I want to solve the following questions:
For $R = \mathbb{Z}[\sqrt{-10}]$ and $I = (\sqrt{-10})$
I ) Is $I$ a prime ideal?
II) Is $I$ a maximal ideal?
This is equivalent to asking if $\frac{R}{I}$ an integral domain and a field
I see that $\mathbb{Z}[\sqrt{-10}] = \{x+y \sqrt{-10} \ | \ x,y \in \mathbb{Z}\}$ and $(\sqrt{-10})= \{-10x+y \sqrt{-10} \ | \ x,y \in \mathbb{Z}\}$. But does it then follow that $\frac{R}{I} \cong \mathbb{Z}$? How can I show it?
Best Answer
Note that, $\mathbb Z[\sqrt{-10}] \cong \mathbb Z[x] / \langle {x^2+10} \rangle$ By homomorphism from $\mathbb Z[x]$ into $\mathbb Z[\sqrt{-10}]$, defined as $ x\mapsto \sqrt{-10}$
So,
$\mathbb Z[\sqrt{-10}]/ \langle{\sqrt{-10}}\rangle \cong \mathbb Z[x] / I$,
Where, $ I=\langle {x,x^2+10} \rangle$
And note that, $x \in I \Rightarrow x^2 \in I \Rightarrow 10\in I$
So we have, $I=\langle {x,10} \rangle$
And, $\mathbb Z[x] / I \cong \mathbb Z_{10}[x]/ \langle{x}\rangle \cong \mathbb Z_{10}$