Discussion of exercises from Algebraic Curves – Chapter 1 – Fulton

Here is a link to the book:

http://www.math.lsa.umich.edu/~w**fulton**/**Curve**Book.pdf

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October 13, 2009 at 4:24 am |

Exercise 1.6Show that any algebraically closed field is infinite.Proof:Let be an algebraically closed field. If is finite, then we can write . We can consider:Since is algebraic closed, there exits a root of in , but for all . It is a contradiction. Hence is infinite.

October 14, 2009 at 4:36 pm |

This one is the last problem in Algebra Qual 09 fall.

October 16, 2009 at 3:09 am |

Exercise 1.7Let be a field, .(a)Show that .Proof:Use induction over . When then , by the division algorithm in we have that , where .Assume that it is true for . Let then by the division algorithm in we obtain:

Using the induction hypothesis in every , we get that:

Hence the assertion is true.

(b)If , show that for some (not unique)Proof:By part (a), implies that when . Consequently, if then there exists such that . Then we can factor out from and regroup the summation. Hence:October 18, 2009 at 3:31 pm |

Exercise 1.8Show that the algebraic subsets of are just finite subsets, together with itself.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.October 18, 2009 at 4:57 pm |

Exercise 1.10Give an example of a countable collection of algebraic sets whose union is not algebricProof: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.October 19, 2009 at 12:42 am |

Fulton 1.14 Let be a nonconstant polynomial in , an algebraically closed field.

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.

October 19, 2009 at 12:44 am |

Exercise 1.16Let be algebraic sets in . Show that if an only if .Proof:For two algebraic sets and , we have (resp. ) if and only if (resp. ), because and . Hence impliesOctober 19, 2009 at 12:54 am |

Fulton 1.20 Claim 1: For any ideal , Rad. Proof: RadRad, trivially.

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.