Algebraic Geometry I Lectures 22 and 23

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1 Algebraic Geometry I Lectures 22 and 23 Amod Agashe December 4, Fibered Products (contd..) Recall the deinition o ibered products g!θ X S Y X Y π 2 π 1 S By the universal mapping property o ibered products, (X S Y )() = {(, g) X() Y () π 1 = π 2 g} For example, consider = Spec k or some ield k. Let X, Y, S be aine varieties over k and let X, Y, S be the corresponding schemes over k i.e X = Spec A(X ) etc. (Note that i k is algebraically closed, then X(k) = X ). Then (X S Y )(k) = {(x, y) X(k) Y (k) π 1 (x) = π 2 (y) in S(k)} In act, in the category o sets, we have X S Y = {(x, y) X Y π 1 (x) = π 2 (y) in S} For example, i we consider S = Spec (k) above, then X S Y = {(x, y) X Y } The construction o the ibered product generalizes many other elementary ones: 1

2 (1) I S contains only one element, then it is just X Y (2) I X π 1 S, Y π 2 S are inclusions, then it is X Y (3) I Y π 2 S is an inclusion, then we get π 1 1 (Y ) (4) I X = Y, it gives the set on which the maps π 1, π 2 are equal i.e the equalizer o π 1, π 2 We want an analogue o (3) or ibers o schemes over points. (In (3) above, i Y = {s} then we get π1 1 (s) i.e the iber o π 1 over s). I S is a scheme and s S (i.e a map {s} S o sets), we want to associate a scheme to s and a map rom that scheme to S. Deinition 1.1. Let x X, where X is a scheme. Let O x be the stalk o O X at x and m x the unique maximal ideal o O x. We deine the residue ield o x on X as k(x) = O x /m x. We now claim that there exists a canonical isomorphism i x : Spec k(x) X whose image is x. To prove this, we begin by noting that U X such that x U and ψ : (U, O X U ) (Spec A, O SpecA ) is an isomorphism or some ring A. Let = ψ(x) be a prime ideal o A. Then ψ induces a local isomorphism O x A whence we have an isomorphism k(x) = O x /m x A / A A A 2

3 Here, the vertical maps are inclusions. This in turn gives rise to Speck(x) Spec(A / A ) SpecA whence we have Thus we see that i x(s) = SpecA ( )(ψ 1 ) U (0) (0) A i x x { pullback o the image o s in O x i x U 0 otherwise Deinition 1.2. Let : X Y be a morphism o schemes and let y Y. We thus have the ollowing commutative diagram φ X ψ Spec k(y) iy Y We deine the iber o over y to be the scheme X y = X Y Spec k(y) Remark 1.3. I is a scheme, then X y () = (X Y Speck(y))() = {(φ, ψ) X() (Spec k(y))() φ = i y ψ = y} 3

4 For example, i =Spec k and X,Y are varieties over k, then X y (k) = {x X(k) (x) = y} So, as a set, X y (k) = iber over y. Example 1.4. See Ex in Hartshorne. Now consider the general situation X s Speck(s) Speck(s) X S s Example 1.5. Consider the map S = SpecR[t] X = SpecR[x, y, t]/(x 2 y 2 t) obtained rom R[t] R[x, y, t]/(x 2 y 2 t) Geometrically, is the projection rom x 2 y 2 = t to the t th co-ordinate. For example, i t = a, then 1 (a) = {(x, y) x 2 y 2 = a}. For t = a = 0, 1 (0) = {(x, y) x 2 y 2 = 0}. The point t = a corresponds to the maximal ideal s = (t a) S. X s k(s) = Spec(R[x, y, t]/(x 2 y 2 t)) (R[t]/(t a)) = SpecR[x, y, t]/(x 2 y 2 t, t a) = SpecR[x, y]/(x 2 y 2 a) I a 0, then this is irreducible. I a = 0, it is reducible. 4

5 1.1 Some Arithmetic Examples Example 1.6. Spec The elements o Spec are the prime ideals o i.e (0), (p) or prime p. The closed subsets are o the orm V ((n)) = ideals containing(n)} {prime { = primes dividing n i n 0 Spec i n = 0 So the closure o (0) is Spec i.e (0) is a generic point. Example 1.7. Consider the map X = Spec([x, y]/(x 2 y 2 5)) S = Spec obtained rom [x, y]/(x 2 y 2 5) Any prime p corresponds to (p) S. Residue ield = (p) /p (p) = quotient ield o /p = /p. Fiber over (p) = Spec([x, y]/(x 2 y 2 5) /p) = Spec([x, y]/(x 2 y 2 5, p)) = Spec(/p[x, y]/(x 2 y 2 5)) So the iber is the curve (x 2 y 2 5) in A 2 F p. Thus we see that schemes unite algebra and geometry (o polynomials). Earlier, polynomial equations were thought o geometrically while now we think o ibers algebraically. The example above also shows that one should think o the map obtained rom X S as a amily o schemes parametrized by points o S. In particular, a scheme over Spec is a amily o curves over Q and F p or primes p. This allows us to pass rom char 0 objects to objects in char p. This process is an example o specialization.the iber over (0) is called the generic iber and that over (p) is called the special iber over p. This motivates the ollowing deinition. 5

6 Deinition 1.8. In the diagram X S T T X S X S T is called the base change o X rom S to T. Example 1.9. (Geometric) Consider the map SpecC[t][x]/(x 2 t) A 1 C = SpecC[t] Fiber over a A 1 C is Spec C[x]/(x2 a). I a 0, then x 2 = a has 2 solutions and it has 1 solution i a = 0. Example (Number Theoretic) Let [i] = {a + bi a, b } C i.e the ring o integers in Q[i]. Consider Spec([i]) Spec obtained rom [i] I p is a prime, then the iber over (p) Spec is Spec([i] (/p)) = Spec([i]/p[i]) So elements o the iber are prime ideals o [i] containing p. Thus schemes give a geometric picture o the behavior o primes in ring extensions o. Schemes thereore incorporate algebraic number theory into algebraic geometry. 6