Results 1 to 5 of 5

Math Help - Commutative Algebra - Going-up theorem

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    11

    Commutative Algebra - Going-up theorem

    I'm trying to prove the Going-Up theorem from Commutative Algebra using a different method to that given in the classic reference Atiyah and Macdonald. There's a couple of parts I'm having trouble with.

    All rings are commutative.

    - Let A be a subring of B
    - Let B be integral over A
    - Let \mathfrak{p} be a prime ideal of A

    1. Let \mathfrak{q} be an ideal of B which is maximal subject to  \mathfrak{q}\cap A\subseteq \mathfrak{p}. Show that
    - \mathfrak{q} is a maximal ideal of B
    - \mathfrak{q}\cap S=\emptyset, where  S=A\setminus \mathfrak{p} (solved)

    NOTE: Not sure if this should be "a maximal ideal of B", or "an ideal of B, maximal subject to \mathfrak{q}\cap S=\emptyset".

    2. Let  \mathfrak{q}'\subsetneq \mathfrak{q} be ideals of B such that \mathfrak{q}'\cap A=\mathfrak{p}=\mathfrak{q}\cap A. Show that
    - \mathfrak{q}' is NOT a prime ideal of B [HINT: Consider integrality in B/\mathfrak{q}'].

    3. Suppose that \mathfrak{q} is a maximal ideal of B, such that \mathfrak{q}\cap A\subseteq\mathfrak{p}. Show that
    - \mathfrak{q}\cap A=\mathfrak{p}. (solved)

    4. Using 1-3, deduce that if  \mathfrak{p}\subseteq \mathfrak{p}' are prime ideals of A and \mathfrak{q} is a prime ideal of B such that  \mathfrak{p}=A\cap \mathfrak{q} then there exists a prime ideal \mathfrak{q}' of B such that
    -  \mathfrak{q}\subseteq \mathfrak{q}'
    -  p'=A\cap \mathfrak{q}'.

    Any help would be greatly appreciated - I have made some progress, for example in Q1 I am fairly sure you have to suppose for a contradiction you can find an ideal $\mathfrak{r}\supsetneq\mathfrak{q}$ and then let $x\in \mathfrak{r}\setminus \mathfrak{q}$ and somehow show $(\mathfrak{q}+Bx)\cap A\subseteq \mathfrak{p}$. Not sure how to do this.

    In 2 it may be helpful to know two theorems:
    - If B is integral over A, q is an ideal of B, and P = A int q, then B/q is integral over A/P.
    - if B is integral over A and both are integral domains, then A is a field if and only if B is a field.
    Last edited by AtticusRyan; July 21st 2010 at 02:57 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by AtticusRyan View Post
    I'm trying to prove the Going-Up theorem from Commutative Algebra using a different method to that given in the classic reference Atiyah and Macdonald. There's a couple of parts I'm having trouble with.

    All rings are commutative.

    - Let A be a subring of B
    - Let B be integral over A
    - Let \mathfrak{p} be a prime ideal of A

    1. Let \mathfrak{q} be an ideal of B which is maximal subject to  \mathfrak{q}\cap A\subseteq \mathfrak{p}. Show that
    - \mathfrak{q} is a maximal ideal of B
    - \mathfrak{q}\cap S=\emptyset, where  S=A\setminus \mathfrak{p} (solved)

    NOTE: Not sure if this should be "a maximal ideal of B", or "an ideal of B, maximal subject to \mathfrak{q}\cap S=\emptyset".

    2. Let  \mathfrak{q}'\subsetneq \mathfrak{q} be ideals of B such that \mathfrak{q}'\cap A=\mathfrak{p}=\mathfrak{q}\cap A. Show that
    - \mathfrak{q}' is NOT a prime ideal of B [HINT: Consider integrality in B/\mathfrak{q}'].

    3. Suppose that \mathfrak{q} is a maximal ideal of B, such that \mathfrak{q}\cap A\subseteq\mathfrak{p}. Show that
    - \mathfrak{q}\cap A=\mathfrak{p}. (solved)

    4. Using 1-3, deduce that if  \mathfrak{p}\subseteq \mathfrak{p}' are prime ideals of A and \mathfrak{q} is a prime ideal of B such that  \mathfrak{p}=A\cap \mathfrak{q} then there exists a prime ideal \mathfrak{q}' of B such that
    -  \mathfrak{q}\subseteq \mathfrak{q}'
    -  p'=A\cap \mathfrak{q}'.

    Any help would be greatly appreciated - I have made some progress, for example in Q1 I am fairly sure you have to suppose for a contradiction you can find an ideal $\mathfrak{r}\supsetneq\mathfrak{q}$ and then let $x\in \mathfrak{r}\setminus \mathfrak{q}$ and somehow show $(\mathfrak{q}+Bx)\cap A\subseteq \mathfrak{p}$. Not sure how to do this.

    In 2 it may be helpful to know two theorems:
    - If B is integral over A, q is an ideal of B, and P = A int q, then B/q is integral over A/P.
    - if B is integral over A and both are integral domains, then A is a field if and only if B is a field.
    in 1 we can only say that q is a prime (not maximal) ideal of B. check the question again!

    in 2, suppose that q' is prime. the map f:A/p \rightarrow B/q' defined by f(a+p)=a+q' is a well-defined injective ring homomorphism and so we may assume that p=q'=\{0\}, \ q \neq \{0\}, and B is an integral domain. now let 0 \neq x \in q. then x^n + a_1x^{n-1} + \cdots + a_n=0, for some a_i \in A and n \in \mathbb{N}. we'll assume that n is as small as possible. clearly a_n \in q \cap A = \{0\}. thus x(x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1})=0, which gives us x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1}=0, because B is a domain and x \neq 0. that contradicts the minimality of n and we're done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    11
    Quote Originally Posted by NonCommAlg View Post
    in 1 we can only say that q is a prime (not maximal) ideal of B. check the question again!

    in 2, suppose that q' is prime. the map f:A/p \rightarrow B/q' defined by f(a+p)=a+q' is a well-defined injective ring homomorphism and so we may assume that p=q'=\{0\}, \ q \neq \{0\}, and B is an integral domain. now let 0 \neq x \in q. then x^n + a_1x^{n-1} + \cdots + a_n=0, for some a_i \in A and n \in \mathbb{N}. we'll assume that n is as small as possible. clearly a_n \in q \cap A = \{0\}. thus x(x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1})=0, which gives us x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1}=0, because B is a domain and x \neq 0. that contradicts the minimality of n and we're done.
    Why may we assume p=q'=\{0\},\; q\neq \{0\}, B an integral domain? Could you give your proof without making this reduction so I can try to see why you made it? Thanks heaps.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by AtticusRyan View Post
    Why may we assume p=q'=\{0\},\; q\neq \{0\}, B an integral domain? Could you give your proof without making this reduction so I can try to see why you made it? Thanks heaps.
    in my solution i replaced A,B with A/p and B/q' respectively. then p,q' in A,B become \{0\} in A/p and B/q'. also the condition q' \subset q in B becomes q \neq \{0\} in B/q'.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2009
    Posts
    11
    For 3:

    The solution I had was only partial. Assume that \mathfrak{q}\cap A\subsetneq \mathfrak{p}. Then it's not too hard to show that Ba \cap A\subset\mathfrak{p} for all a\in \mathfrak{p}\setminus (\mathfrak{q}\cap A).

    How do we then use this to derive a contradiction? Consider the ideal  \mathfrak{q}+Ba\subsetneq\mathfrak{q}. Then since \mathfrak{q} is maximal with respect to the given property we must have (\mathfrak{q}+Ba)\cap A\subsetneq\mathfrak{p}. We want to use this to derive a contradiction (I'm thinking the nonsense  \mathfrak{p}\subsetneq \mathfrak{p}).

    Is it true in this case that  (\mathfrak{q}+Ba)\cap A=\mathfrak{q}\cap A+Ba\cap A?

    And then, having done this, how do we put them all together to get 4.?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Is my non-commutative algebra correct?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 21st 2011, 05:29 PM
  2. Binomial Theorem and Commutative Rings
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 26th 2010, 08:19 AM
  3. commutative C*-algebra
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 8th 2010, 02:03 AM
  4. non-Commutative algebra
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: March 21st 2009, 02:09 AM
  5. To non-commutative algebra, please help!
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 6th 2009, 06:26 AM

Search Tags


/mathhelpforum @mathhelpforum