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I'm computing a class group for an imaginary quadratic field and something seems wrong.

Let $\delta=\sqrt{-29}$ and let $R$ be the ring of integers in $\mathbb Q[\delta]$. From $(1-\delta)(1+\delta)=30=2\cdot 3\cdot 5$, we conclude that 2, 3, and 5 all split (as ideals) in $R$. Let $P_2$ be an ideal of $R$ such that $\overline{P_2}P_2=(2)$ and define $P_3$ and $P_5$ analogously. As 29 is sufficiently small, we know that class group of $R$ is generated by $P_2,P_3,P_5$. We'll also denote by $\langle P\rangle$ the ideal class corresponding to an ideal $P$.

From the following 2 computations: $$(1-\delta)(1+\delta)=2\cdot 3\cdot 5\implies \langle P_2\rangle\langle P_3\rangle\langle P_5\rangle=1$$ $$(4-\delta)(4+\delta)=3^2\cdot 5\implies \langle P_3\rangle^2\langle P_5\rangle=1$$

we have that $1=(\langle P_2\rangle\langle P_3\rangle\langle P_5\rangle)(\langle P_3\rangle^2\langle P_5\rangle)^{-1}=\langle P_2\rangle\langle P_3\rangle^{-1}\implies \langle P_2\rangle=\langle P_3\rangle$. As $\langle P_2\rangle$ has order 2 (2 ramifies in $R$), $\langle P_3\rangle$ has order 2 as well, so this means that $\langle P_5\rangle=1$, in other words, $P_5$ is a principal ideal, in other words, 5 has a proper divisor in $R$. But it's easy to see that 5 does not have a proper divisor, so there seems to be a contradiction. I most likely overlooked something important, but I can't find it. What am I missing?

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If $1+\delta = P_2 P_3 P_5$, then $4+\delta = P_3^2 \overline{P}_5$, not $P_3^2 P_5$. You can see this by noting that $\langle 1+\delta, 4+\delta \rangle$ is an ideal of index $3$, so there cannot be an order $5$ prime dividing both $1+\delta$ and $4+\delta$. Similarly, $4-\delta = \overline{P}_3^2 P_5$.

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  • $\begingroup$ That makes sense! Now everything is consistent with $\langle P_3\rangle=g,\langle P_5\rangle=g^2,\langle P_2\rangle=g^3$ where the class group is isomorphic to $C_6$ and $g$ is a generator. $\endgroup$ – Yongyi Chen Apr 11 '13 at 3:32

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