# Number of solutions for a system of polynomial equations

Consider the given system of polynomial equations, where all the coefficients are in $\mathbb{C}$: $$\begin{cases} y^n=P(x)\\ Q(x,y)=0\end{cases}$$ I would like to establish that either this system has solutions $(x,y)\in\mathbb{C}$ for all $x \in \mathbb{C}$, or either has solutions for countably (or even better, finitely) many $x\in\mathbb{C}$. I am not completely sure this is true but haven't found any counterexamples yet.

So far, I have found that if we have solutions for infinitely many $x$, then we have solutions for infinitely many $y$.

Any ideas?

You can determine the resultant $R(x)=Res_y(y^n-P(x),Q(x,y))$ of both polynomials eliminating $y$. If it is the zero polynomial, then the polynomials have a common factor, which means that for any $x$ you find at least one $y$ so that $(x,y)$ a solution (of the common factor and thus of the system).

If the resultant is not zero, then $R(x)$ is a polynomial. Only at the roots of this polynomial you will then find at least one $y$ so that $(x,y)$ is a solution of the system.

I misunderstood the question and will edit this later with (hopefully) a response.

Given any $x \in \mathbb{C}$, there exists $n$ possible $y \in \mathbb{C}$ such that $y^n = P(x)$ (possibly with multiplicity, if say $P(x) = 0$). However, determining whether the system has a solution (and how many) depends completely on the structure of $Q(x,y)$. Consider the examples below.

If $Q(x,y) = y^n - P(x)$, then any pair $(x,y)$ from above satisfies the system, so we get uncountably many solutions.

If $Q(x,y) = y^n - P(x) +1$, then the first relation forces $y^n - P(x) = 0$, so that $Q(x,y) = 1$ for all $(x,y)$ satisfying the first. Hence the system has no solutions at all in this case.

• Yes, that's precisely my point: I want to establish the duality between "for any $x\in\mathbb{C}$, there is some solution $(x,y)$" and "there are only countably many $x$, such that there is some solution $(x,y)$". You gave examples of these two cases. Can you show there are no other cases? Mar 31 '14 at 19:02
• Oh, I misunderstood. I will edit my response. Mar 31 '14 at 19:08