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Suppose $P(x,y)$ is a polynomial with real coefficients. Is it true that any solution $(x_0,y_0)$ of the system $P(x,y)=P(y,x)=0$ has the property that $y_0 = \overline{x_0}$ (i.e. they are conjugate), provided that the number of solutions is finite?

Notice that if $(x,y)$ solves the equation, then $(\overline{x},\overline{y})$ also solves the system, and hence $(\overline{y},\overline{x})$ does too. Thus, the set of solutions is symmetric w.r.t both reversing order of coordinates and conjugation of both coordinates.

A nice proof would be to just count the number of solutions of the form $(x,\overline{x})$, and just use Bezouts theorem to see that there can't be any more, but I have no success proving that there are sufficiently many of that form.

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up vote 5 down vote accepted

If you pick two numbers $z_1, z_2$, with different minimal polynomials $Q_1(x), Q_2(x)$, then you can build a real polynomial P having $(z_1,z_2)$ and $(z_2,z_1)$ as roots by setting $P(X,Y) = Q_1(X)Q_1(Y) + Q_2(X)Q_2(Y)(X-Y)$

The $(X-Y)$ factor ensures that $P(X,Y)$ and $P(Y,X)$ have no common factor : $P(X,Y) - P(Y,X) = 2Q_2(X)Q_2(Y)(X-Y)$, and none of those is a factor of $P$.

The solutions of the system $P(X,Y) = P(Y,X) = 0$ are the pairs of conjugates $(\sigma z_1, \tau z_2)$, and the $(\sigma z_1,\sigma z_1)$ (modifying the $(X-Y)$ factor into something else changes these solutions)

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You are correct, I should've tried more than a few polynomials before stating the problem. Seems what I am looking at is a sub-set of polynomials, which has the above property. – Per Alexandersson Oct 11 '11 at 15:13

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