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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 $\mathbb Z$

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  • $\begingroup$ What is exactly $d$? I don't want to be a fusspot, but the sign can be very important. And... do you mean $\Bbb Z[\sqrt d]$? $\endgroup$ – ajotatxe Apr 24 '15 at 21:01
  • $\begingroup$ 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.? $\endgroup$ – KCd Apr 24 '15 at 21:09
  • $\begingroup$ @KCd You are right. Sorry I was misinterpreting the question. Yes I mean Z[√d] and yes d is a square-free integer. $\endgroup$ – Krista E Apr 24 '15 at 21:11
  • $\begingroup$ @KCd Basically I just want to know if there is a way to prove infinite primes in Z[√d] $\endgroup$ – Krista E Apr 24 '15 at 21:12
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    $\begingroup$ @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? $\endgroup$ – Krista E Apr 25 '15 at 13:27
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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.

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  • $\begingroup$ 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 $\Bbb Q(\sqrt d)$. $\endgroup$ – Watson Nov 2 '18 at 16:44
  • $\begingroup$ 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 $(p / q) = (5/3) = (2/3) = -1$. $\endgroup$ – Watson Nov 2 '18 at 16:44

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