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It is known that if a single variable polynomial $p$ with degree at most $n$ has at least $n+1$ zeroes then it must be the $0$ polynomial. Is there an easy to use variant of that for multivariate polynomials. Of course in the broadest version, if we can show for a $k$-variate polynomial that for all fixed $k-1$-tuples $x_1,x_2,\ldots x_{k-1}\in\mathbb{C}^{k-1}$ (actually I'm not sure if we would need to consider something broader than the complex numbers?) that $P(x_1,x_2,\ldots x_{k-1},x_k)$ had at least $n+1$ zeroes wrt $x_k$, then $P$ is indeed the $0$ polynomial. This is because we fixed $k-1$ variables and are left with a single variable polynomial in $x_k$.

Let's first look at cases when we are given that $P(x,y)$ has infinite zeroes. I know that having infinite zeroes doesn't necessarily imply it is the zero polynomial because $P(x,y)=x^2+y^2-1$ has infinite zeroes but is not the $0$ polynomial. But maybe an argument on the "dimension" on the known zeroes (e.g. let's say we had a bivariate polynomial $P(x,y)$ that we know was $0$ on the unit disc then since this is an interval of "dimension" $2$, $P(x,y)$ is the $0$ polynomial (at least I think it is since I can't think of a counter example)). This is a bit different as it doesn't really concern the degree of the polynomial, but I think it kind of mirrors how if we know $P(x)$ was $0$ on an interval $[a,b]$ then it must be the $0$ polynomial. I'm not sure if this claim is actually true, but if anyone can prove it, that would be a good start to finding a variant method. Moreover, this would be sufficient for the context I thought of this problem (shared at the bottom).

Now let's instead consider cases when we are only given a discrete set of zeroes (for simplicity I will only consider $P$ as a bivariate polynomial). I think if we were given a lot of zeroes in the unit disk at a lot of radii then I would think we could indeed conclude that $P(x,y)$ was indeed the $0$ polynomial. As for the exact number of zeroes we would need to be given, I'm not sure. But I think if we were given $\left\lfloor \frac{n}{2}\right\rfloor+1$ circles of different radii each with maybe at least $n+1$ zeroes (I'm not really sure about this number) on each circle, then we could conclude that $P(x,y)$ is the zero polynomial (my nonrigorous possibly inaccurate reasoning for this is because one example of a polynomial satisfying this would be $P(x,y)=\prod_{i=1}^{\left\lfloor \frac{n}{2}\right\rfloor+1} (x^2+y^2-r_i^2)$ where $r_i$ is the radii of the circles of zeroes). While this idea might indeed be true, it is not very helpful because it greatly restricts the structure of the zeroes that we are given. I think we might also be able to make arguments for if the zeroes were arranged in a rectangular $(n+1)\times (n+1)$ grid, but the restrictions on the structure are still there.

As for the context of why I brought this up. It came from a HMMT question where you are given the system $$\begin{cases}a^2+ab+b^2=1+i\\b^2+bc+c^2=-2\\c^2+ca+a^2=1\end{cases}$$ and where asked to compute the value of $(ab+bc+ca)^2$. My approach was something like if we instead consider the system $$\begin{cases}a^2+ab+b^2=x^2\\b^2+bc+c^2=y^2\\c^2+ca+a^2=z^2\end{cases}$$ where $x,y,z$ are sides of a triangle (i.e. an element of the certain subset of $\mathbb{R}^3$ that permits this), then we can see that geometrically $a,b,c$ are the lengths of the segments from the fermat point of said triangle to the vertices (I realize now that there is a flaw here in that there exists other solutions for $a,b,c$ to this system where $a,b,c$ cannot geometrically be lengths and hence is assuming that $(ab+bc+ac)^2$ is the same regardless of which solution is chosen for all $x,y,z$). So from this knowledge, we see that $(ab+bc+ca)^2$ is $\frac{16}{3}$ times the square of the area of the triangle, which is $\frac{1}{16}\left(2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4\right)$. Since we are given the values of $x^2,y^2,z^2$ in terms of strictly $a,b,c$, this expression is a polynomial in $a,b,c$, denote it $P_2(a,b,c)$. So this means that $(ab+bc+ac)^2=P_2(a,b,c)$ for values of $x,y,z$ chosen in this particular subset of $\mathbb{R}^3$. And then using the argument I described in the second paragraph of this post, we can conclude it is true for all $(x,y,z)\in\mathbb{C}^3$ and then we finish the problem.

Alan Abraham
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3 Answers3

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As for the context, there is an easier solution, by computing the resultant of each two polynomials associated to the system of equations given in the HMMT question, $$\begin{cases}a^2+ab+b^2=1+i\\b^2+bc+c^2=-2\\c^2+ca+a^2=1\end{cases}$$ We obtain $$ c=\frac{(i+3)b-ia}{3}, $$ and then $$ (ab+bc+ca)^2=- \frac{i^2 + 4i + 12}{3} $$

Dietrich Burde
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Yes, if a polynomial with complex coefficients in $d$ variables is zero on any open ball in $\mathbb{C}^d$, it is zero everywhere.

For two variables: Let the disc's center be $(x_0, y_0)$, $Q(x,y) = P(x+x_0,y+y_0)$ is also a polynomial with the same degree, and $Q$ is zero in the disc centered at the origin with the same radius.

The coefficient of $x^a y^b$ in $Q$ is

$$ a! b! \frac{\partial^{a}}{\partial x^a} \frac{\partial^{b}}{\partial y^b} Q(x,y) $$

evaluated at $(x,y) = (0,0)$. But all these derivatives are zero since the values inside a neighborhood of $(0,0)$ are all zero. So $Q$ is identically zero. This means $P$ is also identically zero.

aschepler
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I'm not sure how useful this might be, but you can treat any polynomial in $\ k\ $ variables as a univariate polynomial in any one of its variables over the field of rational functions in its remaining variables. So, for instance, if $\ p\ $ is a polynomial in $\ k\ $ variables, $\ x_1,x_2,\dots, x_k\ $, of degree $\ n_1\ $ in $\ x_1\ $, and there are $\ n_1+1\ $ distinct rational functions $\ r_1, r_2,\dots, r_{n_1+1}\ $ in $\ k-1\ $ variables such that $$ p\big(r_i(x_2,x_3,\dots,x_{k}),x_2,x_3,\dots,x_k \big) $$ is identically zero for all $\ i=1,2,\dots,n_1+1\ $, then $\ p\ $ must be the zero polynomial.

lonza leggiera
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