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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.

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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. $\square$

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  • $\begingroup$ 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. $\endgroup$ – Neil Strickland May 27 at 8:47
  • $\begingroup$ @NeilStrickland thanks, fixed! $\endgroup$ – R. van Dobben de Bruyn May 27 at 15:55
  • $\begingroup$ 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. $\endgroup$ – Joel Kamnitzer May 27 at 18:24
  • $\begingroup$ 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 $\mathbf A^1$. $\endgroup$ – R. van Dobben de Bruyn May 27 at 19:45
  • $\begingroup$ Ok, I agree, this makes sense, thanks! $\endgroup$ – Joel Kamnitzer May 31 at 19:39

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