Exercise 1.5.7

Question

Find the shortest non-zero vector in the lattice $L = \Z(1,0) + \Z(1/2, \sqrt{3}/2)$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
Write ${\bf b} = (1,0), c = (\frac{1}{2}, \frac{\sqrt{3}}{2})$, and $L = \Z {\bf b} + \Z {\bf c}$.

If $v = x{\bf b}+ y {\bf c}  = (x + \frac{y}{2}, y  \frac{\sqrt{3}}{2}) \in L$, then 
$$||v||^2 = \left(x + \frac{y}{2}ight)^2 + \left( y \frac{\sqrt{3}}{2}ight)^2 = x^2 + xy + y^2.$$
So we must find the minimum of $f(x,y) = x^2 + xy + y^2 >0$ for $(x,y) \in \Z^2 \setminus \{(0,0)\}$.

Since $1 = f(1,0)$, $1$ is this minimum.

The shortest non-zero vector in the lattice $L = \Z(1,0) + \Z(1/2, \sqrt{3}/2)$ is ${\bf b}$ (or ${\bf c}$, or $-{\bf b} + {\bf c}$, or $-{\bf b}$, or $-{\bf c}$, or ${\bf b}+ {\bf c}$). The minimum is reached by 6 vectors.
\end{proof}

Exercise 1.5.7

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Exercise 1.5.7

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