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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.

### HAPPY

#### 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