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.