Rational Solutions to Polynomials, Part II

by t0rajir0u, Mar 30, 2008, 9:25 PM

In the previous post we tried to find rational solutions to (families of) one-variable polynomials, which are a finite set of points. That's easy enough; even without the Rational Root Theorem (which trivializes the problem for any particular polynomial), modular arithmetic (as we have seen) does wonders.

We are doing so within the context of algebraic geometry, which can be defined as the study of algebraic varieties, sets of points defined as the intersection of the solution sets of one or more polynomials. We can talk about affine varieties, which are the ones we're used to, but talking about projective varieties (by embedding things in projective space) turns out to be more useful geometrically and algebraically. Varieties have a notion of dimension; dimension $0$ varieties are collections of points, such as in the previous post. We defined them by a single polynomial in a single variable. (We can also define them by, for example, the intersection of $n$ lines in $n$ variables.)

What happens when we talk about a single polynomial in two variables? Well, we get dimension $1$ varieties, or algebraic curves. For example,
$x^2 + y^2 = 1$
is an algebraic curve that we know as a circle. Of course, any conic section is therefore an algebraic curve. Like the problem of finding rational points among the roots of a polynomial (a variety of dimension $0$ ), we can ask about rational points on algebraic curves, and these points correspond to solutions of Diophantine equations.

Problem 1: Describe the set of rational points on the unit circle $x^2 + y^2 = 1$ .

Method 1: Let $x = \frac {a}{c}, y = \frac {b}{c}$ where $\gcd(a, b, c) = 1$ . Then we are looking at solutions to $a^2 + b^2 = c^2$ ; in other words, primitive Pythagorean triples. It's well-known (but can you prove it?) that the set of primitive Pythagorean triples is generated by
$(a, b, c) = (m^2 - n^2, 2mn, m^2 + n^2)$
which gives $x = \frac {m^2 - n^2}{m^2 + n^2}, y = \frac {2mn}{m^2 + n^2}$ . Letting $t = \frac {n}{m}$ , we can rationally parameterize the set of rational points on the unit circle as
$(x, y) = \left( \frac {1 - t^2}{1 + t^2}, \frac {2t}{1 + t^2} \right)$
for any rational $t$ . We know we have not missed any rational points because we know the complete description of the set of Pythagorean triples. Here information about the solution of a Diophantine equation lets us find the rational points on a curve. But really we have done this backwards.

Method 2: Fix the point $O( - 1, 0)$ and consider the line with slope $t \in \mathbb{Q}$ passing through it. It intersects the circle at another point $P$ . What are the coordinates of $P$ ?

The line in question has equation $y = tx + t$ . Substituting,
$x^2 + t^2 x^2 + 2t^2 x + (t^2 - 1) = 0 \implies$

$x^2 + \frac {2t^2}{1 + t^2} x + \frac {t^2 - 1}{t^2 + 1} = 0.$
This is a quadratic with rational coefficients. Recall that we already know one of its (rational) roots, which is $- 1$ . This tells us that the other root is rational! By Vieta's, we also know that that particular root is $x = \frac {1 - t^2}{1 + t^2}$ . This gives us the same formulas as from Method 1. (The notion of considering a line passing through one rational point and showing that it must pass through a second rational point is rather beautiful, and this isn't the last time you'll see it.)

Here's the fascinating part: given a rational point $P$ on the circle, $PO$ clearly has rational slope, so we have directly proven (without appealing to any facts about Pythagorean triples) a one-to-one correspondence between rational numbers and rational points on the unit circle. The second interesting thing going on here is that Method 2 can be used to prove that our description of Pythagorean triples is correct. In other words, information about rational points on a curve lets us solve a Diophantine equation.

Of course, I have lied a little. We are missing the same point in both descriptions. Note that no rational $t$ maps to the point $( - 1, 0)$ , so our correspondence is not in fact one-to-one!

Fortunately, this is a fixable problem. You can see that as $t \to \infty$ we approach the point $( - 1, 0)$ . We are so close to a one-to-one correspondence that it is tempting to say that $\infty$ maps to $( - 1, 0)$ and be done with it. In fact, geometrically this makes perfect sense: the line with slope $\infty$ is merely the tangent line to $( - 1, 0)$ , which intersects it with multiplicity $2$ . But of course $\infty$ is not a rational number.

What it is, however, is the point at infinity on the rational projective line.

Recall our definition of the rational projective line as the set of equivalence classes of pairs of rationals $(x : y)$ (minus $(0 : 0)$ ) under the equivalence relation $(x : y) \sim (\lambda x : \lambda y), \lambda \in \mathbb{Q}$ . The rational points we have been talking about now can be described as the points $(t : 1)$ (an affine slice), and the point at infinity is then the point $(1 : 0)$ . (Note that we could have chosen to divide out by $x$ instead; affine slices are not unique.) One way we can interpret the projective line is to think about the point $(x : y)$ as the line passing through $(0, 0)$ and $(x, y)$ , defined by its slope; then the point at infinity is the line with infinite slope as we saw earlier.

If we're talking about projective coordinates, we're talking about homogeneous polynomials. How does a two-variable polynomial homogenize?

Definition: The homogenization of a polynomial $f(x, y)$ of degree $n$ is the homogeneous polynomial $f^{*}(X : Y : Z) = Z^n f \left( \frac {X}{Z}, \frac {Y}{Z} \right)$ .

Since the unit circle is the zero set of the polynomial $f(x, y) = x^2 + y^2 - 1$ , its homogenization is given by
$X^2 + Y^2 = Z^2,$
precisely the Pythagorean triple equation. The points on this unit circle (which is a projective variety, not an affine one) are points in the projective plane $P^2$ , denoted by coordinates $(X : Y : Z)$ under the same equivalence relation from before. The rational points on the unit circle are points in the rational projective plane $\mathbb{Q}P^2$ .

Let's take a moment to describe the projective plane. When $Z \neq 0$ we can divide out to write our points as $(x : y : 1)$ , which is essentially a copy of the affine plane (the plane we were working in just now) When $Z = 0$ , we instead write $(X : Y : 0)$ , which is a copy of the projective line called the line at infinity.

The line at infinity does not enter into our discussions because when $Z = 0$ we have
$X^2 + Y^2 = 0$
and so $X = Y = 0$ , which we have said is not a point. The unit circle therefore contains no (real) points at infinity.

I'd like to take a moment to discuss how the notion of points at infinity gives a concrete algebraic reality to intuitive geometric notions. Consider the parabola $y = x^2$ , whose homogenization is
$YZ = X^2$
.
Letting $Z = 0$ we have $X = 0$ and the point at infinity $(0 : 1 : 0)$ . In other words, this parabola does contain a point at infinity that its affine slice does not. If we think of a parabola as an ellipse that's been "stretched" so that its second focus is at infinity, we can consider this point at infinity as the point at which the two sides of the parabola meet.

Here's another example. Remember that affine slices are not unique. If we divide out by $X$ instead, we get the hyperbola $yz = 1$ . The points at infinity here (with respect to this new affine slice) are the points where $X = 0$ , which are the points $(0 : 1 : 0)$ and $(0 : 0 : 1)$ . Thinking about the graph of the affine hyperbola, we can think of these points at infinity as the points at which the branches of the hyperbola meet.

Note that we have also demonstrated that a projective parabola and a projective hyperbola are in some sense the same object; they are merely affine shadows cast by the same projective curve. In fact, this viewpoint generalizes; we find in general that there are fewer different types of curves in projective than in affine space.

Our goal was to "repair" the near-bijectivity of our mapping from the rationals to the rational points on the unit circle. Let's call this map $\phi$ , the projective unit circle $C$ , and its set of rational points $C(\mathbb{Q})$ .

$\phi : \mathbb{Q}P^1 \to C(\mathbb{Q})$ takes the point $(A : B)$ on the rational projective line to the point
$(A^2 - B^2 : 2AB : A^2 + B^2)$
on the rational projective plane (and here we really see how this gives us information about Pythagorean triples). Moreover, this map is one-to-one. If our geometric argument wasn't enough, the inverse map $\varphi : C(\mathbb{Q}) \to \mathbb{Q}P^1$ takes $(X : Y : Z)$ to $(Y : Z - X)$ .

Let's take a moment and think about what kind of a function this is. You are used to functions from the real numbers to the real numbers, which are functions from the affine real line to the affine real line. Here we have a function from a projective variety to another projective variety. In particular, every point on one maps to a unique point on the other; it is a bijection from one variety to the other. Moreover, the bijection has the property that rational points map to rational points. This bijection allows us to compute Pythagorean triples by picking an arbitrary point on the rational projective line and translating it to a point on $C(\mathbb{Q})$ , which in turn lets us describe the set of Pythagorean triples. In the language of algebraic geometry, $C$ is isomorphic to the projective line over $\mathbb{Q}$ , and the above map is an isomorphism of varieties defined over $\mathbb{Q}$ . What this means, for our purposes, is that $C(\mathbb{Q})$ is very easily describable.

Here's where it starts to get interesting: the argument we gave in Method 2 is valid for any conic section with rational coefficients. All we needed was a single rational point. We therefore have the following remarkable fact:

Every rational conic section with a rational point is isomorphic to the projective line over $\mathbb{Q}$ .

We can now solve a pretty broad class of Diophantine equations.

Note that what we've been doing corresponds to solving homogeneous Diophantine equations in three variables such as the Pythagorean triple equation $X^2 + Y^2 = Z^2$ . We cannot use the techniques we have developed to find integer points on a Diophantine equation in two variables such as $x^2 - 3y^2 = 1$ , but that's what Pell's equations are for.

This material is inspired in part by the first chapter of Tate and Silverman's Rational Points on Elliptic Curves, which is a great introduction to the theory if you want to have a go at it yourself. It is also inspired by several problems I've seen on the forums. I've tried to phrase this discussion within the context of competition-style problems so that the "applications" (such as they are) are clearer, but realize that these ideas have very broad scope.

So much for conic sections. Because they're all isomorphic to the projective line, we won't be looking at them further.

Problem 2: (From the previous post) Describe the set of rational points on the curve $x^3 + y^3 = x^2 + y^2$ .

I solved this problem in a comment on the previous post, but here is a more condensed version of that solution that uses the concepts we have discussed above:

Call the curve $E$ and homogenize to $E : X^3 + Y^3 = Z(X^2 + Y^2)$ . Then $Z = \frac {X^3 + Y^3}{X^2 + Y^2}$ and our points take the form
$(X(X^2 + Y^2) : Y(X^2 + Y^2) : X^3 + Y^3).$
This is a morphism from the projective line to $E$ whose inverse map takes $(X : Y : Z)$ to $(X : Y)$ , and it also takes rational points to rational points. We therefore conclude that $E$ is also isomorphic to the projective line over $\mathbb{Q}$ .

It turns out that this isn't a result you should expect; most cubic curves are not isomorphic to the projective line, and are instead elliptic curves, whose sets of rational points have much more interesting structure (and let us solve cubic Diophantines). Cubic curves that are isomorphic to the projective line are singular (or not smooth). $E$ , for example, has a singularity at $(0 : 0 : 1)$ . Singularities include points like cusps, and are essentially points at which the curve behaves "badly."

How can you tell if a cubic curve is singular (and therefore whether you should look for isomorphisms from the rational projective line)? A singularity on the curve $f(X : Y : Z) = 0$ is a point at which the partial derivatives of $f(X : Y : Z)$ with respect to $X, Y, Z$ simultaneously vanish. Now that's fascinating: we are using calculus to help us discern the nature of solutions to a Diophantine equation. This is one of the several deep connections you'll find all throughout algebraic geometry.

What connects a conic section to a singular cubic curve is that they both have genus zero. Don't worry about what this means; suffice it to say that the first really interesting case is genus one, which is an elliptic curve.

Practice Problem 1: Describe the set of rational points on the hyperbola $x^2 - 3y^2 = 1$ .

Practice Problem 2a: Describe the set of rational points on the singular cubic curve $y^2 = x^3$ . (No, this is not a trick question!) Where is its singularity? Graph it.

Practice Problem 2b: Using the results of the previous problem, describe the set of rational points on the singular cubic curve $y^2 = (x - \alpha)^3, \alpha \in \mathbb{Q}$ .

Practice Problem 3a: Describe the set of rational points on the singular cubic curve $y^2 = x^3 - x^2$ . Where is its singularity? Graph it.

Practice Problem 3b: Using the results of the previous problem, describe the set of rational points on the singular cubic curve $y^2 = (x - \alpha)(x - \beta)^2, \alpha, \beta \in \mathbb{Q}$ .

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3 Comments

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Hmm I don't know I think that my formula is pretty close to yours...b/b+2 instead of b or something. Maybe I messed up somewhere.

Anyway, so I'm familiar with that rational number parameterization (I think i posted that some time). I was thinking of a more geometric bijection so like lets say that you have the point (-1,0), and the circle x^2+y^2=1, and x=1. Then we are equivalently doing a central projection from points on that line with a center of (-1,0) onto the circle. (This is basically taking a slope).

So I guess the thing that I noticed is that you can approach the (-1,0) from positive or negative infinity. So how does this work with the projective line theory? In a not very rigorous sense, the positive and negative infinity are both the same?

Also like if you take a unit sphere at the origin, (-1,0,0), and the plane x=1, you can basically do the same thing, however, now you have a 'circle' of points at infinity.

I don't know what my point is, I just think it is an intersting discussion.

by Altheman, Mar 30, 2008, 10:20 PM

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$\mathbb{R}\mathbb{P}^1$ is topologically the circle $S^1$ so approaching from either direction is the same

in general, $\mathbb{P}^n = \mathbb{A}^n + \mathbb{P}^{n-1}$ so a projective space is a normal space plus a projective space of one lower dimension at infinity

combining these two shows that in 2-dimensions, you do indeed have a $\mathbb{P}^1$ , i.e. a circle, of points at infinity

by MysticTerminator, Mar 30, 2008, 11:00 PM

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Indeed. Another example: the complex projective line, interpreted as a surface, is the Riemann sphere (when using words like "line" and "plane" in dealing with the complex numbers, though, we have to be careful to distinguish real and complex dimension!).

by t0rajir0u, Mar 30, 2008, 11:08 PM

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Rational Solutions to Polynomials, Part II

by t0rajir0u, Mar 30, 2008, 9:25 PM

3 Comments