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If $X_1, \ldots, X_n$ are a collection of codimension $\geq 2$ subvarieties of $P^n$, is there an irreducible hypersurface containing them?

I would be satisfied with an answer to : if $x_1, \ldots, x_n$ are points in the plane $P^2$, is there an irreducible curve containing all of them? Okay, this was mostly answered below. Is there a general technique that works for my original question?

Dimension count does not seem to give an answer because being irreducible is an open condition that I don't know how to study using an incidence correspondence.

Elle Najt
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  • Do you perhaps want to add some sort of condition on what sort of irreducible variety you want? Currently, your general condition is satisfied trivially by $P^n$, but your more specific question about points in $P^2$ seems to imply you'd like something of codimension one. – KReiser Oct 25 '15 at 07:19
  • @KReiser Yes, I meant specifically a hypersurface. Thanks. – Elle Najt Oct 25 '15 at 14:29

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I assume you are working over $\mathbb{C}$.

Partial answer: Suppose your points can be mapped into $U_0=\{[1,y,z]\in \mathbb{P}^2\}\subset \mathbb{P}^2 $ by $Aut(\mathbb{P}^2)={PGL}_3(\mathbb{C})$. Say $X_i=[1,y_i,z_i]\in U_0$ are images of your points. Furthermore, We can rotate $U_0$ while fixing $[*,*,0]$. WLOG, we can assume $y_i$ are distinct. Then there exist a polynomial $f$ s.t. $f(y_i)=z_i$ for all $i$. As a result, $z-f(y)$ is an irreducible polynomial in $R[y,z]$. We can homogenize to get $a(x,y,z)=$numerator of ($z/x-f(y/x)=a(x,y,z)/b(x,y,z)$), where $a,b$ are relatively prime in $\mathbb{C}[x,y,z]$. Then $a(x,y,z)$ is an irreducible homogenous polynomial which cut a curve in $\mathbb{P}^2$ and passing through $X_i$.

If your points can't be mapped into $U_0$, I suspect we can do something similar but I haven't figured it out.

Ben
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    We can definitely map it there, because we can find a line that misses all of the points (missing a point is an open condition in the space of lines in $P^2$, which is a $P^2$, and intersecting opens in $P^2$ gives a nonempty open). Then we just translate that line to infinity ($x = 0$ in your notation.) I'm not sure what you meany by rotating $U_0$ - I guess this is where we are using that we are over $C$. So you are saying to rotate about the origin in order to separate the $y$ values? I believe that is possible. What follows that seems reasonable. – Elle Najt Oct 25 '15 at 14:38
  • Oh you are right. We can always maps those point in the affine space. By rotation , I meant separating those y. Your interpretation is correct. Ok. I think this is a complete solution now. – Ben Oct 25 '15 at 14:47
  • I think my method only works in char zero. – Ben Oct 25 '15 at 14:48
  • It's not clear to me that it doesn't work away from char zero, though of course rotation matrices would have to be replaced by some other device. – Elle Najt Oct 25 '15 at 14:48
  • Oh wait. If the field is finite, it is definitely not possible, right? – Ben Oct 25 '15 at 14:51
  • Btw, just curious, why do u come up with this question? – Ben Oct 25 '15 at 14:52
  • It's not possible to separate the y values by a rational automorphism if there are too many points. I'm not sure if it is impossible to find an irreducible curve containing these points - this just means finding an irreducible polynomial vanishing at all of them. This polynomial could vanish at all of the $F_{p^n}$ points of $P^2$, but that doesn't make it $P^2$. (Here you can either think over the algebraic closure, or think "schemey" and consider subvarieties to be ideals in the homogeneous coordinate ring.) – Elle Najt Oct 25 '15 at 14:56
  • I saw a talk recently that mentioned something similar: it is impossible to find a smooth hypersurface containing an algebraic set with a singularity that is too large (Zariski tangent space not codimension 1). This is apparently sharp, in that the local dimension of the tangent spaces are the only obstruction to fitting a smooth hypersurface around that algebraic set. An example of this obstruction is the union of the axis, $(x,y)(x,z)(y,z)$. But in this case there is an irreducible hypersurface containing them ($xz + zy + xy$). Therefore this question. – Elle Najt Oct 25 '15 at 15:01
  • I am hesitating to accept your answer because I really want a technique that will work for that more general question. – Elle Najt Oct 25 '15 at 15:02
  • I know I said I would be satisfied with an answer to this simpler question - but... I guess I do not know myself so well. – Elle Najt Oct 25 '15 at 15:03
  • I still think it is not true for finite dield. Let say consider finite field with 2 elements. Then there are only seven closed points in P^2. Suppose we have a homogenous prime ideal (f) which contains all these points. However, that is not possible by considering parity. – Ben Oct 25 '15 at 15:13
  • I don't follow your parity argument. But there are many more closed points than just those. – Elle Najt Oct 25 '15 at 15:28
  • Here is an algebraic argument that works over an algebraically closed field. Specifically, I want to prove that given a finite set $F$ in $P^2$, we can choose homogeneous coordinates so that each $x \in F$ has a distinct $y$ coordinate. This is the same problem as finding a linear function that takes distinct values on each of the points of $F$. Consider the union $C$ of lines spanned by pairs $x,y \in F$. There is a $p \not \in C$, and projection from the point $p$ onto $P^1$ is injective by our choice of $C$. Now we pull back a suitable linear coordinate from $P^1$. – Elle Najt Nov 04 '15 at 15:52