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laipou

Member

Registered: 2009-09-19

Posts: 24

questions about polynomial rings and prime ideals

(1)
Let B:=K[x,y]/(y^2-x^2-x^3) and A = K[x], where K is an algebraically closed field .
Regard B as an extension over A.
For all prime ideals p in A,determine prime ideals of B lying over p.

This problem appears in my text book in exercises after the section " integral extension".
But I don't have a clue how to solve this... .
I am unfamiliar with problems  concerning " quotient rings of polynomial rings".
Hope someone can explain this in some detail.

(2)
Let B be a commutative ring with identity and

be a subring of B(1 in A).
Let q be a prime ideals in B and

.
Then p is an prime ideal in A.
Consider two local rings B_q and A_p.
I want to identify A_p as a subgroup of B_q by a/s -> a/s,where a in A,s in A\p.
But I can't show that this homomorphism is 1 to 1.
Need some help.

Last edited by laipou (2010-05-10 21:43:56)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: questions about polynomial rings and prime ideals

1. Do you see that B is an integral extension of A?

2. It is unusual to talk about lying over without the assumption that it is an integral extension.  You sure you have the question right?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: questions about polynomial rings and prime ideals

1.Yes,and I know by thm there exists one and if q1,q2 are two of them s,t

,then q1=q2.
   But I dow know how to explicitily write down one for each p.
2.You are right,it is a integral extension.:)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: questions about polynomial rings and prime ideals

1. Hopefully you know that the prime ideals in K[x] are generated by (f) where f is irreducible and monic.  Now when we move to the larger ring K[x,y]/(y^2-x^2-x^3), you really only have to consider multiplication by a single polynomial to get an ideal.  Now all that remains is to check if it's prime.

2. Recall that in a localized ring, an element a/s is zero if and only if there exists a d in R with da = 0.  So we assume there is a d in B with da = 0 for some a in A.  Prove that there is an element c in A with ca = 0 as well.  (Hint: Integral extension)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: questions about polynomial rings and prime ideals

1.Since A is a UFD,prime ideals in A coincide with maximal ideals.
   Also,a maximal ideal in A if of the form (p(x)),where p is irreducible over K.
   Since K is algebraically closed,we have a prime ideal(maximal ideal) in A is of the form (x-a),a in K.
   Suppose q is a prime ideal in B lying over p,by thm we also have q is a maximal ideal in B.
   Above is all I know by now.
   But I still have no idea how to explicitily find one.
   perhaps you can do one example for me?(fix one (x-a)) 

2."An element a/s is zero if and only if there exists a d in R with da = 0."
   I think d should be in S(multiplicative closed set).
   For my case,d should be B\q and

.
   So

.
   How to show

?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: questions about polynomial rings and prime ideals

1. I'll do the easy one for you.

(x) is prime in K[x], and now consider it as a set in K[x,y]/(y^2-x^3-x^2).  Since q must contain (x) (in K[x]), we conclude that yx is in q as well.  Therefore, q contains (x) (in K[x,y]/(y^2-x^3-x^2) ).  Modding out our quotient polynomial ring by (x) leaves the ring K[y]/(y^2).  This shows (if you didn't see it already) that y*y is in (x) but y is not.  To make q prime, it must be that y is in q.  Now q contains (x,y).  This is clearly a maximal ideal and therefore prime.  Therefore q = (x,y).

Try to do the same with x-1.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: questions about polynomial rings and prime ideals

Suppose a_n is in p.  Then so is

We conclude that

This is a problem.  Why?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."