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Let $\Bbbk\in \left\{ \mathbb R,\mathbb C \right\}$. Suppose $\mathbb \Bbbk^n\overset{f}{\to} \Bbbk$ is a homogeneous polynomial map satisfying the following condition: the fiber of $f$ containing the origin is equal to the origin.

Question 1. Is $f$ proper?

Intuition. Let $n=2$. Suppose we localize around the singular fiber over zero. Upon deleting it was obtain at least locally on the source a submersion, which forces the fibers of $f$ to foliate the domain. If the fiber containing the origin is just the origin, then the remaining fibers must foliate the plane minus the origin. It feels like this should make them some sort of loops about the origin, and their alignment should make the whole map proper.

I'm staring at $ax^2+by^2$ vs $ax^2-by^2,xy$ etc and this seems to be the phenomenon. In fact, the same thing seems to happen for $x^4+y^4-xy$, so it seems only the top degree homogeneous summand of a polynomial determines this behavior. This motivates:

Question 2. Suppose $f$ is an arbitrary polynomial mapping whose top degree homogeneous summand satisfies the above condition. Is $f$ proper?

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  • $\begingroup$ For $\mathbb{k}=\mathbb{C}$, note that it is algebraically closed, so $\dim f^{-1}(0)=0$ only if $n\leq 1$, and hence $f$ is proper. $\endgroup$ – user10354138 Jun 10 '19 at 15:01
  • $\begingroup$ @user10354138 I take it you mean the Krull dimension? In that case your comment answers a part of my first question, but it does not (I think) explain why $(x,y)\mapsto x^4+y^4-xy$ is proper. $\endgroup$ – Arrow Jun 11 '19 at 6:18
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The answer to both questions is yes. No differential geometry is needed.

One can use the "boundedness criterion" for properness $\|x\|\to \infty\implies \|fx\| \to\infty$. The following proof works for homogeneous polynomial mappings $\Bbbk^n\to \Bbbk ^\ell$.

If $0\neq x\in f^{-1}(0)$ then by homogeneity $\mathbb Rx\subset f^{-1}(0)$. Since the line $\mathbb Rx$ is unbounded, $f$ is not proper. This shows that if $f$ is proper we must have $f^{-1}(0)= \left\{0\right\}$.

Conversely, suppose $f^{-1}(0)= \left\{0\right\}$. Then $0\notin f(\mathbb S^1_\Bbbk)$ whence by positive-definiteness of norms $0\notin \|f(\mathbb S^1_\Bbbk)\|$. Since $\mathbb S_\Bbbk ^1$ is compact, continuity of $f,\|\cdot\|$ implies $\|f(\mathbb S^1_\Bbbk)\|\subset [0,\infty)$ is compact as well. Since $[0,\infty)$ is Hausdorff, this subset is therefore separated from the origin. This means $\|f(\frac{x}{\|x\|}) \|\geq c>0$ for some $c>0$. By homogeneity $f(x)=f(\|x\|\frac{x}{\|x\|})=\|x\|^\ell f(\frac{x}{\|x\|})\geq \|x\|^\ell c$. This tends to infinity as $x\to \infty$, so $f$ is proper.

Thus the answer to the first question is yes.

Since the top degree homogeneous summand of $f$ dominates its norm, the answer to the second question is also yes.

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