Work out Hartshorne exercises here!

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October 13, 2009 at 2:10 pm |

1.1.1 (a) Let be the plane curve . Show that is isomorphic to a polynomial ring in one variable over .

Proof:

Define by . Then is clearly surjective, and . Thus .

(b) Let Z be the plane curve . Show that is not isomorphic to a polynomial ring in one variable over .

Proof:

Assume to the contrary that . Then there is a homomorphism such that . But every homomorphism from is determined by the images of and $y$. Say and $\phi(y) = g(t)$. Then , since is in the kernel. But then , and so the kernel of must actually contain both . But is a maximal ideal (the quotient is easily seen to be a field), so we have . Clearly , so we have a contradiction.

(c) Let $f$ be any irreducible quadratic polynomial in , and be the conic defined by . Show that or . Which one is it when?

October 13, 2009 at 2:21 pm |

1.1.12 Give an example of an irreducible polynomial whose zero set in .

Example: Let . Then is clearly irreducible, but , which is composed of two irreducible components.

October 17, 2009 at 9:15 pm |

1.1.2

The Twisted Cubic Curve. Let be the set . Show that is an affine variety of dimension 1. Find generators for the ideal . Show that is isomorphic to a polynomial ring in one variable over . We say that is given by theparametric representation.Proof:

Let denote a polynomial ring in one variable over . Define a ring homomorphism by , , which is clearly surjective. For any iff for every . Hence ker and induces an isomorphism . Since is an integral domain, so is . Hence is a prime ideal of . Note that since for any point , one can easily construct a polynomial of the form such that by choosing proper and in . Therefore, is an affine variety. Furthermore, dim = dim = dim . We claim that is generated by and . It is clear that and . For any , by the division algorithm, we have such that and such that . Hence . Since , for every . Therefore, .

October 18, 2009 at 12:22 am |

1.1.3 Let be the algebraic set in defined by the two polynomials and . Show that is a union of three irreducible components. Describe them and find their prime ideals.

Proof: Let . are prime ideals since quotient by them are all easily seen to be isomorphic to which is a domain. Hence are irreducible. By solving the equations , we get which is the unique irreducible decomposition of into three irreducible components.

October 18, 2009 at 12:52 am |

1.1.4 If we identify with in the natural way, show that the Zariski topology on is not the product topology of the Zariski topologies on the two copies of

Proof: The diagonal set is closed in the Zariski topology on . For any base open set of the product topology on , since their complements in is finite while is infinite ( is algebraicly closed). Thus . Hence . Therefore we have proved that any open set in the product topology has nonempty intersection with . Now if is closed in the product topology, then its complement is open in the product topology and they don’t intersect. Contradiction! Thus is not closed in the product topology which proves the statement.

October 18, 2009 at 2:01 am |

1.1.5 Show that a k-algebra is isomorphic to the affine coordinate ring of some algebraic set in , for some , if and only if is a finitely generated k-algebra with no nilpotent elements.

Proof: Suppose , where . Clearly is a finitely generated k-algebra. Now suppose there exists nonzero s.t. , then which implies in . Contradiction! Hence has no nilpotent.

Now suppose is a finitely generated k-algebra with no nilpotent elements. Then for some and ideal in . Now suppose , then in which implies in by assumption. Hence . Thus . Therefore is the coordinate ring of .

October 18, 2009 at 2:30 am |

1.1.6 Any nonempty open subset of an irreducible topological space is dense and irreducible. If is a subset of a topological space , which is irreducible in its induced topology, then the closure is also irreducible.

Proof. Let be open and nonempty in the irreducible space . The case is trival. Now if . Suppose is not dense, then and are nonempty and proper closed subsets of which cover . This contradicts to the fact that is irreducible. Hence is dense. Now suppose , where are closed in . Then there exist closed in s.t. . Note . Hence, say, since is irreducible. This implies and thus is irreducible.

Now suppose with closed in . Note since is closed in , also closed in . However , we get, say, since is irreducible. Therefore which imples . Hence and is irreducible.

October 18, 2009 at 4:39 am |

1.1.9 Let be an ideal which can be generated by elements. Then every irreducible component of has dimension .

Proof: Every irreducible component of is an affine variety. Furthermore, as is a Noetherian ring, is a minimal prime ideal belonging to since if and is a prime ideal, then , which contradicts the fact that is an irreducible component of . Since can be generated by elements, (Atiyah-Macdonald (11.16)). By (1.8A), , and by (1.7), . Hence .

October 18, 2009 at 7:01 am |

1.1.8 Let be an affine variety of dimension in . Let be a hypersurface in , and assume . Then every irreducible component of has dimension .

Proof: Since is a hypersurface in , for some nonconstant irreducible polynomial . Let . We note that . It follows that every irreducible component of is defined by a minimal prime ideal of that contains both and . Let and denote the image of and in under the natural map from to . Then is a minimal prime ideal of containing , and . Hence by (1.8A), . Since , is nonzero in . Hence by (1.11A), . Therefore, .

October 18, 2009 at 3:25 pm |

1.10 (a) If is any subset of a topological space , then .

Proof: Consider any closed irreducible subsets of such that . We claim that every is also irreducible in . Suppose for some nonempty closed subsets and of , then since , and are also closed subsets of , which contradicts the fact that is irreducible in . It follows that every is a closed irreducible subset of and . Hence .

(b) If is a topological space which is covered by a family of open subset , then .

Proof: By (a), . Now consider any closed irreducible subsets of such that . Since is an open cover of , there is such that . We claim that for every . Since is closed and . Suppose . Then where is a nonempty closed subset, which contradicts the fact is irreducible. Therefore, . Hence is irreducible, and we have . It follows that . Hence .

(c) Give an example of a topological space and a dense open subset with .

Example: Let and the open subsets of are and , which clearly defines a topology on . Let . Then is a dense open subset since . It is easy to see that and .

(d) If is a closed subset of an irreducible finite-dimensional topological space , and if , then .

Proof: Suppose . Then by the same argument as in (a), any chain of distinct irreducible closed subsets of is also chain of distinct irreducible closed subsets of . Since is irreducible and , one can add to the end of the chain. It follows that . Since is finite-dimensional, we have , a contradiction.

(e) Give an example of a noetherian topological space of infinite dimension.

Example: Let . Define the closed subsets of to be and , which clearly gives a topology on . Furthermore, each closed subset is clearly irreducible. Since any descending chain of closed subsets is lower bounded by , is noetherian. On the other hand, for any integer , there is a chain of distinct irreducible closed subsets. Hence is infinite dimensional.

October 18, 2009 at 3:36 pm |

1.1.7 (a) Show that the following conditions are equivalent for a topological space :

(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.

October 18, 2009 at 8:37 pm |

1.3.1 (a) Show that any conic in ${latex \bf A^2}$ is isomorphic either to or . (cf Ex. 1.1).

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!

October 22, 2009 at 2:35 pm |

1.2.1

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 .

October 22, 2009 at 2:52 pm |

1.2.2.

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.

October 22, 2009 at 11:15 pm |

1.2.3

(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 =.