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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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    $\begingroup$ 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!! $\endgroup$ Mar 24, 2013 at 14:38
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    $\begingroup$ I'm re-opening and re-closing this post to migrate to main. $\endgroup$ Mar 25, 2013 at 16:28
  • $\begingroup$ math.stackexchange.com/questions/341642/… $\endgroup$ Mar 27, 2013 at 11:08

2 Answers 2

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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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  • $\begingroup$ 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). $\endgroup$ Mar 27, 2013 at 11:10
  • $\begingroup$ @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. $\endgroup$ Mar 27, 2013 at 11:18
  • $\begingroup$ @RagibZaman $\Bbb{Z}[x]/(2,x+1) \cong \Bbb{Z}/2\Bbb{Z}[x]/(x+1) \cong \Bbb{Z}/2\Bbb{Z}$. $\endgroup$
    – user38268
    Mar 27, 2013 at 13:43
  • $\begingroup$ @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. $\endgroup$ Mar 30, 2013 at 11:57

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