2
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Given a finite field extension $L$ of $\mathbb{Q}$ of dimension $n$, there is a natural way to embed it into $\mathbb{R}^n$ such that the image of its ring of integers $\mathcal{O}_L$ is a lattice. If $L=\mathbb{Q}[\zeta_p]$ where $\zeta_p$ is a primitive $p$-th root of unity, and $\sigma$ is the generator for the Galois group, this embedding is defined by $$\alpha \mapsto (\alpha,\sigma(\alpha),...,\sigma^{\frac{p-3}{2}}(\alpha))\in \mathbb{C}^{\frac{p-1}{2}}\cong \mathbb{R}^{p-1}.$$ The ring of integers in this case is just $\mathcal{O}_L=\mathbb{Z}[\zeta_p]$.

I would like to know what is the covering radius for such lattices, and in particular for the case $p=5$. For $p=3$ the lattice is just the tessellation of the plane using equilateral triangles, and the covering radius is just the distance from the point in the middle of the triangle to one of its vertices. In the case of $p=5$ it becomes much more difficult since the lattice is 4 dimensional, and already I do not know the covering radius of it.

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2
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You can use the following Magma code on the online calculator :

for p in [3,5,7] do
    G:=GramMatrix(Lattice(MaximalOrder(CyclotomicField(p))));
    K:=ZeroMatrix(Integers(),p-1,p-1);
    for i in [1..p-1] do
        for j in [1..p-1] do
            K[i,j]:=Round(G[i,j]);
        end for;
    end for;

    CoveringRadius(LatticeWithGram(K));
end for;

What you'll get is

    2/3
    2
    4

i.e. the squared covering radii.

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  • $\begingroup$ Thanks. One more question - when considering p=3, I got that Lattice(MaximalOrder(CyclotomicField(3))) returns the matrix $\sqrt{2}\left(\begin{array}{cc} 1 & 0\\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right)$. The ring of integers is spanned by 1 and $\frac{-1-i\sqrt{3}}{2}$, so do you know from where the $\sqrt{2}$ comes from? $\endgroup$ – Ofir Apr 9 '15 at 12:04
  • $\begingroup$ It is there to make $1_K$ a vector of length $Tr(1_K)$ ... (here, 2). $\endgroup$ – few_reps Apr 9 '15 at 13:34

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