I'm doing some self study on Fulton's Algebraic Curves, and I've done a decent amount. But I'm stuck on a past question that's been bugging me.

Question : Let $K$ be a field, and $P,P'$ be points in $A^{2}(K)$. Suppose $L_1,L_2$ are lines through $P$ and $L_1',L_2'$ are lines through $P'$. Show that there is an affine change of coordinates such that $T(P)= P'$ and $T(L_i)= L_i'$ for $i=1,2$.

Notation : $A^{n}(K)$ is simply the $n$ times Cartesian Product of $K$. $T$ is an affine change of coordinate of $A^{n}(K)$ if $T$ is a bijective map such that for all $(a_1,...,a_n)\in A^{n}(K)$, there exists $F_1,...,F_n\in K[X_1,...,X_n]$ where $F_i$ have degree 1 such that $T(a_1,...,a_n) = T(F_1(a_1,...,a_n),...,F_n(a_1,...,a_n))$.

Well I can't come up with a map that could rotate 2 lines at once. If I could, I would just compose it with the translation map. I thought of reducing the problem to moving any 3 points to any 3 other points and since ACC takes lines to lines, it would yield the result. But that would force me to use polynomials of degree 2.

I also have similiar problems for the projective case in $P^{2}(K)$. I need to find a projective change of coordinates that moves any 3 points not lying on a line to any 3 other points.

If I have the affine case for dim 2, I might be able to do the one for the projective case just by generalizing the method.

I hope someone can help me with this, or give me a hint. ( Oh btw, I don't really have an extended knowledge of maths.. Just a 3rd year maths student so I'm hoping you could help me without any big machinery).

Thanks :)

  • $\begingroup$ "points through $P$" what does this mean? $\endgroup$ – Hoot Feb 5 '16 at 20:39
  • $\begingroup$ sorry, I meant lines.. $\endgroup$ – Demmas Salim Feb 5 '16 at 21:03
  • $\begingroup$ Here is a hint. Two distinct lines $ax+by+c=0, px+qy+r=0$ intersect if an only if $aq-bp\neq 0$. $\endgroup$ – Mohan Feb 5 '16 at 21:17
  • $\begingroup$ I suppose you're saying the determinant of the matrix is non-zero.. I thought about that as well.. But I couldn't get anywhere.. I'm supposed to move lines to lines, I don't see what I could do with that fact.. $\endgroup$ – Demmas Salim Feb 5 '16 at 21:37

Just a quick thought about the affine case: as you mentioned, the problem really reduces to being able to move two lines through a point $P$ to two other lines through $P$ (just apply a translation first otherwise). Applying another translation if necessary, you can assume $P=(0,0)$. Then the problem is the same as being able to move two vectors to two other prescribed vectors, because the span of a vector uniquely determines a line.

If you're happy to do everything concretely in terms of a basis, start with the two "standard basis vectors" for $\mathbb{A}^2$; then it's easy to write down a matrix $A(x,y)$ which takes this pair to any given pair of vectors $x=(x_1, x_2) ,y=(y_1, y_2)$ - namely, the matrix whose columns are $x$ and $y$ written as column vectors. Then the general way to get from the pair $x,y$ to $z,w$ is the matrix product $A(z,w)A(x,y)^{-1}$.

  • $\begingroup$ Thanks.. I think the solution is valid.. and clean. $\endgroup$ – Demmas Salim Feb 6 '16 at 1:30

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