Lecture 8: Product Topology, Complete Varieties

Lecture  18.725 Algebraic Geometry I Lecture 8: Product Topology, Complete Varieties To check that Pn is separated, we used an affine covering of Pn as ∪Ani . Instead, we could have checked that the preimage of ∆ ⊆ Pn × Pn in (An+1 \ 0)2 is closed; this is given by the equation X ∧ Y = 0 (recall that Pn = (An+1 \ 0)/k × . Remark 1. We have that X is Hausdorff if and only if the diagonal in X 2 is closed with respect to the product topology, and not the Zariski topology. Corollary 1. If k = C, then X is separated iff and only if Xcl (which is X with the classical topology coming from C) is Hausdorff. Proof. Let X be a variety over k, and Z ⊆ X be a Zariski locally closed subset. We claim that Z is Zariski closed if and only if it is classically closed. To see this, it suffices to check that if Z is Zariski locally closed and classically closed, then it is Zariski closed. Note that Z is Zariski open in ZZar , and so it is open dense in Zcl , so ZZar = Zcl . Since the diagonal ∆ is Zariski locally closed, we are done. Remark 2. The image of a morphism may not be a subvariety. For example, take the map from A2 to itself 6 0} ∪ {(0, 0)}. It induced by the polynomial mapping k[a, b] → k[x, y], a 7→ x, b 7→ xy. The image is {a = is not a subvariety, but it will be a constructible subset (this is Chevalley’s Theorem, which will be proven later). Suppose X, Y are irreducible and f : X → Y is a morphism. Then either f (X) is contained in a closed subset Z ) Y , or f (X) contains an open dense subset U . Proposition 1. X is separated if and only if for any affine open U, V ⊆ X, U ∩ V is affine and k[U ∩ V ] is generated by k[U ] and k[V ]. Proof. Consider an open U × V ⊆ X × X where U , V are open subsets in X. Since X is separated, the intersection of diagonal with U × V is closed in U × V ; furthermore, this intersection equals U ∩ V . As U × V is affine and U ∩ V is closed, we see that U ∩ V is affine. We also have k[U ] ⊗ k[V ] = k[U × V ]  k[U ∩ V ]. For the converse, the second condition implies that (U × V ) ∩ ∆ is closed in U × V , so ∆ is closed. Example 1. Let X be the affine line with a doubled origin, with the usual affine open covering U ∪ V where U = A11 , V = A21 . Then this covering corresponds to k[t1 , t2 ] 7→ k[t, t−1 ] where t1 , t2 7→ t. This is not surjective. Consider X to now be the affine plane with a doubled origin, with affine open covering U ∪ V where U = A21 , V = A22 . In this case, U ∩ V = A2 \ {0} is not affine. Also, we checked last time that for Y separated, f : X → Y is determined by f |U where U is a dense open subset of X. Proposition 2. (Caternary property). Let X be an algebraic variety, with X = Zn ) Zn−1 ) · · · ) Z0 where each Zi is closed irreducible. If this chain cannot be refined, then dim Zi = i. Proof. Theorem 2.6.7 of [K]. Now we consider “dimension and rate of growth.” Let A be a finitely generated k-algebra. Let V be the space of generators. Set Vn = span{x1 · · · xk : xi ∈ V, k ≤ n} and DV (n) = dim Vn . The asymptotic behavior of DV (n) actually does not depend on V . For if V 0 ⊆ Vd , then DV 0 (n) ≤ DV (nd). Proposition 3. If A = k[X] where X is affine of dimension d, then DV (n) = Θ(nd ); that is, there exist constants c0 , c such that for all n, c0 nd ≤ DV (n) ≤ cnd (*) Proof. Suppose B ⊆ A and A is finite over B. If (*) holds for B, then it holds for A. Given VB to be generators for B, VA = VB ∪ W where W are generators for A as a B-module, note each x ∈ W satisfies an equation of the form xr = br−1 xr−1 + · · · + b0 for bi ∈ B. We can assume without loss of generality that bi ∈ VB . Then DVB (n) ≤ DVA (n) ≤ DVB (n) · c where c = rdim W . Setting B = k[x1 , · · · , xd ], an explicit computation gives a polynomial in n of degree d. 1 Lecture  18.725 Algebraic Geometry I Remark 3. (1) The order of growth function has been used to generalized the concept of dimension to noncommutative algebras, groups etc. in the works of Artin, Gromov and others. (2) In our commutative setting the function DV (n) can in fact be analyzed much more precisely. It turns out that for large n we have DV (n) = P (n) for a certain polynomial P . It is closely related to the so called Hilbert polynomial, to be described in 18.726. Theorem 1.1. Suppose X, Y are irreducible subvarieties in An . Then each component of X ∩ Y has codimension at most codimX + codimY . Proof. Rewrite X ∩ Y = (X × Y ) ∩ ∆An ⊆ An × An . From last time, dim(X × Y ) = dim X + dim Y . The diagonal in affine space is cut out by the n linear equations xi = yi . By a theorem of last time we know that each component of Zf ⊆ X has dimension equal to dim X − 1, so dim(X ∩ Y ) ≥ dim(X × Y ) − n = dim X + dim Y − n. Remark 4. This theorem doesn’t exclude empty intersections. The obvious example is the intersection of subvarieties x1 = 0 and x1 = 1. Theorem 1.2. The previous theorem holds for X, Y ⊆ Pn ; moreover, the intesection X ∩ Y is nonempty if dim X + dim Y > n. Proof. Here is a lemma: the dimension of CX (the cone over X) equals dim X + 1. To see this, note that CX ∩ {xi = 1} is isomorphic to Ui = X ∩ Ani ⊆ X, and from this it is a straightforward exercise to complete the proof of this lemma. Using this, the proof of the theorem goes as follows: dim(X ∩ Y ) = dim CX∩Y − 1 = dim(CX ∩ CY ) − 1 ≥ dim CX + dim CY − (n + 1) − 1 = dim X + dim Y − n. The intersection of cones is nonempty as it contains 0. Complete varieties Definition 1. A variety X is complete if it is separated and universally closed, which means that for all Y , the projection map Y × X → Y sends closed sets to closed sets. We will see that for k = C, X is complete if and only if Xcl is compact. Also, if X is quasiprojective, we will see that complete is equivalent to projective. For the forward direction, suppose ι : X ,→ Pn is locally closed. Then X is in the image of the closed embedding Γι ,→ X × Pn , so X ⊆ Pn is closed. Lemma 1. (i) Suppose Z is closed in X. Then X is complete implies Z is complete. (ii) If f : X → Z is a morphism with Z separated and X complete, then f (X) ⊆ Z is a closed complete subvariety. (iii) If X, Y are complete, then so is X × Y . Proof. (i) We see that Y × Z is closed in Y × X, so by considering the projection to Y , this is clear. (ii) Identify f (X) with Γf in X × Z. As X, Z are separated, so is X × Z. As Γf is a closed subvariety of X × Z, it is also separated (for these facts, see Lemma 3.3.2 of [K]). Hence f (x) is separated. To check that f (X) is universally closed, take a variety Y and closed subset T ⊆ f (X) × Y . It suffices to check that the image of T in Y is closed. Consider the map f × id : X × Y → f (X) × Y , and let Te = (f × id)−1 (T ) ⊆ X × Y . Then it suffices to check that the image of Te under the projection X × Y → Y is closed, which follows from X being complete. 2 Lecture  18.725 Algebraic Geometry I (iii) As X, Y are both separated, so is X × Y (Lemma 3.3.2 of [K]). Let Z be any variety and T ⊆ X × Y × Z closed. As X is universally closed, the image of T in Y × Z is closed. As Y is universally closed, the image of T in Z is closed. Hence, X × Y is universally closed. Proposition 4. Pn is complete. Proof. We know Pn is separated (Lemma 3.3.2 of [K]), so it suffices to check that it is universally closed. We use an “elimination theory” argument. Let Y be any variety and Z ⊆ Pn ×Y be a closed subset. Then e is generated Z comes from a closed subset Ze ⊆ An+1 × Y . Suppose IZe , the ideal of functions vanishing on Z, by some homogeneous polynomials Pi ∈ k[Y ][x0 , · · · , xn ]. For y ∈ Y , let Pi,y = Pi (y, −) ∈ k[x0 , · · · , xn ]d for some d (this is the degree d homogeneous polynomials). Then (Pi,y ) is an ideal of k[x0 , · · · , xn ], so we let Ud = {y ∈ Y : (Pi,y ) ⊇ k[x0 , · · · , xn ]d }. Letting pr(Z) be the image of Z in Pn × Y → Y , we see that y 6∈ pr(Z) iff there is no point (x0 , · · · , xn ) which makes all of the Pi,y vanish, iff it lies in some Ud . Therefore, Y \ pr(Z) = ∪d Ud . It is enough to check that each Ud is open, which is equivalent to checking that theX natural map ⊕i k[x0 , · · · , xn ]d−di → k[x0 , · · · , xn ]d (where di is the degree of Pi ) defined by sending (gi ) 7→ gi Pi,y is surjective. This is equivalent to requiring that some matrix with k[Y ]-entries, when evaluated at y, has maximal rank, which is some condition of non-vanishing of minors. So it is an open condition. So projective varieties are complete, and a quasiprojective variety is complete if and only if it is projective. 3