# Algebraic Geometry: An Introduction (Universitext) by Daniel Perrin

By Daniel Perrin

Aimed essentially at graduate scholars and starting researchers, this publication offers an creation to algebraic geometry that's fairly appropriate for people with no past touch with the topic and assumes merely the normal heritage of undergraduate algebra. it's built from a masters path given on the Université Paris-Sud, Orsay, and focusses on projective algebraic geometry over an algebraically closed base field.

The e-book begins with easily-formulated issues of non-trivial suggestions – for instance, Bézout’s theorem and the matter of rational curves – and makes use of those difficulties to introduce the basic instruments of contemporary algebraic geometry: measurement; singularities; sheaves; kinds; and cohomology. The remedy makes use of as little commutative algebra as attainable through quoting with no evidence (or proving merely in designated instances) theorems whose evidence isn't really useful in perform, the concern being to strengthen an realizing of the phenomena instead of a mastery of the strategy. quite a number routines is equipped for every subject mentioned, and a range of difficulties and examination papers are gathered in an appendix to supply fabric for additional research.

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2. 4. If R0 = k and I is homogeneous and diﬀerent from R, then I is contained in R+ . 5. Let R be a graded k-algebra. An R-module M is said to be graded if it can be written as a direct sum Mn , M= n∈Z where the k-subspaces Mn of M satisfy Rp Mq ⊂ Mp+q for all p ∈ N and q ∈ Z. A homomorphism ϕ : M → N between two graded R-modules is said to be homogeneous of degree d if, for all n, ϕ(Mn ) ⊂ Nd+n . , if x = xn ∈ Ker ϕ, then xn ∈ Ker ϕ for all n. Exercises 1 Homographies Let E be a k-vector space of dimension n + 1 and let P(E) be the associated projective space.

4 (Projective Nullstellensatz). Assume that k is algebraically closed. Let I be a homogeneous ideal of k[X0 , . . , Xn ] and set V = Vp (I). 1) Vp (I) = ∅ ⇐⇒ ∃ N such that (X0 , . . , Xn )N ⊂ I ⇐⇒ (X0 , . . , Xn ) = R+ ⊂ rac(I). 2) If Vp (I) = ∅, then Ip (Vp (I)) = rac(I). Proof. If I = R, then V = Vp (I) = ∅ and 1) is trivially true. Assume therefore that I = R. We apply the aﬃne Nullstellensatz to the cone of V : C(V ) = V (I) ⊂ k n+1 (cf. c). The statement that V = Vp (I) is empty means exactly that C(V ) contains only the origin in k n+1 and hence that rac(I) is equal to R+ , which proves 1).

But then s ∈ RY (D(f )) since RY is a sheaf. QED. 7. In fact, what we are trying to prove is that there is a surjective map of sheaves from OX to OY (cf. § 6 below). We note that if V is an open aﬃne set of Y which is the restriction to Y of an open aﬃne set U in X, then the proof above shows that any regular function on V is the image of a regular function on U . In this case, the problem mentioned above does not arise. On the other hand, we have to be more careful when the open sets are not aﬃne, as the example below shows (cf.