INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

48 Mezzettia) The cuspidal cubic Y = V (x 3 − y 2 ) ⊂ A 2 .We have seen that Y is not isomorphic to A 1 , but in fact Y and A 1 arebirational. Indeed, the regular map φ : A 1 → Y , t → (t 2 , t 3 ), admits a rationalinverse ψ : Y A 1 , (x, y) → y x. ψ is regular on Y \ {(0, 0)}, ψ is dominant andψ◦φ = 1 A 1, φ◦ψ = 1 Y as rational maps. In particular, φ ∗ : K(Y ) → K(X) is a fieldisomorphism. Recall that K[Y ] = K[t 1 , t 2 ], with t 2 1 = t3 2 , so K(Y ) = K(t 1, t 2 ) =K(t 2 /t 1 ), because t 1 = (t 2 /t 1 ) 2 = t 2 2 /t2 1 = t3 1 /t2 1 and t 2 = (t 2 /t 1 ) 3 = t 3 2 /t3 1 = t3 2 /t2 2 ,so K(Y ) is generated by a unique transcendental element. Notice that φ and ψestablish isomorphisms between A 1 \ {0} and Y \ {(0, 0)}.b)Rational maps from P 1 to P n .Let φ : P 1 P n be rational: on some open U ⊂ P 1 ,φ([x 0 , x 1 ]) = [F 0 (x 0 , x 1 ), . . ., F n (x 0 , x 1 )],with F 0 , . . ., F n homogeneous of the same degree, without non–trivial commonfactors. Assume that F i (P) = 0 for a certain index i, with P = [a 0 , a 1 ]. ThenF i ∈ I h (P) = 〈a 1 x 0 −a 0 x 1 〉, i.e. a 1 x 0 −a 0 x 1 is a factor of F i . This remark impliesthat ∀ Q ∈ P 1 there exists i ∈ {0, . . ., n} such that F i (Q) ≠ 0, because otherwiseF 0 , . . ., F n would have a common factor of degree 1. Hence we conclude that φ isregular.c) Projections.Let φ : P n P m be given in matrix form by Y = AX, where A is a(m + 1) × (n + 1)-matrix, with entries in K. Then φ is a rational map, regular onP n \ P(KerA). Put Λ := P(KerA). If A = (a ij ), this means that Λ has cartesianequations⎧a 00 x 0 + . . . + a 0n x n = 0⎪⎨a 10 x 0 + . . . + a 1n x n = 0⎪⎩ . . .a m0 x 0 + . . . + a mn x n = 0The map φ has a geometric interpretation: it can be seen as the projectionof centre Λ to a complementar linear space. First of all, we can assume that rkA = m + 1, otherwise we replace P m with P(Im A); hence dimΛ = n − (m + 1).Consider first the case Λ : x 0 = . . . = x m = 0; we identify P m with thesubspace of P n of equations x m+1 = . . . = x n = 0, so Λ and P m are complementarsubspaces, i.e. Λ ∩ P m = ∅ and the linear span of Λ and P m is P n . Then, forQ ∈ P n \ Λ, φ(Q) = [x 0 , . . ., x m , 0, . . ., 0]: it is the intersection of P m with thelinear span of Λ and Q. In fact, if Q[a 0 , . . ., a n ] then ΛQ has equations{a i x j − a j x i = 0, i, j = 0, . . ., m (check!)so ΛQ ∩ P m has coordinates [a 0 , . . ., a m , 0, . . ., 0].