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I want to show that $(3, \sqrt 15)$ is not a principal ideal in the ring $ R = \mathbb{Z}[\sqrt{15}]$ with norm $N(a + b \sqrt 15) = a^2 - 15b^2$.

My attempt:

Suppose $(3, \sqrt 15) = (x) $

Then $3 = x * r1$ and $\sqrt 15 = x * r2$ , $r1,r2 \in R$.

$N(3) = 9 = N(x) N(r1)$ and $N(\sqrt 15) = -15 = N(x)N(r2)$ so $N(x) = 3$ or -3 or 1 or -1.

Any ideas on how to continue? Thanks.

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If the norm of $x$ is $\pm1$ then $x$ is a unit.

If the norm of $x$ is $\pm3$, you get a contradiction from working modulo $5$.

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Technically, it could be a unit. You need to show it can't be based on the fact that it is expressable in terms of the two generators. – Thomas Andrews Jan 16 '13 at 23:25
@Thomas, I'm not sure what your "it" refers to. – Gerry Myerson Jan 16 '13 at 23:28
$x$. You seem to imply that if $x$ is a unit, you are done, but you need to prove that a unit value of $x$ would be a problem... – Thomas Andrews Jan 16 '13 at 23:29
Yeah, but then you should indicate that there is some more work to do, not just stop, or OP might think he is missing some part of your argument. – Thomas Andrews Jan 16 '13 at 23:35
@Thomas, there is nothing wrong with an answer that leads OP to think. If OP then has a supplementary question, OP can ask it. If you want to write an answer that doesn't require OP to think, please, do so. – Gerry Myerson Jan 16 '13 at 23:39

Here is a proof using some results from algebraic number theory. Suppose your ideal $I = (3,\sqrt{15})$ is a principal generated by some $(\alpha)$ with $\alpha = a + b\sqrt{15}$. Then firstly your ideal $I$ is prime since $$\begin{eqnarray*} R/I &\cong& \Bbb{Z}[x]/(x^2 - 15)/(3,x)/(x^2 - 15)\\ &\cong& \Bbb{Z}[x]/(3,x) \\ &\cong&\Bbb{Z}/3\Bbb{Z}.\end{eqnarray*}$$

Consequently we also have the absolute norm of $I$ denoted $||I||$ being equal to $3$. On the other hand it is well known that $||(\alpha)|| = |N_{\Bbb{Q}(\alpha)/\Bbb{Q}}(\alpha)| = |a^2 -15b^2|.$

Thus it will now suffice to understand why $a^2 - 15b^2 \neq \pm 3$. Now the square of every integer mod $5$ is $0,1$ or $4$. What happens when you reduce mod $5$?

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