Suppose we have two polynomials $p, q \in \mathbb{R}^3$ and we are interested in their simultaneous zeros. Parameter counting tells us that the zero set most probably is going to be a one dimensional curve. But how can I make this statement rigourous? Is there any theorem which gives you the Hausdorff dimension of such a zero set? Indeed, we will have to assume that $p$ and $q$ have no common factors.
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2$\begingroup$ What exactly is a polynomial in $\mathbb{R}^3$? Do you mean in $\mathbb{R} [x_1, x_2, x_3]$? $\endgroup$– Rodrigo de AzevedoCommented Jul 10, 2020 at 21:11
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1$\begingroup$ because you work over the reals, parameter counting will tell you nothing. $\endgroup$– Dima PasechnikCommented Jul 10, 2020 at 22:33
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$\begingroup$ For a postive integer $n$ denote by $P_n$ the vector space of polynomials of degree $\leq n$ in three variables. Sard's theorem shows that if $n$ is sufficiently large, there exists a measure zero subset $Z\subset P_n\times P_n$ such that for $(p,q)\in P_n\times P_n\setminus Z$ the set $\{p=q=0\}$ has dimension $1$ if nonempty. In fact $Z$ itself is semialgebraic. $\endgroup$– Liviu NicolaescuCommented Jul 11, 2020 at 9:40
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1 Answer
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As in my comment denote by $P_n$ the subspace of $\newcommand{\bR}{\mathbb{R}}$ $\bR[x_1,x_2,x_3]$ consisting of polynomials of degree $\leq n$. Consider the map
$$ F: \bR^3\times P_n\times P_n\to\bR^2,\;\;(x,p,q)\mapsto (p(x),q(x)).$$
If $n$ is sufficiently large, then for any $x\in\bR^3$ the map
$$ P_n\times P_n\ni (p,q)\mapsto (p(x),q(x))\in\bR^2 $$
is a submersion. We deduce that the set
$$\Lambda=\{(x,p,q)\in\bR^3\times P_n\times P_n;\;\;p(x)=q(x)=0\} $$
is a submanifold of codimension $2$. We have a natural map
$$\pi:\Lambda\times P_n\times P_n\to P_n\times P_n,\;\;(x,p,q)\mapsto(p,q). $$
Sard's theorem shows that most $(p,q)\in P_n\times P_n$ is a regular value of $\pi$. The fibers of $\pi$ will generically have dimension $1$. Note that the fiber of $\pi$ over $(p,q)$ is the set $\{p=q=0\}$.
answered Jul 11, 2020 at 10:05