Re: Projective closure of variety question
- From: Jannick Asmus <jannick.news@xxxxxx>
- Date: Mon, 09 Oct 2006 12:03:31 +0200
On 30.09.2006 20:17, James wrote:
Dear all,
I have 2 very basic questions about projective closure. If V is an affine variety in C^n, I will denote the projective closure of V by Proj(V). More precisely, to get Proj(V), I take V, embed it into CP^n (P^n for short) by taking some coordinate of P^n to be non-zero, and then I take Zariski closure.
Here the fact that a 1-dimensional variety has a unique "completion"
gives the basic idea of the proofs. In what follows I just give a sketch
of reasoning for (2).
1) Consider the affine variety V = V(I) of C^2 defined by I = x^2 - y. The claim is that the Proj(V) is defined by the ideal generated by x^2 - yz. Why is this? I know the general theorem that if V is an affine variety and I is the radical ideal of polynomials vanishing on V, then the radical homogeneous ideal vanishing on Proj(V) consists of the homogenization of all elements of I. But I would like to know without resorting to this theorem : Well, x^2 - yz defines a projective variety in P^2, and restricting to the open subset z = 1 gives V(I). So clearly V(I) (embedded as a subset of P^2 where we let the z coordinate equal to 1) is contained in the projective variety V(x^2 - yz). Therefore since V(x^2 - yz) is closed in P^2 and V(I) is in V(x^2 - yz), then Proj(V) is in V(x^2 - yz), since Proj(V) is the closure of V(I) in P^2. (I am denoting V(I) sometimes by V, since I would rather not write Proj(V(I)), but rather Proj(V). )
But I need to show that V(x^2 - yz) is contained in Proj(V). Well, we already considered all elements where z = 1, and this is V(I). But there is one more point. If z = 0, then x = 0, so y must be non-zero. So the point [0 : 1 : 0] is in V(x^2 - yz). Now I need to show that [0 : 1 : 0] is in Proj(V).
Note that V is isomorphic to the straight line A^1_k.
The problem is that I do not have a good feel for what Proj(V) looks like. How can I show that [0 : 1 : 0] is in Proj(V)?
Alternatively, if I use the theorem I stated above on what the ideal that defines Proj(V) should be, then I need to find the homogenization of x^2 - y in C[x,y,z]. But x^2 - y is irreducible, so prime, so (x^2 - y) is radical. In particular, I can easily show that the homogenization of the ideal (x^2 - y) is indeed (x^2 - yz) (the homogenization of an ideal J is the ideal generated by the homogenization of all the elements of the original ideal.). Thus by the theorem, V(x^2 - yz) = Proj(V).
But I want to see an easier way, by doing some calculations rather than resorting to general theorems.
2) I'll make this one brief. Let V be the subvariety of C^3 defined by y - x^2 and z - xy. i.e. the twisted cubic (t,t^2,t^3). Embed C^3 in P^3 = {[w : x : y : z]} via the open set w not equal to 0. Consider Proj(V). Now, let us homogenize y - x^2 and z - xy into wy - x^2 and zw - xy. Then look at V(yw - x^2, zw - xy). If w = 1, we recover V. If w = 0, this forces x = 0, and we get the set B = { [0 : 0 : y : z] s.t. one of y,z not zero}. The claim is that B is not contained in Proj(V). Why? (Note : This example is supposed to show that if V is an affine variety defined by I, then Proj(V) is not necessarily defined by taking the homogenization of a set of generators of I).
As you said V is isomorphic to the straight line A^1_k having the
completion P^1_k by adding a *single* point, usually called the point at
infinity.
Now the closed immersion of varieties A^1_k -> V -> P^3_k uniquely
extends to a morphism phi: P^1_k -> P^3_k given by k[X0,...,X3] ->
k[T0,T1], Xi |-> T0^(3-i).T1^i, i=0,...,3. Then phi is a closed
immersion (Hint: Test it on local charts of P^1_k or show that the
homomorphism of graded algebras is essentially surjective). Now compare
the image of P^1_k in P^3_k, Proj(V) and V(X0.X2-X1^2,X0.X3-X1.X2) and
count points.
I thank you very much for your help,
Sincerely,
James
J.
.
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