Is there an extension for the Fundamental Theorem of Algebra for Two or more variables, such in case of polynomials systems:

$ \begin{cases} f(x, y) = 0 \\ g(x, y) = 0 \end{cases} $

For single-variable polynomials, the Theorem states that nth-degree polynomials implies n complex roots. And about two or more variables, there is such extension? Perhaps a sum of degrees or the max number of degrees of the variables?

  • 4
    $\begingroup$ No there isn't. You can't make such a guarantee. However, if you set any one of the variables to be constant, then again you can apply the theorem. $\endgroup$ Sep 7, 2018 at 18:31

1 Answer 1


This is Bézout's theorem. The number of solutions is either infinite or equal to the product of the degrees if you count them in the right way (some solutions may be at infinity). So you need to look for solutions in the complex projective plane, and some solutions may need to be counted multiple times.

An example that shows why you need the projective plane is $f(z,w)=w$, $g(z,w)=w-1$. Then $\{f=0\}$ and $\{g=0\}$ are parallel complex "lines" that do not intersect in $\mathbb{C}^2$, but the projective plane has "points at infinity" where parallel lines intersect.

  • $\begingroup$ Can you tell more about the "right way" for counting ? $\endgroup$
    – EduardoGM
    Sep 7, 2018 at 19:18
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    $\begingroup$ (I am not an algebraic geometer, so I won't give you the formal definition). Basically you count a point once if the intersection of $N_f=\{f(x,y)=0\}$ and $N_g=\{g(x,y)=0\}$ at that point is transversal, i.e. if the tangents to $N_f$ and $N_g$ span $\mathbb{C}^2$. If the tangents are the same the intersection number is at least $2$. If higher order approximations are the same, you have higher intersection number. Some nice examples are in the Wikipedia article I linked. $\endgroup$
    – Kusma
    Sep 7, 2018 at 19:45

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