Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013


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1 Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013 Throughout this lecture k denotes an algebraically closed field Tangent spaces and hypersurfaces For any polynomial f k[x 1,..., x n ] and point P = (a 1,..., a n ) A n we define the affine linear form n f f P (x 1,..., x n ) := (P )(x i a i ). x i The zero locus of f P in A n is an affine hyperplane in A n, a subvariety isomorphic to A n 1. Note that f P (P ) = 0, so the zero locus contains P. Definition Let P be a point on an affine variety V. The tangent space of V at P is the variety T P (V ) defined by the ideal {f p : f I(V )}. i=1 It is clear that T p (V ) is a variety; indeed, it is the nonempty intersection of a set of affine hyperplanes in A n and therefore an affine subspace of A n isomorphic to A d, where d = dim T P (V ). Note that the definition of T P (V ) does not require us to choose a set of generators for I(V ), but for practical applications we want to be able to compute T P (V ) in terms of a finite set of generators for I(V ). The following lemma shows that we can do this, and, most importantly, it does not matter which set of generators we pick. Lemma Let P be a point on an affine variety V. If f 1,..., f m generate I(V ), then the corresponding affine linear forms f 1,P,..., f m,p generate I(T P (V )). Proof. Let g = h i f i be an element of I(V ). Applying the product rule and the fact that f i,p (P ) = 0 yields g P = i ( hi (P )f i,p + h i,p f i (P ) ) = h i (P )f i,p, (1) which is an element of the ideal (f 1,P,..., f m,p ). Thus I(T p (V )) = (f 1,P,..., f m,p ). When considering the tangent space of a variety at a particular point P, we may assume without loss of generality that P = (0,..., 0), since we can always translate the ambient affine space A n ; this is just a linear change of coordinates (indeed, this is the very definition of affine space, it is a vector space without a distinguished origin). We can then view the affine subspace T P (V ) A n as a linear subspace of the vector space k n. The affine linear forms f P are then linear forms on k n, equivalently, elements of the dual space (k n ). Recall from linear algebra that the dual space (k n ) is the space of linear functionals λ: k n k. The orthogonal complement S (k n ) of a subspace S k n is the set of linear functionals λ for which λ(p ) = 0 for all P S; it is a subspace of (k n ), and since k n has finite dimension n, we have dim S + dim S = n. Theorem Let P be a point on an affine variety V A n with ideal I(V ) = (f 1,..., f m ). If we identify A n with the vector space k n with origin at P, the subspace of (k n ) spanned by the linear forms f 1,P,..., f m,p is T P (V ), the orthogonal complement of T p (V ). i Andrew V. Sutherland
2 Proof. This follows immediately from Lemma 17.2 and its proof; the set of linear forms in I(T P (V )) is precisely the set of linear forms that vanish at every point in T p (V ), which, by definition, is the orthogonal complement TP. Moreover, we see from (1) that every linear form in I(T P (V )) is a klinear combination of f 1,P..., f m,p. The vector space T P (V ) is called the cotangent space of V at P. As noted above, as a variety, T P (V ) is isomorphic to some A d, where d = dim T P (V ), and it follows that the dimenstion of T P (V ) as a vector space is the same as its dimension as a variety, since dim A d = d = dim k k d. The dimension of T P (V ) is then n d. Recall from Lecture 13 the Jacobian matrix f 1 f x 1 (P ) 1 x m (P ) J P = J P (f 1,..., f m ) :=... f m f x 1 (P ) n x m (P ) For a variety V with I(V ) = (f 1,..., f m ), we defined a point P V to be smooth (or nonsingular) precisely when rank J P = n dim V. Viewing J P as the matrix of a linear transformation from (k n ) to (k n ) whose image is T P (V ), we obtain the following corollary of Theorem Corollary Let P be a point on an affine variety V A n with I(V ) = (f 1,..., f m ), and let J P = J P (f 1,..., f m ). Then dim T P (V ) = rank J P and dim T P (V ) = n rank J P. In particular, the rank of J P does not depend on the choice of generators for I(V ) and P is a smooth point of V if and only if dim T P = dim V. Remark For projective varieties V we defined smooth points P as points that are smooth in all (equivalently, any) affine part containing P. One can also define tangent spaces and Jacobian matrices for projective varieties directly using generators for the homogeneous ideal of V. This is often more convenient for practical computations. Corollary makes it clear that, as claimed in Lecture 13, our notion of a smooth point P V is well defined; it does not depend on which generators f 1,..., f m of I(V ) we use to compute J P, or even on the number of generators. Now we want to consider what can happen when dim T P (V ) dim V. Intuitively, we would should expect that dim T P (V ) is then strictly greater than dim V ; this is easy to see when V is defined by a single equation, since then J P (f) has just one row and its rank is either 0 or 1. We will prove that we always have dim T P (V ) dim V by reducing to this case. Definition A variety V for which I(V ) is a nonzero principal ideal is a hypersurface. Lemma Every hypersurface in A n or P n has dimension n 1. Proof. Let V A n be a hypersurface with I(V ) = (f) for some nonzero f k[x 1,..., x n ]. We must have dim V n 1, since V A n. Let φ: k[x 1,..., x n ] k[x 1,..., x n ]/(f) be the quotient map. We must have f k, since V, so deg xi f > 0 for some x i, say x 1. If dim V < n 1 then the transcendence degree of k(v ) is less than n 1, therefore φ(x 2 ),..., φ(x n ) must be algebraically dependent as elements of k(v ). Thus there exists g k[x 2,..., x n ] such that g(φ(x 2 ),..., φ(x n )) = 0. But then φ(g) = 0, so g ker φ = (f). But this is a contradiction, since deg x1 g = 0. So dim V = n 1. If V P n, then one of its affine parts V i is a hypersurface in A n, and then dim V = dim V i = n 1.
3 The converse to Lemma 17.7 is true; every variety of codimension 1 is a hypersurface. This follows from the general fact that every variety is birationally equivalent to a hypersurface. Recall that a function field F/k if any finitely generated extension; the dimension of a function field is its transcendence degree. Theorem Let F/k be a function field of dimension n. Then there exist algebraically independent elements α 1,..., α n F and an element α n+1 algebraic over k(α 1,..., α m ) such that F = k(α 1,..., α n+1 ). The following proof is adapted from [1, App. 5, Thm. 1]. Proof. Let γ 1,..., γ m be a set of generators for F/k of minimal cardinality m, ordered so that γ 1,..., γ n is a transcendence basis (every set of generators contains a transcendence basis). If m = n then we may take γ n+1 = 0 and we are done. Otherwise γ n+1 is algebraic over k(γ 1,..., γ n ), and we claim that in fact m = n + 1 and we are also done. Suppose m > n + 1. Let f k[x 1,..., x n+1 ] be irreducible with f(γ 1,..., γ n+1 ) = 0; such an f exists since γ 1,..., γ n+1 are algebraically dependent. We must have f/ x i 0 for some x i ; if not than we must have char(k) = p > 0 and f = g(x p 1,..., xp n+1 ) = g p (x 1,..., x n+1 ) for some g k[x 1,..., x n+1 ], but this is impossible since f is irreducible. It follows that γ i is algebraic, and in fact separable, over K = k(γ 1,..., γ i 1, γ i+1,..., γ n+1 ); the irreducible polynomial f(γ 1,..., γ i 1, x i, γ i+1,..., γ n+1 ) has γ i as a root, and its derivative is nonzero. Now γ m is also algebraic over K, and it follows from the primitive element theorem [2, 6.10] that K(γ i, γ m ) = K(δ) for some δ K. 1 But this contradicts the minimality of m, so we must have m = n + 1 as claimed. Remark Theorem 17.8 holds for any perfect field k; it is not necessary for k to be algebraically closed. Theorem Every affine (resp. projective) variety of dimension n is birationally equivalent to a hypersurface in A n+1 (resp. P n+1 ). Proof. Two projective varieties are birationally equivalent if and only if all their nonempty affine parts are, and the projective closure of a hypersurface is a hypersurface, so it suffices to consider affine varieties. Recall from Lecture 15 that varieties are birationally equivalent if and only if their function fields are isomorphic, and it follows from Theorem 17.8 that every function field arises as the function field of a hypersurface: if k(v ) = k(γ 1,..., γ n+1 ) with γ 1,..., γ n ) algebraically independent, then there exists an irreducible polynomial f in k[x 1,..., x n+1 ] for which f(γ 1,..., γ n+1 ) = 0, and then V is birationally equivalent to the zero locus of f in A n+1. Corollary For any point P on an affine variety V we have dim T P (V ) dim V. Corollary The set of singular points of a variety is a closed subset; equivalently, the set of nonsingular points is a dense open subset. Proof. It suffices to prove this for affine varieties. So let V A n be an affine variety with ideal (f 1,..., f m ), and for any P V let J P = J P (f 1,..., f m ) be the Jacobian matrix. Then Sing(V) := {P: dim T P (V) > dim V} = {P: rank J P < n dim V} 1 As noted in [2], to prove K(α, β) = K(δ) for some δ K(α, β), we only need one of α, β to be separable.
4 is the set of singular points on V. Let r = n dim V. We have rank J P < r if and only if every r r minor of J P has determinant zero. If we now consider the matrix of polynomials ( f i / x j ), the determinant of each of its r r minors is a polynomial in k[x 1,..., x n ], and Sing(V) is the intersection of V with the zero locus of all these polynomials. Thus Sing(V) is an algebraic set, hence closed. Recall the onetoone correspondence between points P = (a 1,..., a n ) in A n and maximal ideals M P = (x 1 a 1,..., x n a n ) of k[a n ]. If V A n is an affine variety, then the maximal ideals m P of its coordinate ring k[v ] = k[a n ]/I(V ) are in onetoone correspondence with the maximal ideals M P of k[a n ] that contain I(V ); these are precisely the maximal ideals M P for which P V. If we choose coordinates so that P = (0,..., 0), then M P is a kvector space that contains MP 2 as a subspace, and the quotient space M P /MP 2 is then also a kvector space. Indeed, its elements correspond to (cosets of) linear forms on k n. We may similarly view m P, m 2 P, and m P /m 2 P as kvector spaces, and this leads to the following theorem. Theorem Let P be a point on an affine variety V. Then T P (V ) m P /m 2 P. Proof. As above we assume without loss of generality that P = (0,..., 0). Then M P consists of the polynomials in k[x 1,..., x n ] for which each term has degree at least 1 (equivalently, constant term 0). We now consider the linear transformation D : M P (k n ) that sends f M P to the linear form f P (k n ). This map is surjective, and its kernel is MP 2 ; we have f P = 0 if and only if f/ x i (0) = 0 for i = 1,..., n, and this occurs precisely when every term in f has degree at least 2, equivalently, f MP 2. It follows that M P /M 2 P (k n ). The restriction map (k n ) (T P ) that restricts the domain of a linear form on k n to T P (V ) is surjective, and composing this with D yields a surjective linear transformation d: M P T P (V ) whose kernel we claim is equal to MP 2 + I(V ) (this is a sum of ideals in k[x 1,..., x n ] that is clearly a subset of M P ). A polynomial f M P lies in ker d if and only if the restriction of f P to T P (V ) is the zero function, which occurs if and only if f P = g P for some g I(V ), since T P the zero locus of g P for g I(V ). But this happens if and only if f g lies ker D = MP 2, equivalently, f M P 2 + I(V ). We therefore have T P (V ) M P MP 2 + I(V ) M P /I(V ) (MP 2 + I(V )/I(V ) = M P /I(V ) MP 2 /I(V ) m P /m 2 P. Corollary The smooth points P on a variety V are precisely the points P for which dim m P /m 2 P = dim V = dim k[v ] The three dimensions in the corollary above are, respectively, the dimension of m P /m 2 P as a kvector space, the dimension of V as a variety, and the Krull dimension of the coordinate ring k[v ]; as noted in Lecture 13, we always have dim V = dim k[v ]. The key
5 point is that we now have a completely algebraic notion of smooth points. If R is any affine algebra, the maximal ideals m of R correspond to smooth points on a variety with coordinate ring R, and we can characterize the smooth maximal ideals as those for which dim k m/m 2 = dim R, where k = R m /m is now the residue field of the localization of R at m. Smooth varieties then correspond to affine algebras R in which every maximal ideal is smooth. References [1] I. R. Shafarevich, Basic algebraic geometry, 2nd edition, SpringerVerlag, [2] B. L. van der Waerdan, Algebra, Volume I, 7th edition, Springer, 1991.
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