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

Proof. Let $[\mathfrak{b}]$ be an ideal class represented by the ideal $\mathfrak{b}$ . Choosing a non-zero $x\in\mathfrak{b}$ then $x\mathfrak{b}^{-1}$ is an ideal of $\mathcal{O}_K$ and, by the condition of the lemma, contains a non-zero $y$ satisfying $\operatorname{N}(y)\le M \operatorname{N}(x\mathfrak{b}^{-1})$ . Then, $\mathfrak{a}\equiv x^{-1}y\mathfrak{b}$ is an ideal representing $[\mathfrak{b}]$ and $\operatorname{N}(\mathfrak{a})=\operatorname{N}(y)/\operatorname{N}(x\mathfrak{b}^{-1})\le M$ .

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

Proof. The proof of this is to be added.

Proof. The proof of this is to be added.

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.