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Prove that there are infinitely many primes in $\mathbb Z[\sqrt{d}]$. I don't know how to prove this, but I think that the proof will be similar to proving that there are infinitely many primes in $\mathbb Z$

Krista E
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  • What is exactly $d$? I don't want to be a fusspot, but the sign can be very important. And... do you mean $\Bbb Z[\sqrt d]$? – ajotatxe Apr 24 '15 at 21:01
  • Surely you do not mean $\mathbf Q$, since $\mathbf Q[\sqrt{d}]$ is a field and contains no primes at all. What ring are you really wanting to work in: $\mathbf Z[\sqrt{d}]$, $\mathbf Z[(1+\sqrt{d})/2]$ if $d \equiv 1 \bmod 4$, why would you take $d$ to be a prime rather than more generally a squarefree integer, what is your definition of a prime in the first place, etc. etc.? – KCd Apr 24 '15 at 21:09
  • @KCd You are right. Sorry I was misinterpreting the question. Yes I mean Z[√d] and yes d is a square-free integer. – Krista E Apr 24 '15 at 21:11
  • @KCd Basically I just want to know if there is a way to prove infinite primes in Z[√d] – Krista E Apr 24 '15 at 21:12
  • But what is a prime in your lexicon? – KCd Apr 24 '15 at 21:13
  • For irreducibles among the integers of $\mathbb{Q}[\sqrt{d}]$ more or less the usual "Euclid" proof will work. But I think for primes the argument is different. – André Nicolas Apr 24 '15 at 21:14
  • @AndréNicolas So is there a way to show it, or no? – Krista E Apr 24 '15 at 21:16
  • Yes, I think so. The argument I have in mind uses the fact (Dirichlet) that in an arithmetic progression $a, a+t,a+2t,\dots$ where $gcd(a,t)=1$ there are infinitely many (ordinary) primes. – André Nicolas Apr 24 '15 at 21:20
  • I have "favourited" the question so that I can get to it when I have time. I need to know the source, in order to know whether irreducible is intended, or prime (the two concepts are equivalent in $\mathbb{Z}$, and for some $d$, but not for all $d$). As mentioned earlier, for irreducible the proof is a variant of the standard Euclid proof. – André Nicolas Apr 25 '15 at 00:00
  • @AndréNicolas I came across it while reading, Elementary Number Theory by stark – Krista E Apr 25 '15 at 00:06
  • @AndréNicolas Prime is intended – Krista E Apr 25 '15 at 00:21
  • Showing this is very nontrivial, and wouldn't be in an elementary number theory book. So I opened up Stark's book. On page 276 (of my version), he defines a prime $\pi$ to be a nonunit such that if $\pi = ab$, then either $a$ or $b$ is a unit. To all of us in the comments, we call such an element irreducible. Showing there are infinitely many irreducibles is very easy. Suppose $\pi_1, ... \pi_k$ are all of them. Then consider $N(\pi_1 \cdot \dots \cdot \pi_k) + 1$ and what must divide it. – davidlowryduda Apr 25 '15 at 00:35
  • @mixedmath What must it divide? – Krista E Apr 25 '15 at 01:21
  • @KristaE: $N(\pi_1\cdot \pi_2\cdots \pi_N+1$ has an irreducible non-unit factor $\alpha$, and $\alpha$ cannot be a unit times one of the $\pi_i$. For if $\alpha$ was a unit times some $\pi_i$, then $\alpha$ would divide $\pi_1\cdots\pi_n$, so $\alpha$ would divide $1$, contradicting the fact $\alpha$ is not a unit. Since the book uses prime as a synonym for irreducible, this finishes things. I wrote out a proof for prime in the modern sense, but it is the mixedmath comment that is the answer to your question. – André Nicolas Apr 25 '15 at 03:18
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    @AndréNicolas So what you wrote: "N(π1⋅π2⋯πN+1 has an irreducible non-unit factor α, and α cannot be a unit times one of the πi. For if α was a unit times some πi, then α would divide π1⋯πn, so α would divide 1, contradicting the fact α is not a unit." does prove that there are infinitely many primes too? – Krista E Apr 25 '15 at 13:27
  • Infinitely many irreducibles. But as mixedmath discovered, in Stark's book the word prime is used to mean what everybody nowadays calls irreducible. Note I had a typo, it should be $N(\pi_1\pi_2\cdots \pi_n)+1$ (there was a parenthesis missing). – André Nicolas Apr 25 '15 at 14:34
  • @AndréNicolas So that's all I need to prove that claim? – Krista E Apr 25 '15 at 14:35
  • To prove that there are infinitely many irreducibles (what Stark calls primes), yes. Of course that is a lot simpler than the proof I wrote down, since there I was proving there are infinitely many primes in the modern sense of prime. – André Nicolas Apr 25 '15 at 14:37
  • @AndréNicolas Thank You – Krista E Apr 25 '15 at 14:40
  • You are welcome. Once I found out that you (Stark) really meant irreducibles, I was hoping someone would flesh out the details of the proof. The theorem has to be stated precisely, particularly for those $d$ in which the ring has infinitely many units. – André Nicolas Apr 25 '15 at 15:03
  • Related: https://math.stackexchange.com/questions/35883/ – Watson Nov 30 '18 at 08:03

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It seems now from the comments that despite the wording of the problem, we are asked to show there are infinitely many irreducibles. This can be done by a straightforward modification of the standard "Euclid" proof, so for fun we show that there are infinitely many primes among the integers of $\mathbb{Q}[\sqrt{d}]$.

If $d=-1$, $2$, or $-2$, there is no problem, since the irreducibles in these cases are prime, and it is easy to show there are infinitely many irreducibles. So we can assume that $d$ has an odd prime divisor. Let $q$ be the smallest odd prime divisor of $d$, and let $a$ be the smallest quadratic non-residue of $q$. Then $a$ is prime.

By Dirichlet's Theorem on primes in arithmetic progressions, there are infinitely many odd primes of the form $a+kq$, and therefore infinitely many such odd primes that do not divide $d$. Any such prime $p$ is a quadratic non-residue of $q$. We will show that $p$ is a prime in the integers of $\mathbb{Q}[\sqrt{d}]$.

So we want to show that if $x$ and $y$ are algebraic integers in $\mathbb{Q}[\sqrt{d}]$, and $p$ divides $xy$, then $p$ divides $x$ or $p$ divides $y$.

Note that $p$ divides the norm $x\bar{x}y\bar{y}$ of $xy$. So $p$ divides one of $x\bar{x}$ or $y\bar{y}$, say $x\bar{x}$. Now the details depend a little on whether $d$ is congruent to $1$ modulo $4$ or not. The argument is a bit simpler if $d\not\equiv 1\pmod{4}$.

In that case, $x$ has the shape $m+n\sqrt{d}$ where $m$ and $n$ are integers. We have that $p$ divides $m^2-dn^2$. If $p$ does not divide $m$, we have contradicted the fact that $p$ is a quadratic non-residue of $q$. So $p$ divides $m$, and since $p$ does not divide $d$, it follows that $p$ divides $n$, so $p$ divides $x$.

If $d\equiv 1\pmod{4}$, it could be that $x=\frac{m+n\sqrt{d}}{2}$ where $m$ and $n$ are odd. But again we get that $p$ divides $m^2-dn^2$, and again we conclude that $p$ divides $m$ and $n$.

Remark: If instead we are working in $\mathbb{Z}[\sqrt{d}]$, then the case $d\equiv 1\pmod{4}$ does not require special treatment.

André Nicolas
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  • There is something wrong with this answer. Let's take $d=3$ and $d'=6$. In both cases, we get the same smallest odd prime divisor, namely $q=3$, and we get $a=2$. We can take $p = 2+1 \cdot 3=5$ as prime. But the Legendre symbols are $(3 / 5) = -1, (6/5)=1$, so $5$ is inert in $\Bbb Q(\sqrt 3)$ but split in $\Bbb Q(\sqrt 6)$. Just choosing one single prime factor $q$ of $d$ cannot be sufficient to determine whether $p$ stays prime in $\Bbb Q(\sqrt d)$. – Watson Nov 02 '18 at 16:44
  • Typically, I don't understand why "we have contradicted the fact that $p$ is a quadratic non-residue of $q$". In my above example, we have $p=5$ dividing $1^2 - 6 \cdot 1^2$, and $5$ does not divide $m=1$, without contradicting $(p / q) = (5/3) = (2/3) = -1$. – Watson Nov 02 '18 at 16:44