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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?

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    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.2011-09-18
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    You seem to be confusing "principal ideal" and "prime ideal".2011-09-18
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    In that case, the claim that the norm of any element of $R$ is prime is just not true.2011-09-18
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    @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?2011-09-18
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    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 time2011-09-18

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