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i've come across a lot of questions recently that ask you whether or not there exist certain kinds of ideal, say; does there exist an ideal$ J $of $\mathbb{Z}[i]$ for which $\mathbb{Z}[i] /J$ is a field of 8 elements?

or for $R ={\{a + bi\sqrt3: a,b \in Z}\} $show there is an ideal $I$ of $R$ so that $R/I$ is isomorphic to $\mathbb{Z}_7,$ but no ideal so that $R/J$ is isomorphic to $\mathbb{Z}_5$?

i am struggling to find examples in such questions, and also in proving them impossible. Is there any intuition to know if such things exist? and any standard approach to answering these questions with ideals? An approach to begin with?

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I've added the algebraic-number-theory tag because the first ring you mention is a ring of integers and the second is an order. This sort of problem, at least for rings of integers, is straight foreward using tools from introductory algebraic number theory, although I assume you want a solution without assuming such knowledge. –  Brett Frankel Dec 26 '12 at 21:42
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2 Answers

up vote 3 down vote accepted

For the case of $\, \Bbb Z[i]\,$: try to prove the nice

Claim: For any $\,a+bi\in\Bbb Z[i]\,\,,\,\,\gcd(a,b)=1\,$ , we have that

$$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}$$

Assuming the above, the question is now whether we can write $\,8=a^2+b^2\,,\,a,b\in\Bbb Z\,,\,(a,b)=1$ , which we can't...

Could you do something similar with $\,\Bbb Z[\sqrt{-3}]\,$ ?

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+1 For an elementary solution of the first problem. –  fpqc Dec 27 '12 at 0:38
    
this isomorphism is just what i was looking for! thank you. –  pad Dec 27 '12 at 15:40
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The ring $\Bbb{Z}[i]$ is the ring of integers of $\Bbb{Q}(i)$. Hence finding an ideal such that the quotient is a field of 8 elements is impossible. This is because such an ideal must be prime, lies over $2\Bbb{Z}$ and is of inertia degree 3 but this is impossible because it violates the formula $\sum e_if_i = n$. We have $e_i$ being the ramification indices of the primes lying over $2\Bbb{Z}$, $f_i$ the inertia degrees and $n = 2$ in this case.

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