Find the primes in $Q[\sqrt{-1}]$ which have norm less than 6 Question: Find the primes in $Q[ \sqrt{-1}]$ which have norm less than $6$. Do the same for the primes in $Q[ \sqrt{-3}]$, and the primes in $Q[\sqrt{-5}]$. Group them according to which ones are associates of each other. As usual, prove that the numbers you list are, indeed, prime, and that there are no others.
My Solution: If $p ⊂ O$ is a prime ideal in the ring of integers of any number field $K/Q$, then it lies over some rational prime $p∈Z$, and the norm of $p$ will be a power of $p$. So in any case, the only possibilities are primes that divide $2$,$3$, or $5$. WE check those cases for your $3$ number fields. For example in $K=Q[i]$, then$(1−i)|2$, with norm $N(1−i)=2$, and this is the only prime lying over $2$. The ideal $(3)$ is prime in $K$, and has norm $N(3)=9>6$, so that doesn't work. And finally, the ideal $(5)$ splits with $(2+i)$ and $(3+i)$ lying over it, both ideals having norm $5$.
Does anyone know of a solution that works without splitting fields? Thanks!
 A: Just test the norms up to the desired bound. For this particular exercise, it's easy enough because the bound, 6, is low, and no negative integers are involved because norms can't be negative in an imaginary ring like $\mathbb{Z}[i]$.
Remember the formula $N(a + b \sqrt{d}) = a^2 - db^2$. Since $d$ is negative, this works out to $a^2 + |d|b^2$. In the case of $\mathbb{Z}[i]$, we're looking for ways to express primes as a sum of two squares, or their squares as a sum of two squares.
In the case of 2, we see that $2 = 1^2 + 1^2$, and hence $(1 - i)(1 + i) = 2$.
There is no such expression for 3, so we look to 9... but that's greater than 6, so 3 is not one of the primes we're looking for.
And then for 5, we see that $5 = 2^2 + 1^2$, corresponding to the primes $2 \pm i$ and $1 \pm 2i$. However, you've been misled about $3 + i$, since $(2 - i)(1 + i) = 3 + i$.
In $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$ you also have to watch out for the possibility of $N(z) = 4p$, since, for example, $$\left(\frac{5}{2} - \frac{\sqrt{-3}}{2}\right)\left(\frac{5}{2} + \frac{\sqrt{-3}}{2}\right) = 7.$$ But clearly that has a norm greater than 6, so you don't have to worry about it in this particular instance.
