It's a couple of days that i'm struggling with this answer, which i'd like very much to understand.

I recall briefely what is the problem: I want to classify all prime ideals of $\mathbb{Z}[i]$. The strategy is the following: for each (non-zero) prime ideal $(q)$ of $\mathbb{Z}$, we want to classify the prime ideals $Q$ of $\mathbb{Z}[i]$ that contain $(q)$. To do this, we first re-express the ring $Z[i]$ in a more convenient form, namely $\mathbb{Z}[i]\cong R:=\mathbb{Z}[x]/(x^2+1)$. Now, the prime ideals of $R$ containing $qR$ are in order-preserving bijection with the prime ideals of $$R/qR=(\mathbb{Z}[x]/(x^2+1))/q(\mathbb{Z}[x]/(x^2+1))\cong \mathbb{Z}[x]/(q,x^2+1)\cong \mathbb{F}_q[x]/(x^2+1).$$

Then we consider various cases for the prime $q$ in $\mathbb{Z}$.

I've many question in all this, main questions are:

1) Consider for example, the case $q\equiv 1\bmod 4$. In this case $x^2+1$ is reducible , so that $$R/qR\cong \mathbb{F}_q[x]/(x-a)(x+a)$$ which has two prime ideals, $(x-a)\mathbb{F}_q[x]/(x-a)(x+a)$ and $(x+a)\mathbb{F}_q[x]/(x-a)(x+a)$. Unwinding our various isomorphisms, this corresponds to a conjugate pair of Gaussian primes $\pi$, $\overline{\pi}$ with norm $q\equiv 1\bmod 4$ in $\mathbb{Z}[i]$. Well, i can't see how exactly to unwind our isomorphisms, who is $\pi$ concretely?

2)Secondly, admitting that $\pi$ and its conjugate $\overline{\pi}$ are the two Gaussian primes for this case, why do i multiply them? I mean, i have two primes in $R/qR$, but only one prime in $\mathbb{Z}[i]$, having norm $q$, why?

A different (last) question is: in this way i classify all ideals of $R$ such that their intersection with $\mathbb{Z}$ is a prime ideal in $\mathbb{Z}$. How can i know whether these ideals are all primes in $R$? I mean, if $P$ is a prime ideal of $R$, certainly $P\cap\mathbb{Z}$ is prime in $\mathbb{Z}$, but could an ideal $Q$ in $R$, not prime in $R$, have $Q\cap\mathbb{Z}$ prime in $\mathbb{Z}$?

Note: also a reference on this classification (not any classification of Gaussian prime ideals, but this particular proof) will be considered an answer.


For your last question, every prime ideal of $\mathbb{Z}$ is contained in a maximal ideal of $R$, which is prime.

For the first question, notice that $F_q [x]/(x^2+1)=(\mathbb{Z}[i])/(q)$. So since the first splits into a ring with two prime ideals, the second does as well, which shows that q is not prime, but in fact is composed of two primes. The ideal $(-x-a)=(x+a)$, so reversing the sign of $x$ switches between the ideals, which means the primes are complex conjugates. Finding the exact primes using this method does not work well; for instance, this method would suggest that the prime factors of 5 and 13 are $2+i,2-i$ and $5+i,5-i$ respectively (since 2 and 5 are the respective square roots of -1). The first case is true, and the second is not; the norm of $5-i$ is 26, not 13. The problem is that $5-i=(1-i)(2+3i)$, where $(2+3i)$ is the actual prime dividing 13.

So one method you could use to find the actual primes is to take $a+i$ (where $a$ is a root of -1 in your finite field), and find its norm; if its norm is not prime, that means you can factor $a+i$. Keep doing so till you get a piece with the right norm.

  • $\begingroup$ I can't see how to use your first line: suppose $Q\triangleleft R$ be an ideal of $R$ such that $Q\cap\mathbb{Z}=:\mathbb{p}$ is a prime ideal of $\mathbb{Z}$. I want to show that $Q$ is a prime ideal. $\endgroup$ – bateman May 30 '13 at 13:43
  • $\begingroup$ But that's false. The principal ideal $Q$ of $R$ generated by $5$ is not a prime ideal, even though its intersection with the integers is a prime ideal. $\endgroup$ – Gerry Myerson May 30 '13 at 13:47
  • $\begingroup$ @GerryMyerson well, so the classification i'm doing is overflowing, since i classify all ideals of $R$ "above" prime ideals of $\mathbb{Z}$, and these ideals are not all prime ideals, while i wanted to classify only prime ideals of $R$ $\endgroup$ – bateman May 30 '13 at 13:52
  • $\begingroup$ Every prime ideal of $R$ lies over some prime ideal of the integers. $(2+i)$ lies over $(5)$ (as does $(2-i)$); $(3)$ lies over $(3)$. They all follow one or the other of those two patterns (except $(1+i)=(1-i)$ lying over $(2)$). $\endgroup$ – Gerry Myerson May 30 '13 at 13:55
  • $\begingroup$ @Gerry Myerson I can't see where do $5$ come from, why $F_q [x]/(x^2+1)=(\mathbb{Z}[i])/(5)$? $\endgroup$ – bateman May 30 '13 at 14:01

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