Roots of rational equation with multiple variables?

Question

Let's say we have a rational polynomial in $k$ variables. We are only interested in rational solutions. If $k = 1$, name the variables ${x}$, if $k = 2$, name them ${x,y}$.

For $k = 1$, it can be done very fast. The Rational Root Theorem gives a set of candidates. But what for $k=2$? How can I split a polynomial into factors in this case?

Example: $x^2-y^2$ should be split to $(x-y)(x+y)$.

Post also available on https://stackoverflow.com/questions/11922956/roots-of-rational-equation-with-multiple-variables.

Answer 1

Polynomials in more than one variable do not generally split into factors, even if you allow complex coefficients. For example, $x^2 + y^2 - 1$ doesn't split in this way.

For quadratic polynomials you can appeal to the Hasse-Minkowski theorem, the Chevalley-Warning theorem, and Hensel's lemma to determine when a solution exists; this argument is described in more detail in the beginning of Cassels' Lectures on Elliptic Curves.

Beyond the quadratic case, this problem is open. Already for cubic polynomials in two variables it is not known whether there exists an algorithm which provably solves this problem, although there appear to be algorithms which work reasonably well in practice. Bjorn Poonen's Computing Rational Points on Curves contains a good discussion of the issues involved. See also, for example, this MO question.

Note that Fermat's Last Theorem can be phrased as the problem of finding rational points on the family of Fermat curves $x^n + y^n = 1$, so there's no reason to expect that this is an easy problem if you believe that Fermat's Last Theorem is difficult.

Answer 2

There is quite pretty geometric proof that generic quadric polynomial over algebraically closed field (this work more generally for any degree $d>1$ and number of variables $n > 2$, but I will stick to quadric in $x, y, z$ for simplicity) $$ ax^2 + bxy + cxz + dy^2 + eyz + fz^2 $$ has no general solution expressible as rational function of coefficient $x, y, z \in \mathbb k(a, b, c, d, e, f)$. To see that consider the incidence variety $$ \Phi = \{([x: y: z], [a:b:c:d:e:f]) \in \mathbb P^2 \times \mathbb P^5 \ |\ ax^2 + bxy + cxz + dy^2 + eyz + fz^2 = 0\} $$ The projection $\Phi \rightarrow \mathbb P^2$ realise $\Phi$ as projective bundle with fibers $E_p$ being the space of conics containing point $p$. It means that $\Phi$ is smooth. The other projection is also proper submersion restricted to the locus of smooth conics with fibers over $F$ isomorphic to smooth conics $V(F) \simeq \mathbb P^1$. This is an example of non-trivial Severi-Brauer variety, fiber bundle with fibers $\mathbb P^n$ that is not a projective bundle. The fastest way to see that uses projective bundle formula (applied to the first projection) to compute the Picard/Chow/cohomology groups, in codimension $1$ generated by pullbacks of hyperplane sections from $\mathbb P^2$ and $\mathbb P^5$. As none of them intersects fiber of $\Phi \rightarrow \mathbb P^5$ in class of a point, this map cannot have a section (you can proof this fact also without invoking homological methods, it is done for example in The geometry of schemes of Eisenbud and Harris)

Finally it remains to observe that the existence of universal solution is equivalent to finding a rational section of this map, which is a contradiction.

In context of your question, what it means is that there never exist an "easy and universal" solution to this problem, no matter of the field. On the other hand over $\mathbb C$ you can find a holomorphic solution locally in analytic topology (or generally in etale topology over different fields) due to Ehresmann's lemma or its algebraic analogue.