Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 20H (Number Fields)

The trace form of a number field isIt is bilinear because the field trace is linear. If $x\ne 0$, choose $y=x^{-1}$. ThenThus no nonzero $x$ lies in the radical, and $B$ is a nondegenerate bilinear form. This is the nondegeneracy of the trace form of a number field.

Let ${\omega }_1,\dots ,{\omega }_n$ be an integral basis of ${\mathcal{O}}_K$. Since each ${\alpha }_j$ is an algebraic integer, there is an integer matrix $A=(a_{ij})$ such thatThe ${\alpha }_j$ form a $\mathbb{Q}$-basis, so $\det A\ne 0$.

Products of algebraic integers are algebraic integers, and their traces are rational algebraic integers, hence ordinary integers. Therefore every entryis integral. Part (i) makes the trace Gram matrix nonsingular, so $\Delta ({\alpha }_1,\dots ,{\alpha }_n)$ is a nonzero integer.

Let $G_{\omega }$ and $G_{\alpha }$ be the two trace Gram matrices. A change of basis givesTaking determinants yields the discriminant-index formula for an integral latticewhere $d_K=\det G_{\omega }$ is the field discriminant. Since $\det A$ is a nonzero integer,Equality holds exactly when $|\det A|=1$, which is equivalent to $A$ being unimodular and to ${\alpha }_1,\dots ,{\alpha }_n$ being a $\mathbb{Z}$-basis of ${\mathcal{O}}_K$. Thus the minimum is the positive integer

Put $\theta =\sqrt[3]{3}$. The given ring of integers has integral basis $1,\theta ,{\theta }^2$. The discriminant of the power basis, equivalently of $T^3-3$, isThe field has signature $(1,1)$: one real embedding and one conjugate pair of complex embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class consequently has an integral representative of norm at most four.

It remains to inspect prime ideals above $2$ and $3$. Modulo $2$,The Dedekind factorization theorem giveswhereButso the principal ideal $(\theta -1)$ is contained in ${\mathfrak{p}}_2$ and has the same norm; henceAlsosoBoth primes above $2$ are principal.

Finally,Thus the unique prime above $3$ is the principal ideal $(\theta )$. By unique factorization of ideals in a number field, every integral ideal of norm at most four is built from these principal prime ideals. Every ideal class is therefore trivial, and the Class group of Q of cube root of three is