Past exam of the number theory course of the University of Oxford / 2008 / B / 6 / Solution

The Dedekind factorization theorem says: if $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and the rational prime $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. Then the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, with integer lifts $f_i$, andThe index condition is part of the theorem and must be checked before applying it at a ramified prime.

For this cubic, $X^3-6$ is Eisenstein and has discriminant $-972$. We first verify ${\mathcal{O}}_K=\mathbb{Z}[\theta ]$. The square of the index divides 972, so only two and three can divide the index. If either prime $p$ divided it, the finite quotient group would contain an element of order $p$, giving an integral element $(a+b\theta +c{\theta }^2)/p$ with digits $0\le a,b,c<p$, not all zero. Its field norm isFor $p=2$, the numerator has valuation zero if $a=1$, valuation one if $a=0,b=1$, and valuation two if $a=b=0,c=1$. For $p=3$ the same three cases, according to the first nonzero digit, give valuations zero, one and two. None gives an integer field norm after dividing by $p^3$, contradicting integrality. Thus the index is one.

Multiplication by $a+b\theta$ in the power basis, or the product over the three conjugates, gives $\boxed{N(a+b\theta )=a^3+6b^3}$. DefineReduction of the cubic gives $X^3$ at two and three, $(X-1)(X^2+X+1)$ at five, and $(X-3)(X-5)(X-6)$ at seven. The quadratic factor at five is irreducible since its discriminant is two, a nonsquare modulo five. HenceThe norms are respectively $2,3,5,25,7,7,7$. Norms and reduction of the elements at those primes giveFor example their absolute element field norms are $6,5,2,21,70$; each indicated prime divides the element by substitution in its residue field, and the product of the indicated ideal norms already equals that absolute field norm. Thus no additional factor or exponent is possible.

In the ideal of the specified quotient every prime exponent cancels:Equality with the unit ideal implies that both $\mu$ and its inverse are integral, so $\boxed{\mu \text{ is a unit}}$. For the real root, $1<\theta <2$. The numerator is negative and has absolute value less than two, while the denominator is greater than five. Thus $-2/5<\mu <0$, proving $\boxed{\mu \ne \pm 1}$. Its field norm is $-1$, consistent with the ideal calculation.