By Richard Bellman (ed.)

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10. If X is an affine algebraic set, and P ∈ X is a nonsingular point, then P is contained in a unique irreducible component of X. In particular, a connected nonsingular algebraic set is automatically a variety. This fits with intuition – a point where two components intersect should not be nonsingular – but it’s not so obvious from either definition. 5. 11. If X is an affine algebraic set, then the set of singular points of X is a nowhere dense closed subset of X. Proof. Let Z ⊆ X be the points which are contained in more than one irreducible component of X; then Z is closed.

Given prevarieties X, Y , we define the product X × Y of X with Y to be the product set X × Y , equipped with the atlas ∼ ϕi × ϕj : Xi × Yj → Ui × Vj , and the topology induced by the atlas. 4. Show the following: (a) The above definition gives a valid prevariety. (b) If X and Y are affine, this definition is consistent with the one we already have (and used in the above) for products of affine varieties. (c) If Y ⊆ X is a subprevariety, then the topology on Y × Y ⊆ X × X is the subset topology.

5 says that TP (X) is the orthogonal complement in k n of the rowspace of J(f1 , . . , fm )(P ), so we have dimk TP (X) = n − r n − r. 2) that dimk TP∗ (X) d, and the definition of nonsingularity. 15) that every component of X has dimension at least n − m, but the hypotheses then imply that if Z is any component of X containing P , then n − m = dimk TP (X) dim Z n − m, so we must have equality and the statement follows. 3. Suppose H = Z(f ) ⊆ Ank is a hypersurface, with f irreducible (so that I(H) = (f )).

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