Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 20H / b / Solution

Because $p\equiv 11(\mathrm{mod}12)$, we have $p\equiv 3(\mathrm{mod}4)$ andFor $\theta =(1+\sqrt{-p})/2$, the minimal polynomial isSince $p\equiv 2(\mathrm{mod}3)$, reduction modulo $3$ gives $f(T)\equiv T(T-1)$. Its two roots are distinct, sowith distinct prime ideals of norm $3$. Thus $3$ splits completely, as recorded in splitting of three in Q of square root minus p.

The ideal class group ${\mathrm{Cl}}_K$ is the group of nonzero fractional ideals of ${\mathcal{O}}_K$ modulo the subgroup of principal fractional ideals. Saying that ${\mathfrak{p}}_1$ has order $n$ means that $n$ is the least positive integer for which ${\mathfrak{p}}_1^n$ is principal.

Suppose for contradiction that $n$ is odd and ${\mathfrak{p}}_1$ has order $n$. Then ${\mathfrak{p}}_1^n=(\alpha )$ for some $\alpha \in {\mathcal{O}}_K$, and taking norms givesWrite $\alpha =(a+b\sqrt{-p})/2$, where $a,b\in \mathbb{Z}$ have the same parity. ThenIf $b\ne 0$, the right-hand side is at least $p$, contradicting $p>3^{n+2}>4\cdot 3^n$. Hence $b=0$, so $\alpha$ is a rational algebraic integer and therefore an integer. But then $3^n=N(\alpha )={\alpha }^2$, impossible when $n$ is odd. This proves the odd-order obstruction for a split prime in an imaginary quadratic field and shows that the order cannot equal $n$.

Solved by gpt-5.6-sol high.