By Ulrich Gortz, Torsten Wedhorn

ISBN-10: 3834806765

ISBN-13: 9783834806765

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**Additional info for Algebraic geometry 1: Schemes**

**Example text**

Proof. Clearly we have Γ(X)f ⊆ OX (D(f )). Let g ∈ OX (D(f )) and set a = { h ∈ Γ(X) ; hg ∈ Γ(X) }. Obviously a is an ideal of Γ(X) and we have to show that f ∈ rad(a). By Hilbert’s Nullstellensatz we have rad(a) = I(V (a)). Therefore it suﬃces to show f (x) = 0 for all x ∈ V (a). , x ∈ D(f ). As g ∈ OX (D(f )), we ﬁnd / mx , with g = gg12 . Thus g2 ∈ a and as g2 (x) = 0 we have x ∈ / V (a). 41. If X is an irreducible aﬃne algebraic set, U ⊆ X open, and f ∈ OX (U ), there do not necessarily exist g, h ∈ Γ(X) with f = hg ∈ K(X) and h(x) = 0 for all x ∈ U .

This certainly holds for prime ideals and therefore for arbitrary intersections of prime ideals as well. That proves (1). 1). A closed set V (b) (for some ideal b) contains Y if and only if b is contained in all prime ideals that belong to Y . This is equivalent to b ⊆ I(Y ). Therefore V (I(Y )) is the smallest closed subset of Spec A containing Y . This shows the second assertion of (2). Part (3) follows from (2). In particular we see that the closure of a set consisting of only one point x ∈ Spec A is the set V (px ) of prime ideals containing px .

Xn ). (2) The set of closed subspaces Z of Pn (k). (3) The set of closed aﬃne cones in C ⊆ An+1 (k) such that C = {0}. If Z ⊆ Pn (k) is a closed subset we denote by I+ (Z) the corresponding homogeneous ideal. Show that I+ (Z) = I(C(Z)) and deduce that the following assertions are equivalent. (i) Z is irreducible. (ii) I+ (Z) is a prime ideal. (iii) C(Z) is irreducible. 22. Let L1 and L2 be two disjoint lines in P3 (k). (a) Show that there exists a change of coordinates such that L1 = V+ (X0 , X1 ) and L2 = V+ (X2 , X3 ).

### Algebraic geometry 1: Schemes by Ulrich Gortz, Torsten Wedhorn

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