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I am stuck with a homework problem.

Let $R=\mathbb{Z}[\sqrt{ -3}]$.

a) Find an ideal $I$ of $R$ such that $(4) \subsetneq I \subsetneq R$. Explain why the inclusions $\subsetneq$ in my example are strict.

b) Now find another ideal $J$ of $R$ such that $(4) \subsetneq J \subsetneq R$ again explain why the $\subsetneq$ are strict and explain why $J \neq I$.

c) Do there exist ideals $I_1, I_2$ of $R$ such that $(4) \subsetneq I_1 \subsetneq I_2 \subsetneq R$? Justify your answer.

Inclusion is not strict because $(4)$ can not be equal to ideal $I$ and to $R$, but I can not find those ideals and have no idea how to do part c).

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Welcome to MSE! Please take some time to see how LaTeX is used here to format questions and answers. –  Andreas Caranti Mar 12 '13 at 14:46
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@AndreasCaranti Why not $\LaTeX$? =) –  Pedro Tamaroff Mar 12 '13 at 22:27
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3 Answers 3

Remember, for any ideal $I \trianglelefteq R$ the ideals above $I$ corresponds to the ideals of $R/I$. Thus it suffices to consider $R/(4)$ in your case.

Since $R \cong \mathbb{Z}[X]/(X^2+3)$ we have $R/(4) \cong \mathbb{Z}[X]/(X^2+3,4) \cong (\mathbb{Z}/(4))[X]/(X^2+3)$. Since $X^2+3 = (X+1)(X-1)$ over $\mathbb{Z}/(4)$ we can conclude by the Chinese remainder theorem that $R/(4) \cong \mathbb{Z}/(4) \times \mathbb{Z}/(4)$. Now it should be really easy to calculate the ideals.

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Wow, that $4$ really runs around, doesn't he :) –  rschwieb Mar 12 '13 at 15:25
    
Is it really easy to find the required ideals by using your isomorphism? Honestly, I doubt this. –  user26857 Mar 12 '13 at 22:05
    
@YACP: Why? $\mathbb{Z}/(4) \times \mathbb{Z}/(4)$ is a finite ring and has only very few ideals. You can easily write down the whole lattice. –  Dune Mar 13 '13 at 7:21
    
Ok. Then please write down the ideals of this finite ring and their corresponding ideals in $R=\mathbb{Z}[\sqrt{ -3}]$. –  user26857 Mar 13 '13 at 10:05
    
Since this is a homework question, I won't write down the whole solution. $\mathbb{Z}/(4)$ has 3 ideals, so the ideals of $\mathbb{Z}/(4) \times \mathbb{Z}/(4)$ are exactly the 9 products. It's not important for the above task to calculate concrete generators of the corresponding ideals in $R$, but if you really want, you have just to apply every isomorphism. That's a bit sophisticated since it involves applying the Chinese remainder theorem, but as I said: there is no need for calculating concrete elements here. –  Dune Mar 13 '13 at 15:15
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For part (a) we have $(4)\subsetneq (2)\subsetneq R.$ See if you can understand that and explain why the containments are strict. Then play around with this basic example and the fact that $\sqrt{-3}$ is in your ring to come up with an example for (b).

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should not be +3 in your last line? I know that and I think from this (4) is coming from by squaring and puting b=1 and a=1 there, but still don't understand how to get an ideal (2), can you say something more, I think I am just missing some idea. –  user66434 Mar 12 '13 at 15:12
    
you keep editing so I don't have chance to comment:) –  user66434 Mar 12 '13 at 15:12
    
@user66434 Rootmath is extremely subtly hinting that you might find another factorization of $4$. –  rschwieb Mar 12 '13 at 15:29
    
I am still not to sure why containments are strict. Is the other ideal equals to 1? –  user66434 Mar 12 '13 at 15:33
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(Very) short answer:

In the ring $R=\mathbb{Z}[\sqrt{-3}]$ we have unique prime factiorization. Therefore all ideals in $R$ are principal Ideals. Thus, your Ideal $I$ must be of the form $\langle a\rangle$ where $a$ is a proper factor of $4$. Due to the famous formula $(a+b)(a-b)=a^2-b^2$ we get

$$ (1+\sqrt{-3})(1-\sqrt{-3})=1-(-3)=4 $$

In fact, this is the prime factorization of 4 in $R$. Perhaps, you should prove this fact. Now you get $I=\langle 1+\sqrt{-3}\rangle$ and $J=\langle 1-\sqrt{-3}\rangle$ and this is the only possible choice up to permutation.

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(-1) This answer is full of mistakes. Learn more from this topic. –  user26857 Mar 12 '13 at 20:57
    
@YACP It would be nice if you pointed them out. –  Pedro Tamaroff Mar 12 '13 at 22:28
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