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I.2.1 can be use to solve I.2.2 and I.2.3 giving the latter two height and volume 2. Similarly I.2.3 can be used to solve I.2.4 giving the it height 2 and volume 3. The rest of the problems have height 1.



  • the hint says it all


  • For (1) ==> (2) use I.2.1
  • Here is (3) ==> (1). For any projective point p, some coordinate of p doesn't vanish, say  p_j \ne 0. Then x_j^d is not zero at p hence  p \not \in V(a)


  • The first three parts are basic algebraic reasoning.
  • For d, if V(a) is nonempty then  a is a proper homogeneous ideal and I.2.1 implies I(V(a)) \subset \sqrt{a}, the other inclusion follows because a field has no nonzero nilpotentns.
  • For e, give parts a and b the same proof that is used in prop. I.1.2 works here.


a) Say Y_i = \overline{Y_i} are algebraic sets such that I(Y_1) = I(Y_2) then using I.2.3e I(Y_i) = I(Y_1) \cap I(Y_2) = I(Y_1 \cup Y_2) etc so  Y_1 \cup Y_2 \subset Y_i which shows  Y_1 = Y_2.
a similar argument works to show V(I_1) = V(I_2) implies  \sqrt I_1 = \sqrt I_2
b) similar to the affine argument given in section 1.1
c) follows from b) because (0) is a prime ideal.


  • The first statement follows because a descending chain of closed sets corresponds to an ascending chain of radical ideals in a noetherian ring.
  • The second statement follows form prop. I.1.5 ( for a noeth. top. sp. every closed can be expressed as finite union of irred. closed sets.)

You'll Want to use proposition I.1.5

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