(a) If are subsets of , then .

(b) If are subsets of , then .

(c) For any two subsets , of , .

(d) If is a homogeneous ideal with , then .

(e) For any subset , .

Pf:

(a) If , then , . However , hence,. Therefore .

(b) If , then , . However , hence,. Therefore .

(c) If , then ,, hence , and ,. Therefore .

If , then , and , hence ,. Therefore .

(d) Since , can not contain any deg 0 polynomial.

Then , with positive degree, we have , by 1.2.1 “homogeneous Nullstellensatz”, .

Therefore .

Conversely, if for some , then , , hence . Therefore .

(e) Use corollary 2.3, is covered by the open sets , , each is homeomorphic to an affine variety. Then for each we have .

Still we have .

So =closure of =.

For a homogeneous ideal , show that the following conditions are equivalent:

(i) ;

(ii) either or the ideal ;

(iii) for some .

(i)(ii): If and deg , for all

since , hence by 1.2.1, , . Hence .

If contains any element of , then it contains the whole since is an ideal, so either or the ideal .

(ii)(iii): Noticed that our .Then for each , there is an integer such that . So for

N =max {}, we have . Therefore for , every monomial of degree d contains a multiple of

for some i, so .

(iii)(i): If for some , then contains for . This implies that

consists of those points which are 0 for all , and hence is empty.

Prove the “homogeneous Nullstellensatz”, which says that if is a homogeneous

ideal, and if is a homogeneous polynomial with deg f > 0, such that for all

in , then for some .

Proof:

Let be a homogeneous ideal and a homogeneous polynomial with

deg f > 0, such that for all

in . Then f vanishes for all representatives

for points in Z(a) in , so for some when applying Theorem

1.3A fot Z(a) as being in . Then for some as homogeneous polynomial and homogeneous

ideal in . ]]>

Conversely, say Then for Rad we have for precisely the such that . We need to show that . It’s very easy to see is a prime ideal. For are such that or or . Thus . We conclue Rad.

Claim 2: Rad. Proof: Suppose . Then for any we have . We conclude Rad.

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*Proof:* For two algebraic sets and , we have (resp. ) if and only if (resp. ), because and . Hence implies

Claim 1: is infinite if . Proof: It’s clear that as is an algebraically closed field is infinite. This implies, of course, that is infinite. Now suppose is finite. Then, there exists such that . This implies , but this only occurs when . As is clearly nonzero, we must have . But is nonconstant by hypothesis. We conclude must be infinite. Claim 2: If , then is infinite. Proof: For every -tuple we have . Because is algebraically closed, there exists such that . Thus we have shown for every -tuple there exists such that is a root of . Thus is infinite.

]]>Proof: By Ex.1.1.1(c), for any conic in , is isomorphic to either or defined in 1.1.1(a) and (b). Since is isomorphic to , a polynomial ring in one variable over , and , by Corollary (3.7). We claim that . Define by , which is clearly bijective. The close subsets of are any finite subsets without including 0. Every close subset of is the intersection of a finite set of polynomials in (since is noetherian) with the curve and hence contains only finite points. It follows that the close subsets of and has a natural correspondence under . Hence is bicontinuous. For any open set and for every regular function , the function is clearly a regular function since a rational function in terms of and is also a rational function in terms of . Hence is a morphism. It is easy to see that is also a morphism. Therefore, is an isomorphism of varieties.

(b) Show that is not isomorphic to any proper open subset of itself. (This result is generalized by (Ex.6.7) below.)

Proof: Let denote a proper open subset of . Then for some . Let . Then is regular on . Suppose . Then as -algebras by Proposition (3.5). However, only elements in are invertible in , while is invertible in , a contradiction!

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*Proof:* Let and for all integers . is an algebraic set of and . Thus is a countable collection of algebraic sets, and . We have that and . By problem , is not algebraic.

(i) is notherian (ii) Every nonempty family of closed subsets has a minimal element (iii) X satisfies the ascending chian condition for open subsets (iv) Every nonempty family of open subsets has a maximal element;

(b) A notherian topological space is quasi-compact

(c) Any subset of a notherian topological space is notherian in its induced topology

(d) A notherian space which is also Hausdorff must be a finite set with the discrete topology

Proof: (a) (i)(ii) Suppose not, then for such a family, we can construct a non-ending descending chain of closed subsets starts any element of this family; (ii)(iii) Suppose not, choose a non-ending ascending chain of open sets, then the complement of this chain has no minimal element; (iii)(iv) Suppose not, then for such a family, we can construct a non-ending ascending chain of open subsets starts any element of this family (iv)(i) Then every family of nonempty closed subsets has a minimal element, thus every descending chain of closed subsets stablizes since the set they form has a minimal element

(b) Suppose is a notherian space which is not quasi-compact, then there exists an open cover of has no finite subcover. Choose from the cover, it doesn’t cover . Hence we can choose from the cover s.t. . Still doesn’t cover , we can choose from the cover with and doesn’t cover . Proceeding in this way, we can construct a non-stationary ascending chain of open sets which contradicts to the fact that is notherian.

(c) Let be a subset of . Let be a descending chain of closed subets of . Now is a descending chain of closed subsets in , hence stationary. Now note , Therefore is stationary. Hence is notherian

(d) Let be a notherian space which is Hausdorff. Suppose has more than two points. is Hausdorff since is. Choose in . Then there exist non-intersect open subsets of s.t. . Then and thus is reducible. We have proved that those irreducible subsets of are one-point subsets. Since is notherian, is a finite union of irreducible components, which in this case, is a finite union of closed points. Hence is a finite set with discrete topology.

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*Proof:* Suppose is an algebraic subset of . If then there exists a non-zero polynomial . has at most a finite number of roots, and thus and are finite. Hence the algebraic subsets of are just finite subsets, together with itself.