Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that you can take any element from $R = \mathbf{Z}[\sqrt{-5}]$ like, say, $a = 1 + \sqrt{-5}$, and it will be a principal ideal, but when I used the norms I got this:

$$N(a) = N(1 + \sqrt{-5})N(1) = 6,$$ so $N(a)$ must be either 1, 2 or 3. $N(a)$ can't be 1 because then $(a) = (1+\sqrt{-5})$ wouldn't be a proper ideal anymore. All that's left is to find out if the equations $a^2 + 5b^2 = 2$ and $a^2 + 5b^2 = 3$ have any solutions. If either one of these has solutions then it confirms that $(1 + \sqrt{-5})$ is a principal ideal, so my question is, are there any?

share|cite|improve this question
I tried to clean up the math -- I hope that's alright. I still don't understand the question, though. $N(a)$ is 6. Why are we trying to show that it isn't? I also don't see what writing $N(a) = N(a)N(1)$ is giving us. – Dylan Moreland Sep 18 '11 at 20:21
You seem to be confusing "principal ideal" and "prime ideal". – Bill Dubuque Sep 18 '11 at 20:22
In that case, the claim that the norm of any element of $R$ is prime is just not true. – Dylan Moreland Sep 18 '11 at 20:24
@crazy student: An ideal $I$ is principal if and only if there exists an $a\in I$ such that $I = (a) = \{ra\mid r\in R\}$. Norms don't enter into it, and your computations above make no sense (you have $N(a)=6$, as it happens). Are you really trying to figure out principal ideals, or are you trying to find principal prime ideals? Or prime ideals? Or irreducible elements? Or something else? – Arturo Magidin Sep 18 '11 at 21:00
I'm totally sorry if it looks confusing. My bad. I do know what we use norms to prove by contradiction that an ideal isn't principal. For example, (3, 1 + sqrt(5)) is not principal becuase there are no solutions to a^2 + 5b^2 = 3. I guess it's not applicable in this situation. Again, soooo sorry people, I am learning all of this stuff for the first time – crazy student Sep 18 '11 at 21:45

Hint: If $b \neq 0$ then $a^2 + 5b^2 > 5$.

share|cite|improve this answer
Hi, I'm not understanding what your hint is trying to say. I am looking to find solutions for two specific equations: a^2 + 5b^2 = 2 and a^2 + 5b^2 = 3. If one of these two equations has a solution then <1 + sqrt(-5)> is a principal ideal. This is my train of thought. – crazy student Sep 18 '11 at 20:13
@crazystudent As some of the others mentioned above, $(1 + \sqrt{-5})$ is principal by its very definition. What you are actually attempting to prove is that it is prime (i.e. it has no proper ideals dividing it). By the hint above you can conclude that if some principal ideal $(a+b \sqrt{-5})$ divides it, you must have $b = 0$. It should be very clear from there. – Brandon Carter Sep 18 '11 at 22:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.