For $I$ and $I^2$ I can directly calculate the product and then apply the norm trick to get a contradiction that if we assume they are principal, but I'm wondering that if there is any other good way to approach $I^3$ rather than calculation (without some excellent simplification), and similar stuff for $I^4.$ I really appreciate it!
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Here's a MathJax tutorial :) – Shaun Apr 10 '23 at 19:49
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1Thanks so much for the tutorial and the guy who converted the original bad latex symbols into readable symbols! – Yeediu Yang Apr 11 '23 at 03:00
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A similar post: https://math.stackexchange.com/questions/4603975/the-ideal-i-3-1-sqrt-23-subseteq-mathcal-o-mathbbq-sqrt-23 – Viktor Vaughn Apr 11 '23 at 14:29
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Ignore $I^3$. Just show $I^4$ is principal, so the least $k \geq 1$ such that $I^k$ is principal has to divide $4$. (This is like showing in a group from $g^4 = 1$ that $g$ has order $1$, $2$, or $4$, but certainly not $3$ without having to bother computing $g^3$.)
Do you know about ideal norms? Since $I$ has ideal norm $3$ and the ideal norm is multiplicative, $I^4$ has ideal norm $81$. If $I^4 = (a+b\sqrt{-14})$ for some integers $a$ and $b$ then taking ideal norms of both sides shows $a^2 + 14b^2 = 81$. The only solutions to this in integers are $(a,b) = (\pm 9,0)$, $(\pm 5, 2)$, and $(\pm 5, -2)$. Therefore if $I^4$ is going to be principal, the only choices are $(9)$, $(5+2\sqrt{-14})$, and $(5-2\sqrt{-14})$. We'll show the first two choices are impossible.
Case 1: If $I^4 = (9)$ then $I^4 = (3)^2$, so $I^2 = (3)$ by unique factorization of ideals, but you already showed $I^2$ is not principal, so we have a contradiction.
Case 2: If $I^4 = (5+2\sqrt{-14})$ then $2\sqrt{-14} \equiv -5 \bmod I$. Also $3 \equiv 0 \bmod I$, so $2\sqrt{-14} \equiv -5 \equiv 1 \bmod I$. From the definition of $I$, $\sqrt{-14} \equiv 1 \bmod I$, so $2\sqrt{-14} \equiv 2 \cdot 1 \equiv 2 \bmod I$, and $2 \not\equiv 1 \bmod I$, so we have a contradiction.
Therefore the only principal ideal that $I^4$ could equal is $(5-2\sqrt{-14})$. Now verify that this choice works: $I^4 = (5-2\sqrt{-14})$.
KCd
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We can directly find $I^2=(9,x-2)$ and $I^4=(2x-5)$ where $x=\sqrt{-14}$. But, computation of $I^4$ is wicked.
Alternative computation: Prime over $2$ is $J=(2,x)$ and $J^2=(2)$. On the other hand, $JI^2=(x-2)$ and thus $J^2I^4=(2)I^4=(4x+10)$. We can conlude as the paragraph above.
Bob Dobbs
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