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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 1

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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