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$1)$ Let $R:=\mathbb Z[w]$, where $w=\frac{1+\sqrt{-15}}{2}$. What is the norm $N_{R/\mathbb Z}(x+yw)$ in terms of $x,y\in\mathbb Z$? Which of the integers $1,\dots,10$ occur as the norm of some element of $R$?

The norm is $N(x+yw)=x^2-y^2w^2=x^2-\frac{y^2}{2}\left(-7+\sqrt{-15}\right)$

Since the squareroot-part vanishes only if $y=0$, the possibilities are $1^2,2^2,3^2$ (because $x,y\in\mathbb Z$)

Is the norm always defined as $N(\alpha)=\alpha\bar{\alpha}$ ?

I've seen somewhere for example in the ring $R$ of algebraic integers of the field $K=\mathbb Q(\sqrt{-19})$ the norm is defined as $N(a+b(1+\sqrt{-19})/2)=a^2+ab+5b^2$

$\bf{EDIT}:$ With the hint of Daniel Fischer I got the norm $x^2+xy+4y^2$, for which the values $1,4,6,9,10$ can occur. (For $2,3,5,7,8$ there is no integer solution.)

$2)$ Which of the integers $1,\dots,10$ occur as the cardinality of $R/\mathfrak p$ for some prime ideal of $R$? May we conclude that $R$ is not a PID?

I don't know how to begin with this exercise, I've seen some cases, where the extension ring fail to be a UFD (which implies that it cannot be a PID), but here I think one has to consider the quotient, and maybe using some properties of a PID, the quotient must have prime cardinality (It is just a guess) and so disprove it in this way.

Any help is appreciated, Thanks.

OBDA
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    The norm is always (in rings of quadratic integers) $N(\alpha) = \alpha\cdot \overline{\alpha}$. But $\overline{x+y\omega} \neq x - y \omega$. For $\overline{\omega} = 1- \omega$. – Daniel Fischer Feb 22 '15 at 13:33
  • @Daniel Fischer So the product becomes $(x-y(1-(1+\sqrt{-15})/2))(x+y(1+\sqrt{-15})/2)$ ? – OBDA Feb 22 '15 at 13:37
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    Nope, $\mathbb{Z}\left[\frac{1+\sqrt{-15}}{2}\right]$ is not an UFD and hence not a PID. e.g. $4 = 2\cdot 2 = \left(\frac{1+\sqrt{-15}}{2}\right)\left(\frac{1-\sqrt{-15}}{2}\right)$ – achille hui Feb 22 '15 at 13:37
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    @OBDA No, it's $$\left(x + y\frac{1+\sqrt{-15}}{2}\right)\left(x+y\frac{1-\sqrt{-15}}{2}\right),$$ you have $\overline{x+y\omega} = x + y\overline{\omega}$. – Daniel Fischer Feb 22 '15 at 13:40
  • @achille hui OK but I think we have to make use of the quotient, for what it is given then ? – OBDA Feb 22 '15 at 13:40
  • See here for some ideas. – Bill Dubuque Feb 22 '15 at 15:52
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    Your mileage may vary, but this ring is way easier to understand if you temporarily forget about defining an $\omega$ and instead think about it containing numbers of the form $$\frac{a + b \sqrt{-15}}{2}$$ (with $a$ and $b$ having the same parity). The norm is then $$\frac{a^2 + 15b^2}{4}.$$ – Robert Soupe Feb 22 '15 at 21:22
  • @Robert Soupe This is much much better, thanks. Otherwise it is difficult to show the impossible cases. – OBDA Feb 22 '15 at 21:42
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    Another way to show this ring is not a principal ideal domain: show that there is no way to express $$\left\langle 2, \frac{1}{2} + \frac{\sqrt{-15}}{2}\right\rangle$$ as a principal ideal (an ideal generated by only one number). – Robert Soupe Jul 24 '16 at 02:29

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