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Suppose $A:=\{f_1,\dots,f_m\}\subset \mathbb{C}[x_1,\dots,x_n]$ with $m>n$ is a set of homogeneous polynomials of equal degree $d>0$. Suppose further that the variety they define consists of a single point (i.e. the zero vector).

Is it true that there exists a subset $g_1,\dots,g_n$ of the $\mathbb{C}$-linear span of $f_1,\dots,f_m$ with the same property? I.e. they have exactly one common zero?

That this is true for $d=1$ is basic linear algebra. For general $d$, my geometric intution (which, I admit, is probably rather bad given my inexperience in algebraic geometry), tells me that any subset $g_1,\dots,g_n$ "in general position" should have the desired property, which is why I believe this statement to be true.

Thus, two "bonus questions":

Is it true that any "sufficiently generic" subset $g_1,\dots,g_n$ does it, and that "almost all" subsets of size $n$ are "sufficiently generic" in some suitable sense?

Finally (this holds for $d=1$, I think):

Can one always choose $g_1,\dots,g_n\in A$?

I wouldn't be surprised if for somebody who is familiar with algebraic geometry, this is a basic exercise, and so I apologize if this is deemed unsuitable for MO. I'd still be grateful for any references, proofs or hints.

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The answer to the last question is no. Consider, $xy, xz, yz, x^2+y^2+z^2$ in $\mathbb{C}[x,y,z]$. The only common zero of these four polynomials is $(0,0,0)$ but removing any one of them enlarges the set of common zeroes.

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First one should be careful how to phrase the condition that $f_1, \ldots, f_m$ define the variety consisting only of the point $0$. For instance, the polynomials $f_1=x^2, f_2=y^2$ have only $\{(0,0)\}$ as zero set in $\mathbb{C}^2$, but in a certain sense they cut out a "fat" (non-reduced) point $(0,0)$. For instance, the coordinate ring of the scheme cut out by these functions is $$\mathbb{C}[x,y]/(x^2,y^2)$$ and it has has dimension $4$ over $\mathbb{C}$, in contrast to the coordinate ring of a (reduced) point, which is only $\mathbb{C}$ itself.

But if the scheme $S=V(f_1, \ldots, f_m)$ cut out by $f_1, \ldots, f_m$ is exactly the reduced point $0$, then we know the tangent space to $S$ at $0$ is the zero-subspace of the tangent space $T_0 \mathbb{C}^n = \mathbb{C}^n$. In general, the tangent space of $V(f_1, \ldots, f_m)$ at a point $p$ is the subspace of $T_p \mathbb{C}^n$ cut out by $df_1|_p, \ldots, df_m|_p \in T_p^* \mathbb{C}^n$.

But this means that the linear forms $df_1|_0, \ldots, df_m|_0$ cut out the zero-subspace in $T_0 \mathbb{C}^n$. Hence, as you mentioned, we can choose $n$ indices $i_1, \ldots, i_n$ from $1, \ldots, m$ such that $df_{i_1}|_0, \ldots, df_{i_n}|_0$ already cut out the zero-subspace. But then $S'=V(f_{i_1}, \ldots, f_{i_n})$ has tangent space $\{0\}$ at $0$. As all $f_i$ are homogeneous, for every $p \in S' \setminus \{0\}$ also the entire line $\mathbb{C} p$ would be contained in $S'$ and hence the tangent direction $p \in T_0 \mathbb{C}^n$ would be contained in $T_0 S'$. This is a contradiction, hence we can indeed cut out the (reduced) point $0$ with exactly $n$ of the $f_i$. The linear algebra argument in the proof above also showas that a generic linear combination of the $f_i$ is sufficient.

Note that the counterexample presented by Tony does not satisfy that the $f_i$ cut out the reduced point $0$, as all differentials $df_i$ vanish at the origin.

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  • $\begingroup$ I intend the classical meaning of the term "variety", i.e. the set of common zeros: $\mathbb{V}(A) = \{ x\in\mathbb{C}^n \mid \forall f\in A: f(x)=0 \}$. And I explicitly need the statement also in the case where the locus is "fat", e.g. if $A=\{ x_1^d, \dots, x_n^d \}$ for any $d>0$ (of course, in that particular case $A$ itself already has size $n$). $\endgroup$ – Max Horn Feb 4 '16 at 10:09
  • $\begingroup$ So, all in all, it seems your answer is saying: "If the scheme is not fat, then the third question has a positive answer." Correct? $\endgroup$ – Max Horn Feb 4 '16 at 10:11
  • $\begingroup$ I'd then still be interested in an answer for the first question, which is the one that is most important to me :) $\endgroup$ – Max Horn Feb 4 '16 at 10:12
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    $\begingroup$ Yes, it's correct that my answer only shows (3) in the special case that the scheme is reduced. $\endgroup$ – JoS Feb 4 '16 at 10:44
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While subsets of $A$ will not work as Tony pointed out, the other questions have a positive answer. There are more general statements, but you situation is particularly vivid in the geometric sense (of course you need an infinite field, but I will assume algbraically closed for simplicity). Your condition implies that the set $A$ gives a morphism $\mathbb{P}^{n-1}\to\mathbb{P}^{m-1}$. If $m>n$, then the image is not all of $\mathbb{P}^{m-1}$ and so choose a point not in the image and project. We get a morphism $\mathbb{P}^{m-2}$. Instead of a point, you could choose a linear subspace of maximal dimension not intersecting the image and then you will end up with a map $\mathbb{P}^{n-1}$ to itself, giving you what you need.

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  • $\begingroup$ Thanks for your reply, but I am not quite sure I understand. Which morphism $\mathbb{P}^{n-1}\to\mathbb{P}^{m-1}$ do we get here? And how does the morphism you obtain after projecting to $\mathbb{P}^{m-2}$ (and finally down to $\mathbb{P}^{n-1}$, I assume) help to find a suitable subset of the span of $A$? (Presumably this is evident as soon as it clear which morphism you have in mind?) $\endgroup$ – Max Horn Feb 4 '16 at 10:05
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    $\begingroup$ I think the morphism $\mathbb{P}^{n-1} \to \mathbb{P}^{m-1}$ should be given by $[f_1, \ldots, f_m]$. The fact that they only vanish simultaneously at $0$ shows that the morphism is well defined. Here, the $f_i$ can be interpreted as sections of $\mathcal{O}(d)$. Taking linear projections means that we take $m-1$ linear combinations of the $f_i$ to define the morphism to $\mathbb{P}^{m-2}$. The fact that they are still well-defined means that these linear combinations still only have $0$ as common zero set. $\endgroup$ – JoS Feb 4 '16 at 10:47

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