# Is this true for a system of polynomial equations as well?

For a system of linear equations with $$n$$ unknowns, if you have as many linearly independent equations as there are unknowns, you can find the only solution that satisfies that system, right? But does this still hold if you have a set of linearly independent polynomial equations in some number $$n$$ unknowns of some degree $$d$$ instead of just linear equations?

In other questions the example $$x^2 + y^2 = 0$$ is given where you only need one equation to solve for both unknowns, but are there cases where you would need more independent equations than unknowns to get to a unique solution?

• Best to take a step back. Consider the three sets of linear eqn-s: [1] $x + y = 6$ and $x - y = 2$ [2] $x + y = 6$ and $2x + 2y = 12$ [3] $x + y = 6$ and $2x + 2y = 11$. Commented Jan 10, 2021 at 3:29
• But in both [2] and [3], the equations aren't linearly independent, right? Commented Jan 10, 2021 at 3:41
• What about $x^2+y^2=2$ and $y=x$? Two equations in two variables, but the solution isn't unique. Commented Jan 10, 2021 at 3:50
• Yes, you're right. In fact, I overlooked your linearly independent constraint. Best to leave my (previous) flawed comment in place as example that constraint required. Commented Jan 10, 2021 at 4:04

This is subtle. It is not at all clear a priori what the correct generalization of "linear independence" is. This is because polynomials can nontrivially divide each other; for example, the system in two variables

$$x - y = 0$$ $$x^2 - y^2 = 0$$

has infinitely many solutions $$\{ (x, y) : x = y \}$$, and the first equation divides the second but they are linearly independent. As a more complicated example, consider the system in three variables

$$x = 0$$ $$y = 0$$ $$xz - yz = 0$$

which still has infinitely many solutions $$\{ (x, y, z) : x = y = 0 \}$$ ($$z$$ is arbitrary). This time none of the equations divide each other but the third one is a linear combination of the first two with polynomial coefficients.

From the modern perspective, coming from commutative algebra and/or algebraic geometry, we can understand a system of polynomial equations by understanding the ideal they generate in the ring of polynomials, and one plausible guess as to the appropriate generalization of "linear independence" is most clearly stated in this language: it's that if we have $$n$$ polynomials $$f_1, \dots f_n$$ then the ideal $$I = (f_1, \dots f_n)$$ they generate cannot be generated by fewer than $$n$$ polynomials. This is related to the notion of a complete intersection except I have only seen complete intersections discussed in projective space. There is also the related notion of a regular sequence which I unfortunately know very little about.

This is still not enough to guarantee a unique solution. Generically one now expects $$(\deg f_1) (\deg f_2) \dots (\deg f_n)$$ solutions, at least over $$\mathbb{C}$$. A very simple example is

$$x - y = 0$$ $$x - y^2 = 0$$

which is equivalent to $$y = y^2$$ and hence which has two solutions $$(1, 1)$$ and $$(0, 0)$$. For more complicated examples see the Wikipedia article on Bezout's theorem: there are nice geometric examples coming from intersections of conic sections.

I don't know off the top of my head if "the ideal is not generated by fewer than $$n$$ polynomials" is enough to guarantee finitely many solutions. This is probably extremely classical and I hope someone else comes along who does know.

Another plausible generalization of "linearly independent" is that the Jacobian of $$f_1, \dots f_n$$ is generically invertible; equivalently, the differentials $$df_1, \dots df_n$$ are generically linearly independent. ("Invertible everywhere" is a quite strong condition.) We should be able to conclude something using the inverse function theorem from this but unfortunately my command of the subject is not quite strong enough to tell immediately what.

• Thank you for the detailed explanation and useful links! I'll have to seriously crack my head over this one for sure. About the last example where you talk about the degrees of the polynomials in the system, that would be the sum of the highest powers of each variable in the polynomials right? Commented Jan 10, 2021 at 12:20
• @JansthcirlU: depending on what you mean by that, yes. I mean the degree of a monomial $\prod x_i^{e_i}$ is $\sum e_i$ and the degree of $f$ is the maximum degree of a monomial in it (with a nonzero coefficient). You can get a sense of why the product of the degrees is the "expected" answer (again, only "generically") by considering systems of the form $x_i - f_i(x_{i+1}) = 0$ where $f_i$ is a polynomial in one variable, or alternately polynomials which are products of linear polynomials. Commented Jan 10, 2021 at 20:12