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I have three ideals, each with two elements of $\mathbb Z[\sqrt{-5}]$. If you show me how to show one of them is maximal (hence prime) then I think I can manage to do the remaining two by myself.

$\langle 2, 1 + \sqrt{-5}\rangle$

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migrated from Mar 25 '13 at 16:28

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I think that this question should be moved to the main site. Also, it wouldn't hurt you to show some of your own efforts. But here's a hint. Show that all the elements are congruent to either $0$ or $1$ modulo that ideal. If you need another hint, show what you have problems with!! – Jyrki Lahtonen Mar 24 '13 at 14:38
I'm re-opening and re-closing this post to migrate to main. – Willie Wong Mar 25 '13 at 16:28

We have

$$\begin{eqnarray*} \Bbb{Z}[\sqrt{-5}]/(2,1+ \sqrt{-5}) &\cong& \Bbb{Z}[x]/(x^2 + 1)/\left((2,1+x)/(x^2 + 1)\right) \\ &\cong& \Bbb{Z}[x]/(2,x+1) \\ &\cong& \Bbb{Z}/2\Bbb{Z}\end{eqnarray*}$$

and so your ideal is maximal.

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Call your ideal $I.$ You want to show that the result of quotienting out by $I$ is a field. It is very easy to see this directly. Since $a+b\sqrt{-5} = a-b + b(1+\sqrt{-5})$ we see that $$ a+b\sqrt{-5}+I = a-b + I = (a-b)\pmod 2 + I$$ so if we can show that $1\notin I$ then $a+b\sqrt{-5}+I$ is either $0+I$ or $1+I$, i.e. the quotient ring is the field with two elements.

Suppose $1 = 2(a+b\sqrt{-5})+ (1+\sqrt{-5})(c+d\sqrt{-5}) = 2a+c-5d +(2b+c+d)\sqrt{-5}.$ Then $c-d=1\pmod 2$ and $c+d=0 \pmod 2.$ But $ 2c=1 \pmod 2$ is nonsense, so the ideal is proper as suspected.

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Actually this only shows that the quotient has at most two elements. One also has to prove that the ideal is proper. But this needs some calculation. Alternatively, use the calculus of presentations etc. (see BenjaLim's answer). – Martin Brandenburg Mar 27 '13 at 11:10
@MartinBrandenburg I suppose it does need some calculation to show the ideal is proper. I actually don't understand the 1st isomorphism in Ben's answer which is why I decided to give my own answer. And is there a quick way to see the third isomorphism? I can only see it using a similar type of argument to the one I used in my answer. – Ragib Zaman Mar 27 '13 at 11:18
@RagibZaman $\Bbb{Z}[x]/(2,x+1) \cong \Bbb{Z}/2\Bbb{Z}[x]/(x+1) \cong \Bbb{Z}/2\Bbb{Z}$. – user38268 Mar 27 '13 at 13:43
@Ragib: These isomorphisms should be really seen as isomorphisms of representing functors. One just "reformulates" the data. No clumsy calculations within the representing objects are necessary. – Martin Brandenburg Mar 30 '13 at 11:57

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