I'm reading over Scheidemann's Intro. to Several Complex Variables book and at the beginning on pg. 9 I got a little stuck on proving that the zero set of a complex poly in several variables is not compact. I'm not entirely sure if my proof is correct up until the point I couldn't finish and I can't find it online. I argue by induction:
Base Case ($n=2$): $$P(z,w)=a_m(w)z^m+ \ldots + a_1(w)z + a_0(w)$$ where $a_i(w)\in\mathbb C[w]$. If $a_i(w)\neq 0$ for some $i>0$ then $\exists w_0\in \mathbb C$ arbitrarily large s.t. $a_j(w_0)\neq 0$ for some $j>0$. Then we have: $$P(z,w_0)\in \mathbb C[z]$$ and is non-constant, so by the FTA there is a root $P(z_0,w_0)=0$. But $w_0$ can be chosen arbitrarily large, so the zero set of $P$ is unbounded and thus not compact.
If $a_i(w)=0 \; \forall i>0$, then assuming $P$ itself is non-constant we have that $P$ does not depend on $z$ and $P(z_0,w_0)=0$ for any $z_0\in \mathbb C$ and any $w_0$ a root of $a_0(w)$. Thus again the zero set of $P$ is unbounded.
Inductive Step: Assume $$\mathcal Z (Q(t_1,t_2,\ldots,t_n))$$ is not compact (and thus unbounded by the fact that the zero set is always closed).
We can write $$P(z_1,z_2,\ldots,z_{n+1})=a_m(z_2,\ldots,z_{n+1})z_1^m+\ldots +a_0(z_2,\ldots ,z_{n+1})$$
where $a_i\in\mathbb C[z_2,\ldots,z_{n+1}]\cong \mathbb C[t_1,\ldots,t_n]$
If $a_0$ is not constant, then by assumption there are arbitrarily large $(z_2',z_3',\ldots,z_{n+1}')\in \mathbb C^n$ s.t. $a_0(z_2',\ldots,z_{n+1}')=0$. This implies: $$P(0,z_2',\ldots,z_{n+1}')=0$$ for arbitrarily large $z=(0,z_1',\ldots,z_{n+1}')\in \mathbb C^{n+1}. \blacksquare$
I do not know how to handle the case when $a_0$ is constant. Thanks.
Edit: If my answer is correct and you want to clean up the proof, I'll accept your answer.
If $a_0$ is constant, then $a_j\neq 0$ for some $j>0$. If $a_j$ is constant, then fix an arbitrarily large $\zeta\in \mathbb C^n$ and we have: $$P(z_1,\zeta)$$ is a non-constant polynomial in $z_1$, thus has a root by FTA.
If $a_j$ is non-constant, then $a_j-1$ is non-constant and has unbounded zero set by assumption. Thus there are arbitrarily large $\zeta\in \mathbb C^n$ with $a_j(\zeta)=1$. So again, $$P(z_1,\zeta)$$ is a non-constant polynomial in $z_1$ and thus has a root by FTA.
Either way we have unbounded zeros.
(If this does not complete the argument I will delete or edit it, whatever the community wants)