Intro:
$K=\mathbb Q(\sqrt{-23})$ be a number field with obvious minimum polynomial.
$\mathcal O_K$ be its ring of integers which is determined as $$\mathcal O_K=\mathbb Z\left[\frac{1+\sqrt{-23}}{2}\right]$$
since $-23\equiv 1 \mod 4$
I want to determine if $I,I^2,I^3$ are principal or not.
I was able to calculate norm of $I$ as: $N(I)=3$.
Then using basic ideas I can show that $I$ is not principal.
$N(I^2)=9=\left(x+y/2\right)^2+\frac{23}4y^2=3^2$
Only solution is $x=\pm 3,y=0$
Why we can't say that $I^2$ is generated by the ideal $(3)$?
I want to show $I^3$ is principal ideal.
Norm is multiplicative so $N(I^3)=27$
So $N(I^3)=27=\left(x+y/2\right)^2+\frac{23}4y^2=3^3$
Has following integer solutions $(x,y)=(-3,2),(-1,2),(1,2),(3,-2)$
And I am stuct to find which one is appropriate candidate, which kind of theorem says/guarantees that?
Best Answer
To solve this problem, it is not necessary to compute powers of $I$ explicitly in terms of generators from knowing generators of $I$. Instead we can focus on the norm of powers of $I$ and reason indirectly by factoring ideals into prime ideals.
The ideal $I$ has prime norm $3$, so it is a prime ideal. (The converse is false: prime ideals need not have prime norm.) So $I^2$ has norm $9$ and $I^3$ has norm $27$.
Are there principal ideals with norm $3$? Let $\alpha = (1 + \sqrt{-23})/2$. For integers $m$ and $n$, $$ {\rm N}(m + n\alpha) = {\rm N}\left(m + n\frac{1+\sqrt{-23}}{2}\right) = \frac{(2m+n)^2 + 23n^2}{4}, $$ so if ${\rm N}(m+n\alpha) = 3$ then $(2m+n)^2 + 23n^2 = 12$. The left side is a sum of nonnegative terms, so the only choice is $n = 0$. Then $(2m)^2 = 12$, which has no integral solution.
Now we turn to $I^2$. It has norm $9$. Can $I^2$ be principal? If $(m + n\alpha)$ has norm $9$ for some integers $m$ and $n$, then $$ 9 = {\rm N}(m + n\alpha) = \frac{(2m+n)^2 + 23n^2}{4}, $$ so $(2m+n)^2 + 23n^2 = 36$. This implies $|n| \leq 1$. If $n = \pm 1$ then $(2m+n)^2 = 36 - 23 = 13$, which is impossible. Thus $n = 0$, so $(2m)^2 = 36$, which tells us $m = \pm 3$. Thus $(m+n\alpha) = (\pm 3) = (3)$.
Can $I^2 = (3)$? No, because we can determine the prime ideal factorization of $(3)$ from the factorization mod $3$ of the minimal polynomial $T^2 - T + 6$ for $\alpha$: $T^2 - T + 6 \equiv T(T-1) \bmod 3$, which is a product of distinct monic irreducibles, so $(3) = \mathfrak p\mathfrak q$ where $\mathfrak p$ and $\mathfrak q$ are distinct prime ideals. So $(3)$ is not the square of the prime ideal $I$ (unique factorization!) and either $\mathfrak p$ or $\mathfrak q$ is $I$.
Now we turn to $I^3$. It has norm $27$. If a principal ideal $(m+n\alpha)$ has norm $27$ then $$ 27 = {\rm N}(m + n\alpha) = \frac{(2m+n)^2 + 23n^2}{4}, $$ so $(2m+n)^2 + 23n^2 = 108$. That implies $|n| \leq 2$, and trying each option gives us the solutions $$ (m,n) = (1,2), (-3, 2), (3,-2), (-1,-2), $$ so $(m+n\alpha)$ is $(1+2\alpha) = (2+\sqrt{-23})$ or it is $(3-2\alpha) = (2-\sqrt{-23})$. We have shown the only principal ideals in $\mathcal O_K$ with norm $27$ are $(2\pm \sqrt{-23})$. Thus we want to factor these into prime ideals to see if either of them is $I^3$.
Since the ideals $(2+ \sqrt{-23})$ and $(2-\sqrt{-23})$ have norm $27$, their only prime ideal factors in $\mathcal O_K$ are prime ideal factors of $3$ (see my post here). We noted above that $(3) = \mathfrak p\mathfrak q$ for distinct prime ideals $\mathfrak p$ and $\mathfrak q$ (where $I$ is one of them). So the only possible ideals in $\mathcal O_K$ with norm $27$ are $$ \mathfrak p^3, \ \ \mathfrak p^2\mathfrak q, \ \ \mathfrak p\mathfrak q^2, \ \ \mathfrak q^3. $$
The ideal $(2+\sqrt{-23})$ is not divisible by both $\mathfrak p$ and $\mathfrak q$, since if it were then it would be divisible by their product $(3)$, and if $(3) \mid (2+\sqrt{-23})$ as principal ideals then $3 \mid (2+\sqrt{-23})$ as elements, but $2+\sqrt{-23}$ is not $3(a + b(1+\sqrt{-23})/2)$ for integers $a$ and $b$. Therefore the only possible prime ideal factorization of $(2+\sqrt{-23})$ is $\mathfrak p^3$ or $\mathfrak q^3$. The exact same reasoning shows $(2-\sqrt{-23})$ is $\mathfrak p^3$ or $\mathfrak q^3$. And since the ideals $(2+\sqrt{-23})$ and $(2-\sqrt{-23})$ are different (they're principal ideals whose generators don't have a ratio that's a unit in $\mathcal O_K$, or in fact even lies in $\mathcal O_K$), their prime ideal factorizations are different. Thus if we let $\mathfrak p$ be the prime ideal such that $(2+\sqrt{-23}) = \mathfrak p^3$, then $(2-\sqrt{-23}) = \mathfrak q^3$.
Since $I$ is $\mathfrak p$ or $\mathfrak q$, we get $I^3 = (2+\sqrt{-23})$ or $I^3 = (2-\sqrt{-23})$. Which one is it?
Recall that $I = (3,1+\sqrt{-23})$, so easily $$ 2-\sqrt{-23} = 3 - (1+\sqrt{-23}) \in I. $$ Thus $(2-\sqrt{-23}) \subset I$, so $I \mid (2-\sqrt{-23})$. Therefore $(2-\sqrt{-23}) = I^3$.