I.2.1-I.2.5

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.

I.2.1

 * the hint says it all

I.2.2

 * 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 $$ p \not \in V(a)$$

I.2.3

 * 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.

I.2.4

 * 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.

I.2.5

 * 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