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By Taylor J.L.

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X1 x2 · · xn 1 1 · · 1 where in the second determinant the f (xi ) replace the j th column of the first determinant. Now, of course, the first determinant is the Vandermonde determinant which has square equal to d by F8. Thus, aj is the product of the Vandermonde and the determinant obtained from the Vandermonde by replacing its j th column with the column formed by the f (xi ). Clearly this product is left fixed by any permutation of the roots x1 , . . , xn since this just amounts to applying the same permutation to the rows in both matrices.

4. Then either j H ∩ I = 0, in which case we are through, or there is a nonzero fj ∈ j H ∩ I. The function fj can be made regular in zj by a linear change of coordinates that involves only the first j coordinates and, hence, does not effect the regularity of the functions chosen previously. The Lemma follows by induction. The notion of an ideal being regular in the variables zm+1 , . . , zn seems to depend on the ordering of these variables. However, the next lemma shows that it depends only on the decomposition Cn = Cm × Cn−m and not on the choice of coordinate systems within the two factors.

The vector space of all tangent vectors is called the tangent space to V and is denoted T (V ). Its dimension is the tangential dimension of V and is denoted tdim V . 50 J. L. 11 Theorem. The vector space T (V ) is naturally isomorphic to the dual of M/M 2 where M is the maximal ideal of V H (V O). Proof. If t ∈ T (V ) then t(1) = 2t(1) and so t kills constants and is, thus, determined by its restriction to M . However, if t is any linear functional on V H (V O) which kills constants and f = 1 + f1 , g = 1 + g1 with f1 , g1 ∈ M then t(f g) = t(f1 ) + t(g1 ) + t(f1 g1 ) = g(0)t(f ) + f (0)t(g) + t(f1 g1 ) from which we conclude that t is a tangent vector if and only if t vanishes on M 2 .

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