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).

But:

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)

...is 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.

was an errorin 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! $\endgroup$5more comments