Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 20H (Number Fields)

For a basis of a module $m_1,\dots ,m_d$ of the full-rank module $M$, define its discriminant of elements of a number field byThe field discriminant is this determinant for an integral basis of ${\mathcal{O}}_K$. A change of integral basis with determinant $\pm 1$ does not change it. Write $m_j={\sum }_i{b}_{ij}{e}_i$ in an integral basis and let $B=(b_{ij})$. The two trace matrices satisfy $T_M=B^{\mathsf{T}}{T}_{{\mathcal{O}}_K}B$. The index of the lattice is $r=|\det B|$, as follows for example from Smith normal form. ThereforeMultiplicative closure makes $M$ an order when it contains one, but the determinant argument itself requires only full finite index, so it also covers the stated module hypothesis without that extra condition.

The cubic has no rational root: its only possible integer roots are $\pm 1$, and neither works. It is therefore irreducible. Let $\alpha$ be its image in $K$; it is an algebraic integer, so $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_K$. The power-basis discriminant isThe index-discriminant formula from part (i) says that the square of $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ divides 59. It must be one. Hence $\boxed{{\mathcal{O}}_K=\mathbb{Z}[\alpha ]}$, with integral basis $1,\alpha ,{\alpha }^2$ and field discriminant $-59$.

Write $f=X^2+bX+c$ and $g=X^2+BX+C$. Since their common square-free discriminant $d$ is congruent to zero or one modulo four, it must be one modulo four; thus $b,B$ are odd. It is not a rational square because a square-free integer square is only zero or one. Both polynomials are irreducible over $\mathbb{Q}$, and their roots generate the same quadratic extension $\mathbb{Q}(\sqrt{d})$. Their integral roots are $(-b+\sqrt{d})/2$ and $(-B+\sqrt{d})/2$, respectively. Each generated order isthe usual ring of integers for square-free $d\equiv 1(\mathrm{mod}4)$. This identifies both quotients with the same ring. Explicitly, if $\xi$ denotes the image of $X$ modulo $g$, send the image of $X$ modulo $f$ to $\xi +(B-b)/2$. Equal discriminants give $f(\xi +(B-b)/2)=g(\xi )=0$, and the opposite translation gives its inverse. The two quotient rings are isomorphic.