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When looking for integral solutions of Diophantine equations there are sometimes trivial solutions. For example, in the Fermat equation $x^n+y^n=z^n$ such a solution is (1,0,1) and in the cubic equation $x^3+y^3=60z^3$ a trivial solution is (1,-1,0).

In a paper these were called "points at infinity" of the corresponding curve. Given a projective variety, is there a general (best algebraic) definition of such points at infinity ?

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up vote 16 down vote accepted

(Sorry if what follows is too elementary and you already knew it! If so, skip to below the horizontal line)

Because the equations are homogeneous, if $(a,b,c)$ is any solution, then so is $(\lambda a,\lambda b,\lambda c)$ for any nonzero constant $\lambda$. Moving to rational solutions, it makes sense then to consider all solutions of the form $(\lambda a,\lambda b,\lambda c)$, $\lambda\neq 0$, to be "equivalent" in a sense (much like with Pythagorean triples, we don't really distinguish between then $(3,4,5)$ triangle and the $(6,8,10)$ triangle when trying to describe all solutions, focusing then on the primitive ones).

This naturally leads you to projective 2-space over the rationals, which is precisely the quotient of the nonzero rational triples, $(\mathbb{Q}^3-\{(0,0,0)\})$, modulo the equivalence relation $\sim$ that is $(a,b,c)\sim(r,s,t)$ if and only if there exists $\lambda\neq 0$ such that $(\lambda a,\lambda b,\lambda c)=(r,s,t)$. (I'm doing this over $\mathbb{Q}$ because we are looking at diophantine problems, but this can be done over any field, and in algebraic geometry you would usually do it over an algebraically closed field, e.g., $\mathbb{C}$).

(An informal way of thinking about projective $2$-space can be found in this answer).

We denote the equivalence class of the point $(a,b,c)$ in Projective $2$-space by $[a:b:c]$.

Projective $2$-space contains copies of the usual ("affine") $2$-space over $\mathbb{Q}$: we can identify $\mathbb{Q}^2$ with all the points of the form $[a:b:1]$. (There are two other copies of the rational plane: the points of the form $[a:1:c]$ and the points of the form $[1:b:c]$). The points that have $c=0$ form the "line at infinity" of projective $2$-space (all representatives of the equivalence class of $[a:b:0]$ will have third coordinate equal to $0$); points on the line at infinity are called "points at infinity".

In the Fermat equation, $x^n + y^n = z^n$, viewed as a projective equation (which we can do because it is homogeneous), the "points at infinity" are the solutions to the equation with $z=0$. In the projective plane, for $n$ odd, the only such point is $[1:-1:0]$. Similarly with $x^3+y^3=60z^3$; these solutions lie on the "line at infinity", so they are "points at infinity" of the curve (which in fact only has one such point).

In the Fermat equation you run into the slight complication that if $n$ is even, then there are no solutions at the "line at infinity" $z=0$, because when $n$ is even the three variables don't play symmetric roles. One can instead look at a different copy of the affine plane, e.g. the $y\neq 0$ subset of the projective plane, which would lead to the line $y=0$ being the "line at infinity", and the only solution that lies at this line at infinity is $[1:0:1]$. If you switch to the $x=0$ line at infinity, you would have $[0:1:1]$ as the "point at infinity".

So given a projective variety, you can pick a copy of the corresponding affine space; the "points at infinity" of the variety will be those that lie in the complement of that copy of the affine space. You can switch viewpoints and select a different affine hyperplane to get a different set of "points at infinity" (that is, the notion of 'points at infinity' depends on the particular affine hyperplane you have decided on, and is not intrinsic to the curve).

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Thanks for your answer. This clarifies what I mattered about. (unfortunately, I can't upvote your answer since I have not enough reputation points) – Ralph Jan 7 '12 at 17:22
@Ralph: Since it is your question, you can accept the answer instead. – Henning Makholm Jan 7 '12 at 17:57
This response helped clarify so many things for me Arturo. Thank you. – Samuel Reid Jan 7 '12 at 19:03

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