proof of Minkowski's bound

The proof of Minkowski's bound will rely on Minkowski's lattice point theorem, but we first need to establish some lemmas.

If the real embeddings of $K$ are denoted by $\sigma_k\colon K\rightarrow\mathbb{R}$ ($k=1,\ldots,r_1$ ) and the complex embeddings are $\tau_k\colon K\rightarrow\mathbb{C}$ together with their complex conjugates $\bar\tau_k$ ($k=1,\ldots,r_2$ ), then we define

Also note that $\mathbb{R}^{r_1}\times\mathbb{C}^{r_2}$ is isomorphic as a real vector space to $\mathbb{R}^{r_1+2r_2}=\mathbb{R}^{n}$ given by the isomorphism
As $f$ and $j$ are linear maps (with respect to the field of rationals $\mathbb{Q}$ ), the combination $f\circ j$ gives a $\mathbb{Q}$ -linear map from $K$ to $\mathbb{R}^n$ . The image will be a lattice, and we can compute its volume.

Proof of Minkowski's bound

For an ideal $\mathfrak{a}$ and any constant $b>1$ , let $L>0$ be given by \begin{equation*} \frac{2^{r_1-r_2}\pi^{r_2}}{n!} L^n=2^nb2^{-r_2}\sqrt{|D_K|}\operatorname{N}(\mathfrak{a}). \end{equation*}Letting $S$ be the set given in Lemma 3 and $\Gamma=f\circ j(\mathfrak{a})$ , Lemmas 2 and 3 give $\operatorname{vol}(S)>2^n\operatorname{vol}(\Gamma)$ . As $S$ is convex and symmetric about the origin, Minkowski's theorem tells us that there is a non-zero $x\in\mathfrak{a}$ with $f\circ j(x)\in S$ .

As the geometric mean is always bounded above by the arithmetic mean, we get the inequality \begin{equation*}\begin{split} \operatorname{N}(x)&=\prod_{k=1}^{r_1}|\sigma_k(x)|\prod_{k=1}^{r_2}|\tau_k(x)|^2\\ &\le n^{-n}\left(\sum_{k=1}^{r_1}|\sigma_k(x)|+2\sum_{k=1}^{r_2}|\tau_k(x)|\right)^n\\ &\le n^{-n}L^n=b M_K\sqrt{|D_K|}\operatorname{N}(\mathfrak{a}) \end{split}\end{equation*}where $M_K=(n!/n^n)(4/\pi)^{r_2}$ . If we choose $b$ such that $bM_K\sqrt{|D_K|}\operatorname{N}(\mathfrak{a})$ is less than the smallest integer greater than $M_K\sqrt{|D_K|}\operatorname{N}(\mathfrak{a})$ , then this gives $\operatorname{N}(x)\le M_K\sqrt{|D_K|}\operatorname{N}(\mathfrak{a})$ and Minkowski's bound follows from Lemma 1.