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## #1 2010-05-10 21:43:21

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)

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## #2 2010-05-11 16:46:56

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?

"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..."

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## #3 2010-05-11 20:45:02

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.:)

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## #4 2010-05-12 07:38:54

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)

"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..."

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## #5 2010-05-12 21:23:34

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
?

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## #6 2010-05-13 05:06:53

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.

"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..."

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## #7 2010-05-13 16:30:05

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?

"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..."

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