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First of all, sorry for the noob question, but it's driving me crazy...

I was reading John Stillwell's "Elements of Number Theory" (Springer, ISBN 0-387-95587-9).

In an exercise on page 225, he gives an example of a prime-yet-non-maximal ideal in the non-UFD ring of integers $\mathbb{Z}[\sqrt{-5}]$: the prinicipal ideal generated by 2.

The excercise prompts the reader to find the non-zero elements of the residue class ring $\mathbb{Z}[\sqrt{-5}] / (2)$ and check none of them is a zero divisor (thus confirming the residue class ring is in fact an integral domain, as required by a prime ideal).


1) If the ideal is indeed prime and its residue class ring is an integral domain, because said integral domain is finite (it has only four elements: $0$, $1$, $\sqrt{-5}$ and $1+\sqrt{-5}$), it must be a field. But then, if the residue class ring were a field, the ideal would not only be prime but in fact maximal (by the definition of maximal ideal). And it is clearly not maximal, because it is strictly contained in the larger ideal $(2, 1+\sqrt{-5})$ (which, incidentally, is maximal).

2) I further calculated the multiplication table of the residue class ring, and found that:

$(1+\sqrt{-5})^{2} = (1^2 + 2\sqrt{-5} + \sqrt{-5}^2) = -4 + 2\sqrt{-5} \equiv 0 \mod{(2)}$

(since both terms on the left side are divisible by 2) in fact nil-potent, right?

So, which is it? Is the ideal (2) in $\mathbb{Z}[\sqrt{-5}]$ a prime ideal or not?

...Or did he mean to show that "A principal ideal generated by an irreducible (not quite 'prime', as are primes in $\mathbb{Z}$ or other UFDs ) element in this non-UFD, fails to be maximal"?

Sorry if it's just the technicality, but it didn't make sense to me. I should add that I like Stillwell's books (this one included) very much.

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closed as too localized by Andrés Caicedo, Alex Bartel, Kevin Buzzard, Daniel Litt, JBorger May 1 '11 at 8:37


This question is undergraduate-level so probably not appropriate for this site. Try on . The numerical calculations you do in the question seem fine. In the ring of integers of a number field, the maximal ideals are the same as the non-zero prime ideals; hopefully this will help you in your confusion. I don't have access to the book you refer to (which makes the question much harder to make sense of) so I can't really say anything more precise about where the misunderstanding has been. – Kevin Buzzard May 1 '11 at 8:01
The ideal (2) is not prime: (1 + sqrt(-5))^2 = -4 + 2sqrt(-5) is a multiple of 2 in Z[sqrt(-5)], so Z[sqrt(-5)]/(2) has a zero divisor other than 0, namely 1 + sqrt(-5) mod (2). – KConrad May 1 '11 at 8:05
I looked the book up on Google; there is definitely an error. The exercise literally asks you to show that the quotient ring is a finite domain, then to show that the ideal is not maximal. – Kevin Ventullo May 1 '11 at 9:51
+1, to balance the downvote. Fine, it's not the most advanced question, and perhaps slightly too basic for MO. But, afterall since it seems there was an error in the book, I think to get confused and to ask for clarification is perfectly natural. And, I'd say there is plenty more basic stuff to be found on MO, in particular in NT. So, closure maybe, but downvote without comment, no! – quid May 1 '11 at 11:49
Nicolás, I apologise for confusing you with those exercises. I truly do not know what I was thinking when I wrote such nonsense, but I'm glad it has been sorted out now. – John Stillwell May 13 '11 at 2:17