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This seems intuitive, but I'm having trouble coming up with an exact matrix for the problem.

Let $\{L_1, \ldots, L_N\}$ be a set of lines through the origin $(0,0,0)$ in the affine space $\mathbb{A}^3$ (over an algebraically closed field). Show that after a linear change of coordinates in $\mathbb{A}^3$, we may assume $L_i$ does not lie in the plane $z=0$ for any $i$ and that $L_i$ is in the span of $(x_i,y_i,1)$ where the $x_i$ are pairwise distinct.

I see the second condition basically says that the lines do not lie directly over each other. It seems intuitive that we can rigidly rotate the axes so that these conditions are held, and as rotation is a linear transformation, the statement follows. I want to try to get a more solid, less handwaving proof.

Haskell Curry
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Math2012pc
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  • A single point $(x_i,y_i,1)$ does not determine a line. Not sure what you mean by "span of" there. – Thomas Andrews Feb 21 '13 at 18:52
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    Take any hyperplane through the origin that doesn't contain any of the lines, and use a linear transformation that shifts that hyperplane exactly to the plane $z = 0$. – bzc Feb 21 '13 at 18:54
  • @BrandonCarter Yeah, the existence of that hyperplane is, I think, the point of the proof, I don't think you can skip over it. :) In particular, the existence is why there is the requirement that $k$ is algebraically closed. – Thomas Andrews Feb 21 '13 at 18:58
  • It seems like the problem would want the $L_i$ to be colinear, because in that case, the first step leads to proving the second step. But if $L_i$ is not in the hyperplane $z=0$ but can be in $z=c$ for some $c$, then $L_i$ cannot be written as having (directional) vector $(x_i,y_i,1)$, since $z$ cannot change in that line. – Thomas Andrews Feb 21 '13 at 19:07
  • Well, the fact that there are finitely many lines means that there has to be some hyperplane that doesn't contain any of them, correct? – Math2012pc Feb 21 '13 at 19:08
  • @Math2012pc Well, if $k$ is a finite field, then that really isn't true :) There is definitely something to prove. It seems obvious, but I think proving this step is the crux of this problem, and you can't just assert it. – Thomas Andrews Feb 21 '13 at 19:14
  • But $k$ is an infinite field in our case. Oh, I see I forgot to add that in the problem. Apologies. Any help appreciated! – Math2012pc Feb 21 '13 at 19:17
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    Let $H_i$ be the set of hyperplanes that contain $L_i$. Then you want $\cup_{i=1}^N H_i$ to not cover the complete set of hyperplanes. Yes, $N$ is finite, but the $H_i$ are infinite, and it requires proof that there are hyperplanes outside that union. My point about finite fields was just that it wasn't true for arbitrary fields. (Also, you implicitly have that $k$ is infinite, since no finite field is algebraically closed, so you don't need to add that condition.) – Thomas Andrews Feb 21 '13 at 19:24
  • You still haven't answered my question about what you mean by "span," since a single point does not determine a line. Are the $L_i$ all supposed to go through $(0,0,0)$? – Thomas Andrews Feb 21 '13 at 19:27
  • Yes, all $L_i$ are through the origin. I tried setting up a system to solve a matrix. – Math2012pc Feb 21 '13 at 19:46

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You need a plane $H$ through the origin containing none of the lines. So let's first show that one exists.

Without loss of generality, we can assume that none of the lines is the $z$-axis, i.e. the line corresponding to the equations $x=y=0$. Consider the family of hyperplanes cut out by $x + ay = 0$, where $a \in k$ . Any two such planes intersect only at points of the form $(0,0,b)$. Since none of the lines is the $z$-axis, we conclude that each $L_i$ lies on at most one hyperplane in this family. Since there are infinitely many planes and finitely many $L_i$, we can find a plane containing none of the lines. QED.

Now that you know this plane exists, by changing coordinates you can assume that it is the plane $z =0$. Since each line goes through the origin and none are contained in $z=0$, it follows immediately that every $L_i$ must contain a point of the form $(x_i,y_i,1)$. If some of the $x_i$ are the same, then distinctness of the lines guarantees that the corresponding $y_i$ are different. Since we have finitely many lines, we can then choose a $C \in k$ such that the automorphism $x \rightarrow x + Cy$, $y \rightarrow y$, $z \rightarrow z$ brings our original lines to ones with the desired pairwise distinctness of the first coordinate.

hot cheetos
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  • What do you mean by "cut out by $x +ay = 0$?. I know that the lines contain a point of that form, but why is it that the $x_i$ are pairwise distinct? – Math2012pc Feb 21 '13 at 21:06
  • "Cut out by" just means that the equation $x + ay$ defines the hyperplane. As for the $x_i$ being pairwise distinct, I think you can guarantee that by a coordinate transformation of the form $x \rightarrow x + ay$, $y \rightarrow y$, $z \rightarrow z$. – hot cheetos Feb 21 '13 at 21:43