I have an elementary question concerning zeros of polynomials, which must be well-known.


Fix a base field $ k$ (can assume to be characteristic zero if it makes a difference). Consider the affine space $ P\_n \times \mathbb A^1\_k $, where $ P\_n $ denotes the space of polynomials of degree $ n $ over $ k$ (so $ P\_n $ is an affine space of dimension $ n+1 $ given by the coefficients).


Given $ p \in P\_n $ and $ z \in k $, let $ ord\_z(p) $ denote the order of vanishing of the polynomial $ p $ at $z $.


Fix $ m \in \mathbb N $ and consider the closed subset of $ X\_m \subset P\_n \times \mathbb A^\times $ given as
$$
X\_m := \{ (p,s) : ord\_s(p) + ord\_0(p) \ge m \}$$
In other words, we study those polynomials whose vanishing orders at $ 0 $ and $ s $ add up to at least $ m$ (where $ s $ is non-zero). Then I take $ \overline{X\_m} \subset P\_n \times \mathbb A $.


**Question:
Is it true that $ \overline{X\_m} \cap P\_n \times \{0\} $ consists of those polynomials which vanish to order at least $ m $ at $ 0$?**


This is just a set-theoretic question. Scheme-theoretically, this statement seems to be false.



Closure of the locus of polynomials vanishing to a given order at two points
For a commutative ring $R$, write $P\_{n,R}$ for the affine $(n+1)$-space of polynomials of degree $\leq n$ over $R$. In other words, its $S$-points for an $R$-algebra $S$ are given by $S[x]\_{\leq n}$. Note that $P\_{n,R} = P\_{n,\mathbf Z} \times \operatorname{Spec} R$.


**Lemma.** *Let $R$ be a domain, and $g \in R[x]$ a monic polynomial of degree $d$. Then the map*
\begin{align\*}
\phi \colon P\_{n,R} &\to P\_{n+d,R} \\
f &\mapsto fg
\end{align\*}
*is a closed immersion. For any $R$-algebra $S$, the $S$-points of the image of $\phi$ are the polynomials in $S[x]$ that are divisible by $g$.*


*Proof.* Both $P\_{n,R}$ and $P\_{n+d,R}$ are affine spaces over $R$, and $\phi$ is a linear map with fibrewise trivial kernel. Such a map is a closed immersion (exercise), and the second statement is obvious. $\square$


Applying this to $R = k[s]$ and $g = x^i(x-s)^j$ for $i+j = m$, we see that the locus
$$Z\_{i,j} := \left\{(f,s) \in P\_n \times \operatorname{Spec} R : x^i(x-s)^j \mid f\right\}$$
is closed, hence the same goes for $Z\_m = \bigcup\_{i+j=m} Z\_{i,j}$. The restiction of $Z\_m$ to $P\_n \times \operatorname{Spec} R[1/s]$ is $X\_m$, which shows $\bar X\_m \subseteq Z\_m$. But if $Z = \{f \in P\_n : x^m \mid f\}$, then $Z\_m \cap \big(P\_n \times \{0\}\big)$ is just $Z \times \{0\}$. Since $Z \times \operatorname{Spec} R[1/s] \subseteq X\_m$, we get $Z \times \{0\} \subseteq \bar X\_m$, hence $Z\_m \subseteq \bar X\_m$. We conclude that $Z\_m = \bar X\_m$, which proves the required statement. $\square$


