Avoiding norm calculation in quadratic number field To show that the ring $A=\mathbb{Z}[\sqrt{-5}]$ is not principal,  we show that the ideal $I=(1+\sqrt{-5},1-\sqrt{-5})A$ is not principal (but $I^2= \langle 2 \rangle$ is). This amounts not to find  $\alpha=a+a'\sqrt{-5}$ and $\beta=b+b'\sqrt{-5}$  such that $\alpha\beta=2$.
One normally uses the norm  machinery,  which applied to the equation $\alpha\beta=2$ gives the following equation to be discussed: $$(a^2+5a'^2)(b'^2 +5b'^2)=\alpha\alpha'\beta\beta'=N(\alpha)N(\beta)=N(2)=4$$ where  $\alpha'=a-a'\sqrt{-5}$ and $\beta'=b-b'\sqrt{-5}$ are the conjugates.
Question. Suppose you know nothing yet about norms or don't want to use them, how can we get the result above, namely that $\alpha'\beta'=2$?
Edited: if $a+a'\sqrt{-5}$ divides some $n$, say $\alpha\beta=(a+a'\sqrt{-5})(b+b'\sqrt{-5})=n$, this means that $a'b+ab'=0$ so $\alpha'\beta'=(a-a'\sqrt{-5})(b-b'\sqrt{-5})=n$ as well.
Then $\pm a\pm a'\sqrt{-5}$ divides $n$ in the same way.
 A: It seems excessive to me to call the norm function "machinery," but it does enable us, with some adjustments, to do certain things that we take for granted in $\mathbb Z$ in any ring of algebraic integers, be it real or imaginary.
At least in an imaginary ring like $\mathbb Z[\sqrt{-5}]$, there is one nice little fact from $\mathbb Z$ that we can carry over: if $n$ is a composite nonzero number, then its nontrivial divisors are closer to 0 than $n$ itself. For example, $-42$ is 42 away from 0, whereas its divisors (like $-14$ and 21) are much closer to 0.
And so in $\mathbb Z[\sqrt{-5}]$, we see that 6 is 6 away from 0, while its divisors like 2 and 3 are 2 and 3 respectively away from 0, and $1 + \sqrt{-5}$ is roughly 2.45 away from 0.
By the way, we can restrict our search to the positive-positive quadrant, because if $a + b \sqrt{-5}$ is a divisor of $n$, then so are $a - b \sqrt{-5}$, $-a + b \sqrt{-5}$ and $-a - b \sqrt{-5}$.
So, if $1 + \sqrt{-5}$ is a divisor of 6, what are the divisors of $1 + \sqrt{-5}$? The obvious candidates are 1 (a trivial divisor) and $\sqrt{-5}$, since they are clearly closer to 0 than $1 + \sqrt{-5}$ is. And there is 2, which is just slightly closer to 0 than $1 + \sqrt{-5}$.
But $(-\sqrt{-5})(\sqrt{-5}) = 5$, which is farther away from 0 than $1 + \sqrt{-5}$, and the same is also true of $2 \sqrt{-5}$. If we multiply other pairs of numbers which are farther away from 0, the results will be farther away from 0 still.
This means that $1 + \sqrt{-5}$ is irreducible. Similar calculations will show that 2 and 3 are also irreducible. Since $6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5})$, neither $\langle 2 \rangle$, $\langle 3 \rangle$,  $\langle 1 - \sqrt{-5} \rangle$ nor $\langle 1 + \sqrt{-5} \rangle$ are prime ideals.
Except for $\langle 3 \rangle$,  each of these ideals is properly contained in $\langle 2, 1 + \sqrt{-5} \rangle$, which is definitely not a principal ideal. And $\langle 3 \rangle$ is properly contained in both $\langle 3, 1 - \sqrt{-5} \rangle$ and $\langle 3, 1 + \sqrt{-5} \rangle$.
So, one way not to use norms is to visualize the complex plane. But this method is no good when the numbers of the ring are all on the real number line, e.g., $\mathbb Z[\sqrt{10}]$. In a ring like that, norms save the day.
