I hope this question isn't trivial. Let $L$ be a lattice in $\mathbb{C}$ generated by two complex numbers $w_1,w_2$ which are linearly independent over $\mathbb{R}$. Let $\gamma\in\mathbb{C}$ be a root of unity such that $\gamma\notin\mathbb{R}$ and $\gamma L=L$, and let $l\in L\backslash\{0\}$ be an element of shortest length. Is it true that the set $\{l,\gamma l\}$ generates $L$? It seems almost trivial, but I have been unable to prove it. Thanks!
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2 Answers
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Since $\gamma L=L$, we have $\gamma w_1=aw_1+bw_2$, $\gamma w_2=cw_1+d w_2$ with integer $a,b,c,d$, which immediately implies that $\gamma$ is a root of a polynomial of degree $2$ with integer coefficients, so the root of unity in question can be of degree $3,4,6$ only (otherwise the cyclotomic polynomial is irreducible over $\mathbb{Z}$ and is of degree greater [or smaller] than $2$). In each of these cases the lattice generated by $l$ and $\gamma l$ satisfies the following property: for each point $A$ inside the fundamental parallelogram of that lattice there is a vertex $X$ of that parallelogram such that the distance between $A$ and $X$ is less than the length of $l$. Now, if our lattice is not $L$, take a point not in $L$ and bring it into the fundamental parallelogram by subtracting an integral combination of $l$ and $\gamma l$, thus finding a shorter element of $L$.
answered May 17, 2011 at 14:04
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$\begingroup$ Should be AX! Sorry, corrected now. $\endgroup$ May 17, 2011 at 14:26
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$\begingroup$ I don't see why that should be... What if $l=1$ and $\gamma=i$. Then $x=0.8+0.8i$ (for example) is in the fundamental parallelogram, but $|xi|=|x|=1.28$, $|x\cdot1|=1.28$ and $|x(1+i)|=2.56$; none of these are greater than 1, which is the length of $l$. $\endgroup$– rfauffarMay 17, 2011 at 14:36
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$\begingroup$ Okay, there is an ambiguity I will correct now - by AX I mean the segment connecting A and X, not the product. Let me change the wording of the answer... (Just in case, in your example above the geometric statement is: for every point inside the unit square there exists a vertex such that the distance from our point to that vertex is less than 1, - this guarantees that Z[i] is an Euclidean domain, you know.) $\endgroup$ May 17, 2011 at 14:41
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$\begingroup$ Ok of course! Now, how can you be sure that you can always bring an element inside the fundamental parallelogram (assuming the lattice isn't L)? $\endgroup$– rfauffarMay 17, 2011 at 15:14
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Since L actually must be the Gaussian integers or Eisenstein integers, up to a scalar factor, this is true and easy to see. There is a little algebra involved, behind scenes, in that statement. I guess you need something equivalent to class number one to be sure (a non-principal ideal in the Gaussian integers would give a different situation).
answered May 17, 2011 at 14:00