Why are the number of roots of the characteristic equation of a PDE infinite? My teacher was doing analogies between ODEs and PDEs. At some moment he talked about ODE characteristic equations, which is simply a polynomial for which we have some roots. With these roots, I can find solutions for the ODE. 
For PDE's, it's almost the same thing, but we get a characteristic equation in the form
$$\sum_{|\alpha| \le m}a_{\alpha}\zeta^{\alpha} = 0$$
for $\zeta = (\zeta_1, \cdots, \zeta_N)\in \mathbb C^N$.
And it says that if $N\ge 2$, then the set of solutions of the chracteristic equation is infinite.
Here, $\alpha$ is a multi-index element.  
As an example of a characteristic equation, we could have, I guess:
$$5\zeta_1\zeta_2 + 7\zeta_3\zeta_4\zeta_2 + \zeta_3  + 9 = 0$$
And we need to find the roots for these equations. Why would they be infinite? Could somebody give me an intuition?
Also, could somebody link me somethig that talks about the characteristic equation for a PDE? The only thing I can find is about the method of characteristic for PDE, which is something completely different.
UPDATE:
I'm trying to formalize a proof that these roots are infinite (and more: uncountably infinite).
We have that $$\sum_{|\alpha| \le m}a_{\alpha}\zeta^{\alpha} = 0$$
so for some multiindex $\alpha_0$ and $\alpha_1$ such that $a_{\alpha_0}$ and $a_{\alpha_1}$ are not $0$. We can do $$\zeta^{\alpha_0} = - \frac{1}{a_{\alpha_0}}\sum_{|\alpha| \le m, \alpha\neq\alpha_0}a_{\alpha}\zeta^{\alpha}$$
then set all $\zeta^\alpha=0$ for $\zeta^{\alpha}\neq\zeta^{\alpha_0}$ and $\zeta^{\alpha}\neq\zeta^{\alpha_1}$ so we have:
$$\zeta^{\alpha_0} = \frac{a_{\alpha_1}}{a_{\alpha_0}}\zeta^{\alpha_1} \tag{1}$$
or simply 
$$a_{\alpha_0}\zeta^{\alpha_0} + a_{\alpha_1}\zeta^{\alpha_1} = 0. \tag{1}$$
Now in (1) I have a multiplication of several $\zeta$ variables on the left, and also on the right. I can choose all of them to be $1$ except for two of them, to end with something like this:
$$\zeta_i = \frac{a_{\alpha_1}}{a_{\alpha_0}}\zeta_k$$
And thus I can choose $\zeta_k$ freely over the reals (which are uncountable) to find $\zeta_i$ that will solve the equation with the root $\zeta = (1,\cdots,0,\cdots,z_i,\cdots,z_k,\cdots,1)$.
Is this a valid proof?
 A: In case of ODE you have one equation and one unknown. If the equation is linear there is only one root. If the equation is not linear they are a finite number of roots.
In case of PDE, there is still one equation but several variables. Thus they are an infinity of roots. In other words the roots are not independent. If they are $n$ variables, one variable is function of the $n-1$ other variables.
For example in case of :
$$5\zeta_1\zeta_2 + 7\zeta_3\zeta_4\zeta_2 + \zeta_3  + 9 = 0$$
$\zeta_3$ is function if the other variables :
$$\zeta_3=-\frac{5\zeta_1\zeta_2  + 9}{1+7\zeta_4\zeta_2}$$
You have as many solutions that $\zeta_1,\zeta_2,\zeta_4$ are arbitrary defined : So, an infinity of solutions.
Of course this is until no more equation is specified. In general, the PDE is associated to some conditions (boundary, initial,...) which can be expressed as complementary equations.
A: Proof of uncountability:
Assume WLOG that the $n$-variable polynomial $\sum_{\lvert \alpha \rvert \le m} a_\alpha \zeta^\alpha \in \mathbb C[\zeta_1, \dotsc, \zeta_n]$ has degree at least one in $\zeta_1$, and consider it as a polynomial $\sum_{k=0}^m P_k(\zeta_2, \dotsc, \zeta_n) \zeta_1^k \in (\mathbb C[\zeta_2, \dotsc, \zeta_n])[\zeta_1]$. The set $$A = \bigl\{(\zeta_2, \dotsc, \zeta_n) \in \mathbb C^{n-1} \mid \text{there exists at least one $k \geq 1$ with $P_k(\zeta_2, \dotsc, \zeta_n) \neq 0$}\bigr\}$$ cannot be countable, because otherwise by continuity the $P_k$ would be identically zero for all $k \geq 1$. For each $(\zeta_2, \dotsc, \zeta_n) \in A$, the fundamental theorem of algebra guarantees the existence of at least one $\zeta_1 \in \mathbb C$ such that $\sum_{\lvert \alpha \rvert \le m} a_\alpha \zeta^\alpha = 0$. ∎

As for the attempted proof in the edit to the question, the error is that you essentially assumed that


*

*for each $\alpha \notin \{ \alpha_0, \alpha_1 \}$, either $a_\alpha = 0$, or there exists some $i \in \{1, \dotsc, n\}$ such that $\alpha_i > 0$ and $(\alpha_0)_i = (\alpha_1)_i = 0$ (by setting $\zeta^\alpha = 0$);

*for each $j \in \{0, 1\}$ and $i \in \{1, \dotsc, n\}$, either $(\alpha_j)_i = 0$ or $(\alpha_j)_i = 1$.


These assumptions do not hold for arbitrary polynomials: For example, the PDE $\partial_1^5 u + \partial_1^2 \partial_2^3 u + \partial_2^6 u - 3u = 0$ has the characteristic equation $\zeta_1^5 + \zeta_1^2 \zeta_2^3 + \zeta_2^6 - 3 = 0$. As a matter of fact, we can't find a general formula for $\zeta_i$ using only the other components of $\zeta$, constants, and elementary algebraic operations, due to Abel's impossibility theorem and Galois theory. So we have to resort to an abstract existence theorem like the FTA.
