a - i \rangle \langle p, a + i \rangle$. Conversely, $p(a + i) + p(a - i) = 2 ap$ and $\gcd(2a, p) = 1$, so $p$ is in the ideal generated by $2ap$ and $p^2$. Similarly, if $b$ is a square root of $-3$, then $\langle p \rangle = \langle p, b + \sqrt{-3} \rangle \langle p, b - \sqrt{-3} \rangle$ and, if $c$ is a square root of $3$, then $\langle p \rangle = \langle p, c + \sqrt{3} \rangle \langle p, c - \sqrt{3} \rangle$.


Having given the simple answer, let me address two issues raised in the comments and other answers: Is $\mathbb{Z}[i, \sqrt{-3}] = \mathbb{Z}[\zeta_{12}]$? Should we care? And where does Chebotaryov come in?

In fact, we have $\mathbb{Z}[i, \sqrt{-3}] \subsetneq \mathbb{Z}[\zeta_{12}]$. To prove the containment, note that $i = {\zeta_{12}}^3$ and $\sqrt{-3} = 2 {\zeta_{12}}^2 -1$. However, we do not have equality. We have $$\mathbb{Z}[i, \sqrt{-3}] = \{ a+bi+c\sqrt{-3} + d \sqrt{3} : a,b,c,d \in \mathbb{Z} \}$$ and each element of $\mathbb{Z}[i, \sqrt{-3}]$ has a unique expression of this form. But $$\zeta_{12} = \tfrac{1}{2} i + \tfrac{1}{2} \sqrt{3}$$ so it is not in $\mathbb{Z}[i, \sqrt{-3}]$. In fact, working a little harder, we have $$\mathbb{Z}[\zeta_{12}] = \left\{ \tfrac{1}{2} (a+bi+c\sqrt{-3} + d \sqrt{3}) : a,b,c,d \in \mathbb{Z},\ a \equiv c \bmod 2,\ b \equiv d \bmod 2 \right\}.$$ So $\mathbb{Z}[i, \sqrt{-3}]$ is an order in $\mathbb{Z}[\zeta_{12}]$ of index $4$.

Should we care? For some purposes, this is important. The ring $\mathbb{Z}[i, \sqrt{-3}]$ is not a Dedekind domain and the semigroup of fractional ideals is not a group. This leads to problems such as that $\langle 2, 1 + \sqrt{-3} \rangle^2 = \langle 2 \rangle \langle 2, 1 + \sqrt{-3} \rangle$ (exercise!) but $\langle 2, 1 + \sqrt{-3} \rangle \neq \langle 2 \rangle$.

However, to answer the question about asymptotic density, we don't care. The rings $\mathbb{Z}[i, \sqrt{-3}]$ and $\mathbb{Z}[\zeta_{12}]$ become equal after inverting $2$. So all odd primes factor the same way in both rings, and we can discard the prime $2$ for asymptotic purposes.

Where does Chebotaryov come in? We have the quality of fields $\mathbb{Q}(i, \sqrt{-3}) = \mathbb{Q}(\zeta_{12})$. The Galois group over $\mathbb{Q}$ is $C_2 \times C_2$, where $C_k$ is the cyclic group of order $k$. We can think of this group as either switching the signs in $\pm i$, $\pm \sqrt{-3}$, or as the group of units in $\mathbb{Z}/12$. Either way, a prime $p$ remains inert in a Galois extension if and only if the Frobenius at that prime generates the Galois group; since this Galois group is not cyclic, no primes remain inert.

The deep part of Chebotaryov is that the $4$ elements of $\mathrm{Gal}(\mathbb{Q}(i, \sqrt{-3})/\mathbb{Q}))$ occur equally often as Frobenius elements, but, since none of them correspond to remaining inert, we don't actually need this. (Also, in this case, we can explicitly write down the Frobenius at $p$ as a function of $p$ modulo $12$, so Chebotaryov density is simply Dirichlet's theorem on density of primes in arithmetic progressions in this case.)

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An important property of a ring $R$ is that if $a$ and $b$ are numbers in it, then $a + b$ and $ab$ are also in $R$.

Clearly $i + \sqrt{-3}$ is in this ring that you're looking at. Big deal, no one cares. Multiplication is a bit more productive (pardon the pun) in this case: $$i \sqrt{-3} = \sqrt{-1} \sqrt{-3} = \sqrt{-1 \times -3} = \sqrt 3.$$

If you had the slightest doubt that 3 ramifies in this ring, doubt no more now.

Then 5 splits on account of $(2 - i)(2 + i)$, and 7 splits on account of $(2 - \sqrt{-3})(2 + \sqrt{-3})$, or, if you prefer, $$\left(\frac{5}{2} - \frac{\sqrt{-3}}{2}\right)\left(\frac{5}{2} + \frac{\sqrt{-3}}{2}\right).$$

Next, 11 is a little trickier: it splits on account of $(1 - 2 \sqrt 3)(1 + 2 \sqrt 3) = -11$. Or how about $(8 - 5 \sqrt 3)(8 + 5 \sqrt 3)$? That's also equal to $-11$. If that's acceptable to you, we can move on.

Actually, that's where I'm going to have to leave it at for tonight...

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Based on Euler results and Chebotarev's density theorem, we have that, asymptotically:

i) One half of all rational primes split in $Z[i]$, and these are the primes $p = 1$ mod $4$. Let's call this set $Split(Z[i])$ and,

ii) One half of all rational primes split in $Z[\sqrt{-3}]$, and these are the primes $p = 1$ mod $3$. Let's call this set $Split(Z[\sqrt{-3}])$

Therefore, the rational primes that split in $Z[i,\sqrt{-3}]$ are the union of $Split(Z[i])$ and $Split(Z[\sqrt{-3}])$. We have then, by set rules, that:

$Split(Z[i,\sqrt{-3}]) = Union$ ($Split(Z[i])$,$Split(Z[\sqrt{-3}])$) := {primes| $p = 1$ mod $4$ OR $p = 1$ mod $3$}.

And the rational primes that remain inert in this ring $Inert(Z[i,\sqrt{-3}])]$ are in the complement set $C[Split(Z[i,\sqrt{-3}])]$, and by set rules:

$Inert(Z[i,\sqrt{-3}])]$$C[Split(Z[i,\sqrt{-3}])] = C[Union (Split(Z[i]),Split(Z[\sqrt{-3}]))] :=$ {primes | NOT ( $p = 1$ mod $4$ OR $p = 1$ mod $3$) } = {primes| NOT ($p = 1$ mod $4)$ AND NOT ($p = 1$ mod $3$) } = {2,3,primes| $p = 3$ mod $4$ AND $p = 2$ mod $3$ }

Finally, by Chebotarev's density theorem and probability rules, asymptotically, the probability $P$ that a prime $p = 3$ mod $4$ AND $p = 2$ mod $3$ is equal to the product of the probabilities for each condition:

$P$ (prime $p = 3$ mod $4$ AND $p = 2$ mod $3$) = $P$ (prime $p = 3$ mod $4$) * $P$ (prime $p = 2$ mod $3$) = $1/2 * 1/2 = 1/4$

Since the primes that ramify in $(Z[i,\sqrt{-3}])$ form a finite set, we have that $P(p$ is inert in $Z[i,\sqrt{-3}]$)=$P(p$ doesn't split in $Z[i,\sqrt{-3}]$), therefore $1/4$ of all rational primes remain prime in $Z[i,\sqrt{-3}]$

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    $\begingroup$ Nope. A prime $p>3$ splits in $\Bbb{Q}(i)$ if and only iff $\left(\dfrac{-1}p\right)=1$. It splits in $\Bbb{Q}(\sqrt{-3})$ iff $\left(\dfrac{-3}p\right)=1$. And it splits in $\Bbb{Q}(\sqrt3)$ iff $\left(\dfrac3p\right)=1$. So for the prime to be totally inert we need $$\left(\dfrac{-1}p\right)=\left(\dfrac{-3}p\right)=\left(\dfrac3p\right)=-1.$$ How many primes satisfy all three of these conditions? Remember that the Legendre symbol is multiplicative. $\endgroup$ Jan 20, 2019 at 6:15
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    $\begingroup$ Or, in other words, if $p$ is inert in $\Bbb{Z}[i]$ and inert in $\Bbb{Z}[\sqrt{-3}]$ it does not follow that it would be inert in $\Bbb{Z}[i,\sqrt{-3}]$. $\endgroup$ Jan 20, 2019 at 6:17
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    $\begingroup$ More precisely, it would not be totally inert in $\Bbb{Z}[i,\sqrt{-3}]$. The prime $p=11$ is an example. It is inert in $\Bbb{Z}[i]$ as well as in $\Bbb{Z}[\sqrt{-3}]$. But it splits in the subring $\Bbb{Z}[\sqrt3]$ because $3\equiv5^2\pmod p$. $\endgroup$ Jan 20, 2019 at 6:29
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    $\begingroup$ As I understand it, the Chebotarev density theorem gives an estimate, not a precise number. In any case, I think you'd do well to ask: what are the algebraic integers of $\textbf Q(i, \sqrt{-3})$? I asked a similar question about $\textbf Q(i, \sqrt 2)$. $\endgroup$
    – David R.
    Jan 24, 2019 at 22:32
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    $\begingroup$ @DavidR. Would that be math.stackexchange.com/questions/2176672/… ? $\endgroup$
    – Lisa
    Feb 3, 2019 at 21:38

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