I've been trying to read through this number theory paper by Bombieri and Pila, and I'm struggling with seeing why the following facts hold (used in their proof of Theorem 1):

Let $f(x)$ be a real analytic function defined on the interval $[0, 1]$ and suppose $f(x)$ is not algebraic.

  1. The graph of $f$ will intersect any particular algebraic curve only finitely many times.
  2. If we fix an integer $d$, then there is some finite constant $\gamma(f, d)$ such that $f$ will intersect any algebraic curve of degree $d$ at most $\gamma(f, d)$ times.

I have not studied algebraic geometry so any references or elementary explanations would be appreciated. I suspect the first result holds as the result of the identity theorem for real analytic functions, but I am struggling to see how to handle the cases where the algebraic curve is multivalued or non-smooth. As for the second fact, the authors of the paper mentioned that this holds as a consequence of the fact that the space of algebraic curves is compact, but I don't see how this immediately follows.

  • $\begingroup$ It does not seem easy with the method indicated by Bombieri to produce any explicit bound, even for a given function like $f(t) = \exp t$. Worth thinking about. $\endgroup$
    – orangeskid
    Apr 1 at 20:56

2 Answers 2


Your guess that 1) holds due to the identity theorem is correct. Let $p(x,y)$ be the polynomial cutting out an algebraic curve $C$. Then $$C\cap \Gamma_f= \{(x,f(x))\mid p(x,f(x))=0\},$$ where $\Gamma_f=\{(x,f(x))\mid x\in[0,1]\}$ is the graph. The set $\{x\in[0,1]\mid p(x,f(x))=0\}$ cannot have an accumulation point: if it did, then we would have $p(x,f(x))=0$ identically on $[0,1]$ by the identity theorem, implying that $f(x)$ is algebraic. But any infinite subset of $[0:1]$ has an accumulation point, so we can only have finitely many intersections.

For 2), in order to exploit compactness one should find some way to cover the space of algebraic curves by open subsets where the number of intersections is bounded on each open subset. Then by compactness, finitely many of these sets suffice to cover, and you can pick the largest bound among these finitely many open sets.

One does have to exercise a little care in proving this - you'd love to say that the open sets are just "curves where there are at most $e$ intersections" as $e$ varies, but that's not an open set in the usual topology: if our two curves are $y=x$ and $y=x^2$, there's a single intersection point at $(0,0)$, but any neighborhood of $y=x^2$ includes things like $y=x^2-\varepsilon$ with two intersections. Instead, one should take as the criteria "sum of intersection multiplicity is at most $e$", and this does the right thing.


Consider $\phi \colon I \to \mathbb{R}^k$ analytic curve ( $I$ compact) with image not contained in any algebraic hypersurface. Then for every $d>0$ there exists $N = N(\phi, d)$ such that for every $P\ne 0$ polynomial on $\mathbb{R}^k$ of degree at most $d$, the set $$I(P) \colon =\{ t \in I \ | \ P(\phi(t)) = 0\}$$ has at most $N$ elements.

Proof: Fix $d$. Consider the set $S_d$ of polynomials $P(x) = \sum a_{\alpha} x^{\alpha} $ of degree $\le d$, with $\sum |a_{\alpha}|^2 = 1$ ( a compact set). Assume that for every $n$ there exists a polynomial $P_n$ in $S_d$ with $|I(P_n)| \ge n$. Using the compactness of $S_d$ and of $I$, we conclude that there exists a sequence $P_n$ of polynomials in $S_d$ convergent to $P$, with $$\{t_{n1}, t_{n 2} , \ldots, t_{n n} \} \subset I(P_n)$$ and $(t_{n k})_{n\ge k} $ convergent to $t_k \in I$ for all $k\ge 1$.

Now we distinguish two cases.

  1. The set $\{ t_k \} \subset I$ is infinite. We conclude $P(\phi(t_k)) = 0$ for all $k$, so the analytic function $P(\phi(t))$ has infinitely many zeroes, contradiction.

  2. The set $\{t_k\}$ is finite. This is like the case of all of the $t_k$ are equal to some $t_{*}$. Now, note that $P_n(\phi(\cdot))$ has $n$ zeroes around $t_{*}$. That means its derivative up to order $n-1$ have zeroes around $t_{*}$. We conclude that $P(\phi(t))$ has all of the derivatives at $t_{*}$ equal to $0$. Since the function is analytic, we get $P(\phi(t)) \equiv 0$, again contradiction.


We only used the normality of the family $S_d$: every sequence in $S_d$ contains a convergent subsequence (convergence together with all the derivatives, on compacts subsets).

If we only had $\phi$ $C^{\infty}$ then there are again two possibilities: the existence of a number $N = N(\phi, d)$ for the family $S_d$ , or the existence of $P \in S_d$, and $t_{\star} \in I$ a zero of infinite order of the function $P\circ \phi$. As an example, consider the $C^{\infty}$ function $\phi(t) = (t, \exp(-\frac{1}{t^2}) \sin(\frac{1}{t}) )$ on $[-\epsilon, \epsilon]$

Perpahs the most important, we can consider the set of zeroes of $P(\phi(t))$ including multiplicities ( so $I(P)$ becomes a multiset) Again we have a bound $|I(P)|\le N(\phi, d)$ for all $P\in S_d$, or we have a $P$ with $|I(P)|$ infinite

We could replace the image of $\phi$ with a compact analytic set of dimension $1$. Locally it has some parametrizations, so we could reduce it to the known problem. The multiplicity of intersection is harder to figure out, so just leave it at the set theoretic level.

Things become more delicate ( but doable) when we consider an analytic set intersecting algebraic sets of complementary dimensions.


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