Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 20H (Number Fields)

Write $K=\mathbb{Q}(\alpha )$ with minimal polynomial of degree $n$. An embedding is determined by the image of $\alpha$, and each of the $n$ distinct complex roots gives one embedding, so there are exactly $n$.

For an integral basis ${\omega }_1,\dots ,{\omega }_n$, the field discriminant isOrder the embeddings with all real embeddings first and each nonreal embedding adjacent to its conjugate. Replacing each conjugate pair of rows by their real and imaginary parts extracts one factor $2i$ from the determinant. The remaining determinant is real and nonzero. Squaring showswhere $s$ is the number of conjugate pairs.

The polynomial $X^3+2X^2+1$ is irreducible by the rational-root test. The power basis $1,\theta ,{\theta }^2$ has discriminantFor the order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_L$,Since $59$ is squarefree, the index can only be one. Hence

If the embedding with unit modulus is real, it sends $\alpha$ to $\pm 1$; irreducibility of the minimal polynomial then forces $\alpha =\pm 1$. Otherwise $\overline{{\sigma }_j(\alpha )}=1/{\sigma }_j(\alpha )$ is another conjugate. Hence ${\alpha }^{-1}$ is conjugate to $\alpha$, so their monic minimal polynomials coincide. The minimal polynomial is therefore reciprocal up to its constant sign, and its constant term has absolute value one. Taking the product of all conjugates gives

If $\alpha \in {\mathcal{O}}_K$, its norm is an integer. The preceding result gives norm $\pm 1$. The characteristic polynomial of multiplication by $\alpha$ then expresses ${\alpha }^{-1}$ as an integral polynomial in $\alpha$, so ${\alpha }^{-1}\in {\mathcal{O}}_K$. Therefore

Take $K=\mathbb{Q}(i)$ andIts two embeddings give $(3\pm 4i)/5$, both of modulus one. But $alpha\notin \mathbb{Z}[i]={\mathcal{O}}_K$, since its real and imaginary parts are not integers. This is the required example.