Algebraic geometry 1: Schemes by Ulrich Gortz, Torsten Wedhorn

By Ulrich Gortz, Torsten Wedhorn

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Xn ]d → { g ∈ R[T0 , . . , Ti , . . , Tn ] ; deg(g) ≤ d }, f → f (T0 , . . , 1, . . , Tn ). ) Proof. We construct an inverse map. Let g be a polynomial in the right hand side set d and let g = j=0 gj be its decomposition into homogeneous parts (with respect to T for = 0, . . , n, = i). Define 27 d Xid−j gj (X0 , . . , Xi , . . , Xn ). Ψi (g) = j=0 It is easy to see that Φi and Ψi are inverse to each other (as both maps are R-linear, it suffices to check this on monomials). The map Φi is called dehomogenization, the map Ψi homogenization (with respect to Xi ).

B) Let V = V (X 2 − Y Z, XZ − X) ⊆ A3 (k). Show that V consists of three irreducible components and determine the corresponding prime ideals. 6. Let f ∈ k[X1 , . . , Xn ] be a non-constant polynomial. Write f = i=1 fini with irreducible polynomials fi such that (fi ) = (fj ) for all i = j and integers ni ≥ 1. Show that rad(f ) = (f1 · · · fr ) and that the irreducible components of V (f ) ⊆ An (k) are the closed subsets V (fi ), i = 1, . . , r. 7. Let f ∈ k[T1 ] be a non-constant polynomial.

69 it suffices to show that two isomorphic quadrics have the same dimension. Let Q ⊆ Pn (k) be a quadric of rank r. We show that the transcendence degree over k of the function field (of an irreducible component) of Q is always equal to the dimension. As isomorphic prevarieties have isomorphic 2 ). function field, this shows the corollary. We may assume that Q = V+ (X02 + · · · + Xr−1 n−1 ∼ ∼ (k), and thus K(Q) = k(T1 , . . , Tn−1 ) and For r = 1 we have Q = V+ (X0 ) = P trdegk K(Q) = n − 1. For r = 2 the two irreducible components Z1 and Z2 of Q are given by a linear equation and thus are hyperplanes in Pn (k).

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