# What is the closure of space of polynomials in a dense subspace along with a marked point equal to?

EDIT

Let $\mathbb{C}^{m*}$ be the space of non zero polynomials of degree at most $d$ in two variables. So an element of this space is essentially $$f:=f_{00} + f_{10} x + f_{01} y + \ldots f_{0d} y^d$$ where not all the coefficients are zero. Hence, this space can be identified with $\mathbb{C}^{m*}$, where $m = \frac{d(d+3)}{2}+1$.

Let $\psi: \mathbb{C}^{m*} \times \mathbb{C}^2 \rightarrow \mathbb{C}$ be the evaluation map, ie $$\psi(f, x,y) = f(x,y) = f_{00} + f_{10} x + f_{01} y + \ldots f_{0d} y^d.$$ I have two questions:

1) Given $(f ,x_0,y_0) \in \mathbb{C}^{m*} \times \mathbb{C}^2$ such that $\psi(f, x_0,y_0) =0$ and a polynomial $f_t$ that is sufficiently "close" to $f$, does there exist a point $x_t, y_t$ close to $x_0, y_0$ such that $$\psi(f_t, x_t, y_t) =0 ~~?$$

Note that this would be true if $(x_0, y_0)$ was a smooth point of $f$. But I am not making this assumption.

2) Let $\mathcal{S}$ be a dense subspace of $\mathbb{C}^{m*}$. Define the space $\mathcal{F}$ as
$$\mathcal{F} := \{ (f, x,y) \in \mathcal{S} \times \mathbb{C}^2: \psi(f, x,y) =0 \}$$
Is it true that $$\overline{\mathcal{F}} = \{ (f, x,y) \in \mathbb{C}^{m*} \times \mathbb{C}^2: \psi(f, x,y) =0 \}$$ where $\overline{\mathcal{F}}$ denotes the closure of $\mathcal{F}$ inside $\mathbb{C}^{m*} \times \mathbb{C}^2$?

The basic idea being that if $\psi(f, x,y) =0$ but $f \notin \mathcal{S}$ we can choose a sequence $f_n \in \mathcal{S}$ converging to $f$. And by 1), there exists a sequence (possibly non unique) $(x_n, y_n) \in \mathbb{C}^2$ such that $$\psi(f_n, x_n, y_n) =0$$ This sequence $(f_n, x_n,y_n)$ should converge to $(f, x,y)$.

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I don't get it: $0 \cdot 1=0$ but $xy \ne 0$ for small $x \ne 0$ and $y$ close to $1$. –  Ben McKay Jul 3 '13 at 11:27
Thank you Ben for this counter example. I apologize for the question. The answer to the question as stated could not be yes. I have therefore stated the specific example of $\psi$ I had in mind. –  Ritwik Jul 3 '13 at 12:37

(1) Yes. For notational simplicity, assume $(x_0, y_0)=(0,0)$. After a linear change of coordinates, we may assume that $f(x,0)$ is not identically zero; say $f(x,0) = x^n g(x)$ with $g(0) \neq 0$. Choose a small disc $D$ around $0$ so $f(x,0)$ has no zeroes for $x \in D \setminus \{ 0 \}$.

Choose $f_t$ close enough to $f$ that $f$ is nonvanishing on $\partial D$. Then $z(t) := \frac{1}{2 \pi i} \int_{\partial D} \frac{\partial f_t(x,0)/\partial x}{f_t(x,0)} dx$ is a continuous function of $t$. But $z(t)$ is the number of roots of $f_t(x,0)$ inside $D$, so it is integer valued and constant. So $f_t(x,0)$ has $n$ roots inside $D$ for all $t$ sufficiently close to $0$.

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The justification for (1) seems perfect (plus you also need $f$ to be non zero, which accounts for the counter example Serge gave). For (2) I am a bit confused. The set $\mathcal{F}$ is $\psi^{-1}(0)$ intersected with $\mathcal{S} \times \mathbb{C}^2$. This is not closed in $\mathbb{C}^{m*} \times \mathbb{C}^2$. –  Ritwik Jul 3 '13 at 13:33
I think I understand the cause for the confusion. I have now indicated clearly what I hope $\overline{\mathcal{F}}$ should be. –  Ritwik Jul 3 '13 at 13:41
OK. I'll delete my answer to (2) then. With the new version, the answer to (2) is morally yes, but I need to figure out whether you have omitted a technical condition somewhere. –  David Speyer Jul 3 '13 at 15:12

If I understood the first question right, it sounds as follows: suppose that $\psi$ is a polynomial in $N>1$ variables over $\mathbb C$; can its set of zeroes in $\mathbb C^N$ have an isolated point?

The answer is 'no': connected components of a, say, affine algebraic set in Zariski topology and in the classical topology are the same, so this isolated point would be an irreducible component of the set of zeroes of $\psi$. However, each irredicuble component of the zero set of a non-zero polynomial in $\mathbb C^N$ must have dimension $N-1$.

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I wasn't asking if the zeros in $\mathbb{C}^N$ can be isolated. I was asking a stronger question. As Ben has pointed out, in general the answer to that question can not be yes. The answer may be yes in the specific example I am asking now (I have edited the question). –  Ritwik Jul 3 '13 at 12:41
OK, the answer to the edited question is also negative: put $f\equiv 0$, $f_t=t(x+y+100)$, where $t$ tends to zero, and $x_0=y_0=0$. What exactly are you asking? –  Serge Lvovski Jul 3 '13 at 13:04
I am sorry, I need $f$ to be not identically zero, ie I am looking at the space of non-zero polynomials. –  Ritwik Jul 3 '13 at 13:35