This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 128 LECTURE 13. TANGENT SPACE Proof. Obviously, it suffices to find an open subset U of X where dim K T ( X ) x = dim X for all x U . Replacing X by an open affine set, we may assume that X is isomorphic to an open subset of a hypersurface V ( F ) A n ( K ) for some irreducible polynomial F (Theorem 4.10 of Lecture 4). This shows that we may assume that X = V ( F ) . For any x X , the tangent space T ( X ) x is given by one equation F Z 1 ( x ) b 1 + . . . + F Z n ( x ) b n = 0 . Clearly, its dimension is equal to n 1 = dim X unless all the coefficients are zeroes. The set of common zeroes of the polynomials F Z i is a closed subset of A n ( K ) contained in each hypersurface V ( F Z i ) . Obviously, F Z i 6 ( F ) unless it is equal to zero (compare the degrees). Now the assertion follows from Lemma 13.13 . Obviously, the assertion of the previous theorem is not true for a reducible set. To see this it is sufficient to consider the union of two sets of different dimension. It issee this it is sufficient to consider the union of two sets of different dimension....
View Full
Document
 Spring '09
 wei
 Algebra, Geometry

Click to edit the document details