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May 13, 2019 Problem 17 Let K Be a Fi Algebraic Geometry II Prof. Dr. Ulrich G¨ortz SS 2019 Felix Gora Problem sheet 5 Due date: May 13, 2019 Problem 17 Let k be a field. If f 2 k[T1;:::;Tn] and f = fr +fr+1 +···+fd is its decomposition ∗ into homogeneous polynomials fi of degree i with fr 6= 0, then f := fr is called the ∗ ∗ leading term of f. For an ideal a of k[T1;:::;Tn] let a be the ideal generated by f for f 2 a. Now let X be a scheme locally of finite type over k. Let x 2 X(k) and let m = mx ⊂ L d d+1 OX;x be the maximal ideal at x. Then A := d≥0 m =m is a k-algebra (where we define the product of f 2 md=md+1, g 2 me=me+1 as fg 2 md+e=md+e+1 and extend this bilinearly). The affine scheme TCx(X) := Spec A is called the tangent cone of X in x. (a) For every open affine neighborhood U of x there exists a closed immersion n i: U,! Ak for some n ≥ 0 with i(x) = 0 and such that i yields an isomor- ∼ ∼ phism U ! V (a) for an ideal a ⊆ k[T1;:::;Tn] (why?). Show that TCx(X) = ∗ Spec k[T1;:::;Tn]=a . 2 2 2 (b) Let X = V (Y − X (X − 1)) ⊂ Ak. Compute the tangent cone of X in the 2 origin of Ak. 2 3 2 2 (c) Let X = V (Y − X ) ⊂ Ak. Compute the tangent cone of X in the origin of Ak. Remark. One can show that the dimension of the tangent cone (as a k-scheme) is equal to supi dim Zi, where the Zi are those irreducible components of X which contain x. One way to do this is to invoke the non-trivial commutative algebra result that the dimension of a local noetherian ring agrees with the dimension of its associated graded ring, see e.g. [Ma] Thm. 13.9. See also [Mu] III.3 for an extensive discussion of the tangent cone, including a proof of the result on its dimension, and n loc. cit. III.4 for the relationship with the tangent space: For x = 0 2 U ⊆ Ak as 1 n above, the tangent space Tx(X), seen as a linear subspace of Ak in the natural way, is the smallest linear subspace that contains the tangent cone as a scheme. Problem 18 Let X be a topological space. (a) Let Y ⊂ X be a subspace. Then dim Y ≤ dim X. If X is irreducible, dim X is finite, and Y ( X is a proper closed subset, then dim Y < dim X. S (b) Let X = i Ui be an open covering. Then dim X = sup dim Ui: i 1I.e., as a closed subscheme defined by linear equations. (c) Now let X be a scheme. Show that dim X = sup dim OX;x: x2X Problem 19 (a) Prove the following fact which was stated in the lecture: Let f : X ! Y be a morphism of schemes, and let x 2 X be a smooth point of relative dimension d for f. There exists an open neighborhood U of x, such that f is smooth of relative dimension d in every point of U. (b) Let f : X ! Y be smooth of relative dimension d at x 2 X, and Y 0 ! Y be an 0 0 arbitrary morphism. Prove that the morphism X ×Y Y ! Y obtained by base 0 change is smooth of relative dimension d at all points of X ×Y Y which project to x. (c) Prove that open immersions are smooth of relative dimension 0. Problem 20 d d n Let k be a field and let X = V+(T0 + ··· + Tn ) ⊆ Pk be the Fermat hypersurface of degree d. Show that if p := char(k) does not divide d, then X is smooth over k. If p divides d, show that X is not reduced, determine Xred and show that Xred is smooth over k. References [Ma] H. Matsumura, Commutative Ring Theory, Cambridge University Press [Mu] D. Mumford, The Red Book on Varieties and Schemes, Springer LNM 1358..
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