# hw3 - You can assume that this gives a well-deﬁned...

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Math W4051: Problem Set 3 due Wednesday, September 26 Reading: Munkres, Ch.2, § 18-22. 1. Munkres, Ch.2 § 17, exercises 6, 7, 8, 14, 15, 20 on p. 100-102. 2. Munkres, Ch.2 § 19, exercises 3, 7 on p. 118. 3. Munkres, Ch.2 § 20, exercise 5 on p. 127. 4. Recall the deﬁnition of the Zariski topology on C n given in class: a subset A C n is closed in the Zariski topology if and only if there exists a ﬁnite collection of polyno- mial functions f 1 ( z 1 , . . . , z n ) , f 2 ( z 1 , . . . , z n ) , . . . , f m ( z 1 , . . . , z n ) in n complex variables (with complex coeﬃcients), such that A is their common zero set A = { ( z 1 , . . . , z n ) C n | f k ( z 1 , . . . , z n ) = 0 for all k = 1 , . . . , m } .
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Unformatted text preview: You can assume that this gives a well-deﬁned topology. (a) Show that the standard topology on R 2 n ∼ = C n is ﬁner than the Zariski topology. ( Note: By “ﬁner”, I always mean “at least as ﬁne”; see the Deﬁnition in Munkres, p. 77.) (b) Consider the subset A ⊂ C n consisting of all points of the type ( s, , . . . , 0) with s ∈ Z . Prove that A is not closed in the Zariski topology. Conclude that the standard topology is strictly ﬁner than the Zariski topology. Challenge (extra credit): Munkres, Ch.2 § 17, exercise 21 on p. 102....
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