# proof of Minkowski’s bound

The proof of Minkowski’s bound will rely on Minkowski’s lattice point theorem (http://planetmath.org/MinkowskisTheorem), but we first need to establish some lemmas.

###### Lemma 1.

Let $M$ be a real number and suppose that for every non-zero ideal $\mathfrak{a}$ of the ring of integers $\mathcal{O}_{K}$ there exists a non-zero $x\in\mathfrak{a}$ with norm $\operatorname{N}(x)\leq M\operatorname{N}(\mathfrak{a})$.

Then, every ideal class of $\mathcal{O}_{K}$ has a representative $\mathfrak{a}$ satisfying $\operatorname{N}(\mathfrak{a})\leq M$.

###### 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 $\operatorname{N}(y)\leq M\operatorname{N}(x\mathfrak{b}^{-1})$. Then, $\mathfrak{a}\equiv x^{-1}y\mathfrak{b}$ is an ideal representing $[\mathfrak{b}]$ and $\operatorname{N}(\mathfrak{a})=\operatorname{N}(y)/\operatorname{N}(x\mathfrak% {b}^{-1})\leq M$. ∎

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

 $\displaystyle j\colon K\rightarrow\mathbb{R}^{r_{1}}\times\mathbb{C}^{r_{2}},$ $\displaystyle j(x)=(\sigma_{1}(x),\ldots,\sigma_{r_{1}}(x),\tau_{1}(x),\ldots,% \tau_{r_{2}}(x)).$

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

 $\displaystyle f\colon\mathbb{R}^{r_{1}}\times\mathbb{C}^{r_{2}}\rightarrow% \mathbb{R}^{n},$ $\displaystyle f(x_{1},\dots,x_{r_{1}},y_{1},\ldots,y_{r_{2}})=(x_{1},\ldots,x_% {r_{1}},\Re(y_{1}),\ldots,\Re(y_{r_{2}}),\Im(y_{1}),\ldots,\Im(y_{r_{2}})).$

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.

###### Lemma 2.

If $\mathfrak{a}$ is a non-zero ideal of $\mathcal{O}_{K}$, then $\Gamma=f\circ j(\mathfrak{a})$ is a lattice in $\mathbb{R}^{n}$ (http://planetmath.org/LatticeInMathbbRn). Its fundamental mesh has volume

 $\operatorname{vol}(\Gamma)=2^{-r_{2}}\sqrt{|D_{K}|}\operatorname{N}(\mathfrak{% a}).$
###### Proof.

The proof of this is to be added. ∎

###### Lemma 3.

For any $L>0$, let $S$ be the set in $\mathbb{R}^{r_{1}}\times\mathbb{C}^{r_{2}}$ consisting of points $(x_{1},\ldots,x_{r_{1}},y_{1},y_{r_{2}})$ satisfying

 $\sum_{k=1}^{r_{1}}|x_{k}|+2\sum_{k=1}^{r_{2}}|y_{k}|\leq L.$

Then, $f(S)$ has volume $(2^{r_{1}-r_{2}}\pi^{r_{2}}/n!)L^{n}$.

###### 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

 $\frac{2^{r_{1}-r_{2}}\pi^{r_{2}}}{n!}L^{n}=2^{n}b2^{-r_{2}}\sqrt{|D_{K}|}% \operatorname{N}(\mathfrak{a}).$

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

 $\begin{split}\displaystyle\operatorname{N}(x)&\displaystyle=\prod_{k=1}^{r_{1}% }|\sigma_{k}(x)|\prod_{k=1}^{r_{2}}|\tau_{k}(x)|^{2}\\ &\displaystyle\leq n^{-n}\left(\sum_{k=1}^{r_{1}}|\sigma_{k}(x)|+2\sum_{k=1}^{% r_{2}}|\tau_{k}(x)|\right)^{n}\\ &\displaystyle\leq n^{-n}L^{n}=bM_{K}\sqrt{|D_{K}|}\operatorname{N}(\mathfrak{% a})\end{split}$

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 $\operatorname{N}(x)\leq M_{K}\sqrt{|D_{K}|}\operatorname{N}(\mathfrak{a})$ and Minkowski’s bound follows from Lemma 1.

Title proof of Minkowski’s bound ProofOfMinkowskisBound 2013-03-22 18:33:41 2013-03-22 18:33:41 gel (22282) gel (22282) 5 gel (22282) Proof msc 11R29 msc 11H06 MinkowskisTheorem MinkowskisConstant IdealClass