128LECTURE 13.TANGENT SPACEProof.Obviously, it suffices to find an open subsetUofXwheredimKT(X)x=dimXfor allx∈U. ReplacingXby an open affine set, we may assume thatXisisomorphic to an open subset of a hypersurfaceV(F)⊂An(K)for some irreduciblepolynomialF(Theorem4.10of Lecture 4).This shows that we may assume thatX=V(F). For anyx∈X, the tangent spaceT(X)xis given by one equation∂F∂Z1(x)b1+. . .+∂F∂Zn(x)bn= 0.Clearly, its dimension is equal ton-1 = dimXunless all the coefficients are zeroes.The set of common zeroes of the polynomials∂F∂Ziis a closed subset ofAn(K)containedin each hypersurfaceV(∂F∂Zi). Obviously,∂F∂Zi∈(F)unless it is equal to zero (comparethe degrees). Now the assertion follows from Lemma13.13.Obviously, the assertion of the previous theorem is not true for a reducible set. Tosee this it is sufficient to consider the union of two sets of different dimension. It iseasy to modify the statement to extend it to the case of reducible sets.
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