Quotient of a quotient ring 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?
 A: The ring is $$\mathbb{Z}[\sqrt{47}]/(2,1+\sqrt{47}) \cong(\mathbb{Z}[X]/(X^2-47))/(2,1+X) \cong\mathbb{Z}[X]/(X^2-47,2,1+X)\\ \cong  \mathbb{Z}_2[X]/(X^2-47,1+X)=  \mathbb{Z}_2[X]/(X^2-1,1+X) =\mathbb{Z}_2[X]/(1+X)\cong \mathbb{Z}_2$$
where $\mathbb{Z}_2 = \mathbb{Z}/(2)$.
How did you come to $(\Bbb Z[X]/(X^2 - 47))/(2, X^2 - 2X - 46)$ ? 
It is $\cong \mathbb{Z}[\sqrt{47}]/(2,(1+\sqrt{47})^2)=\mathbb{Z}[\sqrt{47}]/(2) \cong \mathbb{Z}_2[X]/(1+X)^2$
A: 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.
