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Question. Why a polynomial on $\mathbb{C}^n$ with $n>1$ has infinitely many distinct zeros?

I don't see this. By example, in the case $n=2$ with $P(D)\exp{i\xi\cdot x}=\partial_{x_2}^{2}\partial_{x_1}\exp{i(\xi_1x_1+\xi_2x_2)}=-\xi_1\xi_2^2\exp{i\xi\cdot x}$ then $p(\xi=(\xi_1,\xi_2))=-\xi_1\xi_2^{2}$, but, if $p(\xi)=0$, then $\xi_1=0$ or $\xi_2=0$ (only two roots).Where are the other infinitely many roots?

eraldcoil
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    $\xi_1\xi_2^{2}=0$ when $(\xi_1,\xi_2)$ is of the form $(0,c)$ or $(c,0)$ with $c$ arbitrary. – geetha290krm May 04 '23 at 05:58
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    A polynomial on $\mathbb R^2$ defines a locus in the plane, which can have infinitely many points - think for example $y=x$ for the main bisector, or $x^2+y^2=1$ for the unit circle. Replace $\mathbb R^2$ with $\mathbb C^2$ and "can have" becomes "always has". – dxiv May 04 '23 at 06:14
  • Because its restriction to every line has at least one root, and because there are infinity of lines.. – Thomas May 04 '23 at 06:16
  • Thanks. Now, i see my mistake. $\xi_2\xi_2^2=0$ then $\xi_1=0$ or $\xi_2=0$, but, $\xi=(\xi_1,\xi_2)=(0,c)$ or $(c,0)$. – eraldcoil May 04 '23 at 22:04

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The polynomial $p(\xi_1, \xi_2) = -\xi_1\xi_2^2$ has infinitely many roots, because either $\xi_1$ or $\xi_2$ can be $0$, there is no requirement that both are $0$. So, the set of roots of $p$ is $$\{(\xi_1, \xi_2)|\xi_1=0\lor \xi_2=0\} =(\mathbb C\times \{0\})\cup (\{0\}\times \mathbb C)$$

which is a union of two infinite sets, and therefore infinite.

In general, a complex polynomial in $n$ variables has infinitely many roots, assuming the following:

  • $n > 1$
  • The degree of the polynomial is not $0$ (i.e. the polynomial is not constant)

You can see that for yourself if you think of the polynomial as a parametrized polynomial of one variable.

Let's stick to $n=2$ for now. Take a polynomial $p(\xi_1,\xi_2)$. We can assume WLOG that the polynomial is not constant in $\xi_1$. Now, define $p_{\xi_2}$ to be the polynomial of one variable, defined as $p_{\xi_2}(z)=p(z, \xi_2)$.

For all but a finite selection of $\xi_2$, the polynomial $p_{\xi_2}$ is now a complex polynomial, and as such it has a root. In other words, $p_0$ has a root, $p_1$ has a root, $p_i$ has a root and so on. So, let's call $z_{\xi_2}$ the root of $p_{\xi_2}$. Then the set $$\{(z_{\xi_2}, \xi_2)|\xi_2\in\mathbb Z\}$$ is both infinite and a subset of the set of all roots of $p$.

If $n>2$, then you can start with a polynomial $q:\mathbb C^n\to \mathbb C$ and first define the polynomial $p:\mathbb C^2\to \mathbb C$ as $p(\xi_1,\xi_2)=q(\xi_1,\xi_2,0,\dots,0)$ and use the same argument as above (assuming, again WLOG, that $\xi_1$ does not have degree $0$ in $q$).

5xum
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