-2

Let $R=\mathbb Z[\sqrt{-5}]$. I want to show $P=3\,R+(1+\sqrt{-5})\,R$ and $Q= 3\,R+(1-\sqrt{-5})\,R$ are prime ideals of $R$.

Ian Coley
  • 6,000
sahra
  • 77
  • $I\subset R$ is a prime ideal iff for every $a,b\in R$ such that $ab\in I$, either $a\in I$ or $b\in I$. How can you go from there? – Ian Coley Jun 13 '13 at 14:47

1 Answers1

2

This can be done using the classifcation of prime ideals of $\Bbb Z[x]$ and an isomorphism theorem for rings.

First note that your ring $R\cong \frac{\Bbb Z[x]}{(x^2+5)}$, and that the ideals you are talking about correspond to $\frac{(3,1+x)}{(x^2+5)}$ and $\frac{(3,1-x)}{(x^2+5)}$. The containments are justified since $x^2+5=(x+1)(x-1)+2\cdot 3$.

An isomorphism theorem says $\frac{\Bbb Z[x]}{(x^2+5)}/\frac{(3,1+x)}{(x^2+5)}\cong \frac{\Bbb Z[x]}{(3,1+x)}$, so $\frac{(3,1+x)}{(x^2+5)}$ is prime in $R$ iff $(3,1+x)$ is prime in $\Bbb Z[x]$.

A similar argument can be formulated for $(3,1-x)$.

Community
  • 1
rschwieb
  • 153,510
  • @rschweib thanks alot. why that ideals I am talking about correspond to$\frac{(3,1+x)}{(x^2+5)}$ and $\frac{(3,1-x)}{(x^2+5)}$ – sahra Aug 15 '13 at 04:43
  • @sahra $\Bbb Z[x]/(x^2+5)$ is isomorphic to your ring by thinking of x as $\sqrt{-5}$. Clearly the first ideal is generated by 3 and 1+x. – rschwieb Aug 15 '13 at 11:07
  • Thanks so much for your hint. – sahra Aug 15 '13 at 15:12