# Closure of the locus of polynomials vanishing to a given order at two points

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.

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. $$\square$$

• I don't think that $\phi$ is a closed immersion if $R=\mathbb{Z}$ and $g=2$. If $g$ is monic then $\phi$ is a closed immersion. – Neil Strickland May 27 at 8:47
• @NeilStrickland thanks, fixed! – R. van Dobben de Bruyn May 27 at 15:55
• Thanks for this proof! I followed the individual steps, but I am somehow still confused. It seems that you have replaced my space $P_n \times \mathbb A^1$, whose $k$ pts were pairs of a polynomial and a number, with $P_{n, k[s]}$ whose $k[s]$ pts are polynomials with coefficients in $k[s]$. I understand that $P_n \times_{Spec k} Spec R = P_{n,R}$, but somehow I am still confused. – Joel Kamnitzer May 27 at 18:24
• I actually confused myself about this for a bit. But the point is that $k[s]$-points of $P_{n,k[s]}$ are sections of $P_{n,k[s]} \to \mathbf A^1$, not just points of $P_{n,k[s]}$ in any classical way. Unlike geometry over algebraically closed fields, in this relative setting you don't just want to look at sections over $R$, but over any $R \to S$ (the functor of points point of view). Maybe my main observation is that $P_n \times \mathbf A^1$ really wants to be a relative gadget over $\mathbf A^1$. – R. van Dobben de Bruyn May 27 at 19:45
• Ok, I agree, this makes sense, thanks! – Joel Kamnitzer May 31 at 19:39