Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 20H (Number Fields)

By unique factorization of ideals in a number field, each nonzero integral ideal has a unique expressionwith only finitely many nonzero exponents. Since the ideal norm is multiplicative, formally expanding one geometric series for each prime ideal givesUniqueness of ideal factorization says that every term occurs exactly once. This proves the Euler product for the Dedekind zeta function as a formal identity.

For each rational prime $p$, defineThese are the three cases in splitting of rational primes in a quadratic field.

Set $x=p^{-s}$. In the split case the local factor of the Dedekind zeta function is $(1-x{)}^{-2}$; in the inert case it is $(1-x^2{)}^{-1}$ because the prime ideal has norm $p^2$; and in the ramified case it is $(1-x{)}^{-1}$. On the other hand, the product of the local factors of ${\zeta }_{\mathbb{Q}}(s)$ and $L(\chi ,s)$ iswhich gives exactly those three expressions when $\chi (p)=1,-1,0$. Multiplying over all rational primes proves the quadratic Dedekind zeta factorizationformally.

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 $n$.