# Lattice generated by elements of minimal length

I'm following Miranda's book on algebraic curves and Riemann surfaces. In the section where he talks about the automorphisms group of the complex tori he claims the following:

Let $L$ be a lattice in $\mathbb{C}^2$ of rank 2. Let $\gamma$ such that $\gamma L = L$. Then $\gamma$ is a root of unity. Now choose $\ell \in L$ a nonzero element of minimal length. If $\gamma \neq 1, -1$ then $\ell$ and $\gamma \ell$ generate the lattice $L$. It follows that $\gamma^2 \ell \in L$ and $\gamma$ is the root of a polynomial of degree at most 2.

Well, it is clear that $\gamma$ has to be a root of unity because it must map an element of minimal length to another one of the same length, otherwise we couldn't have $\gamma L = L$. It is also clear that $\gamma^n = 1$ for some $n$ because there is a finite amount of elements of minimal length. Now I see that $\ell$ and $\gamma \ell$ must be two nonzero elements of $L$ of minimal length linearly independent over $\mathbb{R}$. But I don't see any reason a priori for them to generate $L$. The rest of affirmations are clear. Any help would be greatly appreaciated.

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## 1 Answer

That $\gamma$ preserves length does not imply it's a root of unity, only that it has modulus $1$. When you establish $\gamma^n=1$, that's when you have established that $\gamma$ is a root of unity.

Now consider the sublattice $M$ of $L$ generated by $\ell$ and $\gamma\ell$. If it isn't all of $L$ then there's some other element $\ell'$ in $L$. Now $\ell'$ is contained in some rhombus of $M$, and its distance from (at least) one of the corners of this rhombus is less than the length of $\ell$. That gives rise to an element of $L$ of length less than that of $\ell$, contradiction.

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