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Question:

Show that $3$ is not a prime in $\mathbb Q [\sqrt 7] $.

To show this, should I start by assuming that $3 = ab$ where $a$ and $b$ are integers in $\mathbb Q[\sqrt{7}]$ and then try to show they are not units?

What should I do to show that? Or is there another better way to do this problem?

Aaron Maroja
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Keyang Zhang
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    $3= (\sqrt{7}-2)(\sqrt{7}+2)$. – copper.hat Jun 04 '15 at 18:42
  • If $p$ is a prime and $u$ is a unit, is $up$ a prime? I don't know but supect... – Will Jagy Jun 04 '15 at 18:43
  • @copp, right, $-3$ is represented by $x^2 - 7y^2.$ Good to confirm my impression – Will Jagy Jun 04 '15 at 18:44
  • @WillJagy: :-). – copper.hat Jun 04 '15 at 18:45
  • @cop, I spent a month preparing http://math.blogoverflow.com/2014/08/23/binary-quadratic-forms-over-the-rational-integers-and-class-numbers-of-quadratic-%EF%AC%81elds/ and one thing I had to stomach was how, when the principal form ( here $x^2 - 7 y^2$) does NOT represent $-1,$ the field version of things is a two-to-one mapping, primitive forms $A x^2 + B xy + C y^2$ and $-A x^2 - B xy - C y^2$ are sent to the same ideal. Which I find annoying, but there it is. This suggested to me here that $-3$ could be substituted for $3$ – Will Jagy Jun 04 '15 at 18:52
  • @WillJagy: Wow. – copper.hat Jun 04 '15 at 18:58
  • @cop, my dedication to my craft is absolute. – Will Jagy Jun 04 '15 at 18:59
  • @cop, I also love dogs. And they me. – Will Jagy Jun 04 '15 at 19:00
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    @WillJagy: I guess I should have written bow wow... – copper.hat Jun 04 '15 at 19:04
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    Dumb question: Aren't all nonzero integers units in $\mathbb{Q}$, and not primes? – NovaDenizen Jun 04 '15 at 21:29
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    @NovaDenizen You're absolutely right; I think everyone has just been presuming that the question is actually about $\mathbb{Z}[\sqrt{7}]$ since otherwise it's entirely trivial. – Steven Stadnicki Jun 04 '15 at 21:41
  • @copper.hat, making sure, in one of these rings of integers in a number field, some $p$ is a prime if and only if $-p$ is prime? – Will Jagy Jun 05 '15 at 18:56
  • @WillJagy: I am still searching for more even primes... – copper.hat Jun 05 '15 at 18:58
  • @cop, positive rational primes by $x^2 - 7 y^2$ 2 29 37 53 109 113 137 149 193 197 233 277 281 317 337 373 389 401 421 449 457 541 557 569 613 617 641 653 673 701 709 757 809 821 877 953 977 – Will Jagy Jun 05 '15 at 19:05
  • @cop, pos rat primes by $7 x^2 - y^2$ 3 7 19 31 47 59 83 103 131 139 167 199 223 227 251 271 283 307 311 367 383 419 439 467 479 503 523 563 587 607 619 643 647 691 719 727 787 811 839 859 887 971 983 – Will Jagy Jun 05 '15 at 19:06
  • @WillJagy: I am out of depth. – copper.hat Jun 05 '15 at 20:03
  • @cop, Oh, same here I guess. I can instantly tell you the class number, find ideal representatives, all my own software, but I've never actually studied the number field viewpoint. I have plenty of books that discuss the beginnings, though, I can try to look this up. – Will Jagy Jun 05 '15 at 20:09

4 Answers4

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If 3 is prime in this ring, the equations $x^2 + bx \pm 3 = 0$ would have no solutions for $x \in \mathbb Z[\sqrt 7]$, where $b \in \mathbb Z$.

To use the quadratic formula, we have $a = 1$ and $c = \pm 3$.

$$x = \frac{-b \pm \sqrt{b^2 \pm 12}}{2}$$

We could hope that $b$ is small, and in this case it is, but it's not very smart to try several potential values by brute force.

However... since we know we want the solution to have $\sqrt 7$, we just need to solve $b^2 \pm 12 = 28 (= 4 \times 7)$ in integers. The only solutions are $b = \pm 4$ for $b^2 + 12 = 28$ (since 28 + 12 = 40, which is not a square).

Then $$x = \frac{\pm 4 \pm \sqrt{28}}{2} = \pm 2 \pm \sqrt 7$$ (since $\sqrt{28} = 2 \sqrt 7$). None of these numbers are units, because if they were, they'd be solutions to $x^2 \pm 4x \pm 1 = 0$, which clearly, by the foregoing, they're not.

Lastly, just to make sure we haven't made any silly arithmetic mistakes along the way, we check that $$(2 - \sqrt 7)(2 + \sqrt 7) = -3.$$

Robert Soupe
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    There are a number of other answers which seem to convincingly demonstrate that 3 is not prime in this ring. Can you explain where the other answers have gone wrong? – Xander Henderson Apr 04 '19 at 16:04
  • @Xander Oh geez. I came to this question through a duplicate. Maybe the asker of the duplicate will be satisfied by the answers to this question, maybe not, or maybe he won't even notice. It's not like I don't have anything better to do than answer questions on this website. – Robert Soupe Apr 04 '19 at 16:21
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$$3 = (\sqrt 7 - 2)(\sqrt 7 + 2)$$

And $2^2 - 1^2 \dot \, 7 \neq \pm4 $ or $2^2 - 1^2 \dot\, 7 \neq 1$.

Edit: A fundamental unit in $\mathbb Q[\sqrt 7]$ is $8 + 3\sqrt 7$. Therefore the units are of the form $(8 + 3 \sqrt 7)^n , n\in \mathbb Z$.

user26857
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Aaron Maroja
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Recall that in a commutative ring $R$, an ideal $I$ is prime iff $R/I$ is a domain. Also note that $\mathbb{Z}[\sqrt{7}] = \frac{\mathbb{Z}[x]}{(x^2-7)}$. Then \begin{align*} \frac{\mathbb{Z}[\sqrt{7}]}{(3)} &\cong \frac{\mathbb{Z}[x]/(x^2-7)}{(3,x^2-7)/(x^2-7)} \cong \frac{\mathbb{Z}[x]}{(3,x^2-7)} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x^2-7)} = \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x^2-1)}\\ &=\frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)(x+1)} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)} \times \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x+1)} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \end{align*} where the second isomorphism holds by the Third Isomorphism Theorem, and the second-to-last by the Chinese Remainder Theorem. Since $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is not a domain, then $(3)$ is not prime in $\mathbb{Z}[\sqrt{7}]$.

Moreover, we can recover a factorization of $3$ from this isomorphism. Since $$ x^2 - 7 \equiv x^2 - 1 = (x+1)(x-1) \equiv (x-2)(x+2) \pmod{3} $$ replacing $x$ by $\sqrt{7}$ yields the factorization $3 = (\sqrt{7}-2)(\sqrt{7}+2)$.

Viktor Vaughn
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I believe 3 is small enough that you can just use the continued fraction digits of $\sqrt{7} = [2;\overline{1,1,1,4}]$.

The approximations are $2, 3, \frac{5}{2}, \frac{8}{3},\frac{37}{14},\dots$ Indeed $37^2 - 7\cdot 14^2 = -3$ and even $2^2 - 7*1^2 = -3$.

These lead to factorizations $3 = (\sqrt{7}-2)(\sqrt{7}+2) =(14\sqrt{7}-37)(14\sqrt{7}+37) $

cactus314
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