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Let $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$ an ideal generated by $2$ and $1+\sqrt{-5}$. Show that $I$ is a maximal ideal.

So I tried to prove that if $a \notin I$ then $(I,a)$ must be the whole ring or the $I$ itself. That is, we can find a combination $ax+2y+(1+\sqrt{-5})z=1$ or $0$ for some $x,y,z \in \mathbb{Z}[\sqrt{-5}]$. Is my thought correct? How should I find such combination?

user112358
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    How about just showing that $R/I$ has two elements, so is a field? – Thomas Andrews Feb 05 '16 at 03:51
  • Could you explain why having two elements implies it's a field? Thanks – user112358 Feb 05 '16 at 03:53
  • Basically, show that if $r\in R$, then either $r\in I$ or $r-1\in I$ – Thomas Andrews Feb 05 '16 at 03:54
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    If a ring has exactly two elements, the only nonzero element is $1$, so every nonzero element is invertible. – Eric Wofsey Feb 05 '16 at 03:54
  • There's only one ring with two elements and it is a field. – Thomas Andrews Feb 05 '16 at 03:54
  • You can also use the fact that $\mathbb Z[\sqrt -5]$ is the ring of integers of $\mathbb Q[\sqrt -5]$ and thus is a Dedekind Domain and there is unique factorization of ideals, then calculate the norm of the ideal that is $2$ which is prime. By the way, If I don't go wrong, in a Dedekind Domain an ideal is maximal iff it is prime which is easier to check. This is just another idea on how to approach, of course the other comments are much easier! – Maffred Feb 05 '16 at 05:31
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    Thanks to all for your comments. I am trying to derive that $R/I$ has only two elements while I found this:http://math.stackexchange.com/questions/1157461/why-is-2-1-sqrt-5-not-principal. But I don't quite follow the answer in the link. In general, I think I am still confused by the concept of quotient rings and finding their properties. I would appreciate any suggestion on how to learn this thing! – user112358 Feb 05 '16 at 06:41

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