I'm trying to understand the quotient $\Bbb Z[\sqrt{47}]/(2, 1 +\sqrt{47})$, in order to find out whether or not $(2, 1 + \sqrt{47})$ is a prime ideal in $\Bbb Z[\sqrt{47}]$. I think it is but my calculations seem to be giving me something that doesn't agree with this, so either it isn't a prime ideal or I'm doing something very wrong.
Since $\Bbb Z[\sqrt{47}] \cong \Bbb Z[X]/(X^2 – 47)$, I'm writing
$$ (\Bbb Z[X]/(X^2 – 47))/(2, X^2 – 2X – 46) $$
where $X^2 – 2X – 46$ is a monic irreducible polynomial with $1 + \sqrt{47}$ as a root. Is this step correct? If so, then I think it follows that
$$ (\Bbb F_2[X]/(X^2))/(\overline{X^2 – 2X – 46}) $$
where $\overline{.}$ denotes the reduction map. The problem is, this is the zero-ideal in $\Bbb F_2[X]/(X^2)$, and this finite ring is not an integral domain, so the conclusion is that $(2, 1 + \sqrt{47})$ is not a prime ideal.
Which steps here (if any) are correct? Is any body able to show me how they might do it if it is not correct?
Best Answer
Lets denote $R=\mathbb{Z}\left[\sqrt{47}\right]$, $I=\left<2\right>$ and $J=\left<1+\sqrt{47}\right>$. By the third isomorphism theorem
$$\mathbb{Z}\left[\sqrt{47}\right]/\left<2,1+\sqrt{47}\right>=R/\left(I+J\right)\cong\left(R/J\right)/\left(\left(I+J\right)/J\right)$$
As @Quasicoherent pointed out, I have made a mistake saying that $\mathbb{Z}\left[\sqrt{47}\right]/\left<1+\sqrt{47}\right>\cong\mathbb{Z}$. In fact
$$R/J=\mathbb{Z}\left[\sqrt{47}\right]/\left<1+\sqrt{47}\right>\cong\mathbb{Z}_{46}$$
since $\left<1+\sqrt{47}\right>\ni-\left(1-\sqrt{47}\right)\left(1+\sqrt{47}\right)=-\left(1-47\right)=46$ and
$$\left(a+b\sqrt{47}\right)\left(1+\sqrt{47}\right)=\left(a+47b\right)+\left(a+b\right)\sqrt{47}$$
is an integer iff $a=-b$ iff it is $46b$ for some $b$. Thus
$$\mathbb{Z}\left[\sqrt{47}\right]/\left<2,1+\sqrt{47}\right>\cong\mathbb{Z}_{46}/\left<2\right>=\left(\mathbb{Z}/46\mathbb{Z}\right)/\left(2\mathbb{Z}/46\mathbb{Z}\right)\cong\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_{2}$$
and $I+J=\left<2,1+\sqrt{47}\right>$ is prime.