INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

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14 MezzettiIt is a map a : K[x 0 , x 1 , . . ., x n ] → K[y 1 , . . ., y n ] such thata (F(x 0 , . . ., x n )) = a F(y 1 , . . ., y n ) := F(1, y 1 , . . ., y n ).Note that a is a ring homomorphism.(ii) homogeneization of polynomials with respect to x 0 .It is a map h : K[y 1 , . . ., y n ] → K[x 0 , x 1 , . . ., x n ] defined by.h (G(y 1 , . . ., y n )) = h G(x 0 , . . ., x n ) := x deg G0G( x 1x 0, . . ., x nx 0)h G is always a homogeneous polynomial of the same degree as G. The map his clearly not a ring homomorphism. Note that always a ( h G) = G but in generalh ( a F) ≠ F; what we can say is that, if F(x 0 , . . ., x n ) is homogeneous, then ∃r ≥ 0such that F = x r 0( h ( a F)).Let X ⊂ U 0 be closed in the topology induced by the Zariski topology ofthe projective space, i.e. X = U 0 ∩ V P (I) where I is a homogeneous ideal ofK[x 0 , x 1 , . . ., x n ]. Define a I = { a F | F ∈ I}: it is an ideal of K[y 1 , . . ., y n ](because a is a ring homomorphism). We prove that φ 0 (X) = V ( a I). For:let P[x 0 , . . ., x n ] be a point of U 0 ; then φ 0 (P) = ( x 1x 0, . . ., x nx0) ∈ φ 0 (X) ⇐⇒P[x 0 , . . ., x n ] = [1, x 1x 0, . . ., x nx0] ∈ X = V P (I) ⇐⇒ F(1, x 1x 0, . . ., x nx0) = 0 ∀ a F ∈a I ⇐⇒ φ 0 (P) ∈ V ( a I).Conversely: let Y = V (α), α ideal of K[y 1 , . . ., y n ], be a Zariski closed setof A n . Let h α be the homogeneous ideal of K[x 0 , x 1 , . . ., x n ] generated by the set{ h G | G ∈ α}. We prove that φ −10(Y ) = V P ( h α) ∩ U 0 . In fact: [1, x 0 , . . ., x n ] ∈φ −1 (Y ) ⇐⇒ (x0 1 , . . ., x n ) ∈ Y ⇐⇒ G(x 1 , . . ., x n ) = h G(1, x 1 , . . ., x n ) = 0 ∀ G ∈α ⇐⇒ [1, x 1 , . . ., x n ] ∈ V P ( h α).□From now on we will often identify A n with U 0 via φ 0 (and similarly with U ivia φ i ). So if P[x 0 , . . ., x n ] ∈ U 0 , we will refer to x 0 , . . ., x n as the homogeneouscoordinates of P and to x 1x 0, . . ., x nx0as the non–homogeneous or affine coordinatesof P.Exercises to §3.1*. Let n ≥ 2. Prove that, if K is an algebraically closed field, then in A n Kboth any hypersurface and any complementar set of a hypersurface have infinitelymany points.2. Prove that the Zariski topology on A n is T 1 .3*. Let F ∈ K[x 0 , x 1 , . . ., x n ] be a homogeneous polynomial. Check that itsirreducible factors are homogeneous. (hint: consider a product of two polynomialsnot both homogeneous...)4. The ideal of an algebraic set and the Hilbert Nullstellensatz.
Introduction to algebraic geometry 15Let X ⊂ A n be an algebraic set, X = V (α), α ⊂ K[x 1 , . . ., x n ]. The ideal αdefining X is not unique: for example, let X = {0} ⊂ A 2 ; then 0 = V (x 1 , x 2 ) =V (x 2 1, x 2 ) = V (x 2 1, x 3 2) = V (x 2 1, x 1 , x 2 , x 2 2) = . . . Nevertheless, there is an ideal wecan canonically associate to X, i.e. the biggest one. Precisely:4.1. Definition. Let Y ⊂ A n be any set.The ideal of Y is I(Y ) = {F ∈ K[x 1 , . . ., x n ] | F(P) = 0 for any P ∈ Y } ={F ∈ K[x 1 , . . ., x n ] | Y ⊂ V (F)}: it is formed by all polynomials vanishing on Y .Note that I(Y ) is in fact an ideal.For instance, if P(a 1 , . . ., a n ) is a point, then I(P) = 〈x 1 − a 1 , . . ., x n − a n 〉.Indeed all its polynomials vanish on P, and, on the other side, it is maximal.The following relations follow immediately by the definition:(i) if Y ⊂ Y ′ , then I(Y ) ⊃ I(Y ′ );(ii) I(Y ∪ Y ′ ) = I(Y ) ∩ I(Y ′ );(iii) I(Y ∩ Y ′ ) ⊃ I(Y ) + I(Y ′ ).Similarly, if Z ⊂ P n is any set, the homogeneous ideal of Z is, by definition,the homogeneous ideal of K[x 0 , x 1 , . . ., x n ] generated by the set {G ∈K[x 0 , x 1 , . . ., x n ] | G is homogeneous and V P (G) ⊃ Z}. It is denoted I h (Z).Relations similar to (i),(ii),(iii) are satisfied. I h (Z) is also the set of polynomialsF(x 0 , . . ., x n ) such that every point of Z is a projective zero of F.Let α ⊂ K[x 1 , . . ., x n ] be an ideal. Let √ α denote the radical of α, i.e. theideal {F ∈ K[x 1 , . . ., x n ] | ∃r ≥ 1 s.t. F r ∈ α}. Note that always α ⊂ √ α; ifequality holds, then α is called a radical ideal.4.2. Proposition.1) For any X ⊂ A n , I(X) is a radical ideal.2) For any Z ⊂ P n , I h (Z) is a homogeneous radical ideal.Proof. 1) If F ∈ √ I(X), let r ≥ 1 such that F r ∈ I(X): hence if P ∈ X, then(F r )(P) = 0 = (F(P)) r in the base field K. Therefore F(P) = 0.2) is similar, taking into account that I h (Z) is a homogeneous ideal (seeExercise 4.7.).□We can interpret I as a map from P(A n ), the set of subsets of the affine space,to P(K[x 1 , . . ., x n ]). On the other hand, V can be seen as a map in the oppositesense. We have:4.3. Proposition. Let α ⊂ K[x 1 , . . ., x n ] be an ideal, Y ⊂ A n be any subset.Then:(i) α ⊂ I(V (α));(ii) Y ⊂ V (I(Y ));(iii) V (I(Y )) = Y : the closure of Y in the Zariski topology of A n .
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Page 45 and 46: Then v n,d (X) ≃ X: it is the set
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