Math 720

Fall 2000

Exam 2

 

The purpose of these exercises is to show that the dimension of a polynomial ring does not get “much bigger” than the dimension of its coefficient ring. In this exam you may assume that all rings are commutative with identity and S is a multiplicatively closed set not containing 0. The result that we will obtain is that if R is a commutative ring with 1 such that dim(R)=n, then the dimension of R[x] is between n+1 and 2n+1. Remember to start counting at 0 where Krull dimension is concerned.

 

1.        Let be an ideal and let  be a ring containing We define the extension of I to T to be the ideal  (that is, the ideal generated by  in  

a)        Show that  Is the analogous result true for power series?

b)       Show that if  is a prime ideal, then the ideals (respectively ) are prime in (respectively ).

c)        Show that the following are isomorphic:

·         

·         

2.        Let be a multiplicatively closed subset of Show that  Is the analogous result true for power series?

 

We now assume that there are (at least) three prime ideals of R[x] in a chain lying over a fixed prime ideal of R. That is, we have  such that  for i=0,1,2.

 

 

3.        (Reduction of problem I) Consider the situation above:

a)        Use 1a) to show that  must contain

b)       Continue to use the above results to show that if  is lain over by at least three primes in a chain from  then you can assume that

c)        Use parts a) and b) and problem 1 to show that you can reduce the problem to the case where is an integral domain. More specifically, we can reduce to the statement below.

 

We have reduced to the case where R is an integral domain. In this case we have (at least) three prime ideals of R[x] in a chain such that  for i=1,2.

 

4.        (Reduction of problem II) In our new situation prove the following:

a)        Show that if is the set of nonzero elements of then  is a principal ideal domain.

b)       Show the set has empty intersection with the ideals

c)        Show that in  there must be a chain consisting of at least three prime ideals. Obtain a contradiction.

 

Finishing touches.

 

5.        Given a commutative ring with identity such that  show the following:

 

a)        Given a chain of primes of length  in exhibit a chain of length  in

b)       In problems 3 and 4 we showed that there cannot be three primes of in a chain lying over any fixed prime in  Use this information to show that