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Math Help - Subrings and Ideals Questions

  1. #1
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    Subrings and Ideals Questions

    1) Which of the following subsets of M2(R) are subrings?
    a) S = all matrices of the form
    (0 b)
    (c d)
    (Sorry, I don't know how to do matrices. These are all each one matrix)
    b) S = all matrices of the form
    (a 0)
    (c d)
    c) S = GL(2, R)
    d) S = all matrices of the form
    (a b)
    (b a)

    2) Find a maximal ideal in
    a) Z6
    b) Z12
    c) Z18

    3) Let R = {q in Q s.t. q = a/b, a,b in Z and b is odd). Show that R has a unique maximal ideal.

    4) Let R be a ring and I an ideal of R. Show that
    a) if R is commutative, so is R/I
    b) if R has a unity, so does R/I

    5) Let R be a commutative ring.
    a) Show that the set of nilpotent elements in R forms an ideal
    b) Show that the quotient of R by this ideal has no nonzero nilpotent elements

    6) Let R be an integral domain. We call R a principal ideal domain (PID) if every ideal of R is of the form aR for some a in R. Show that in a PID every nontrivial proper prime ideal is maximal.
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  2. #2
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    Can someone at least tell me what I should be looking for? Not only do I not know how to do these, but I do not even know what I should be looking for...

    OK, so I get #1. I know what a maximal ideal is as well, but I don't get how I show that one exists for #3.

    And 5 and 6 I don't really get, these are some of the more challenging questions we were asked. Any help there would be appreciated.
    Last edited by mr fantastic; April 23rd 2009 at 06:37 PM. Reason: Merged posts
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  3. #3
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    For 4, I'm guessing it has something to do with R/I being a field? Am I right on this?

    5 and 6 I'm still rather confused. 6 I don't really get at all. 5 I've been playing around with, but not quite sure.
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  4. #4
    Super Member Gamma's Avatar
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    R/I is a field iff I is a maximal ideal. So this is not the way to go.

    Just think about how multiplication works in R/I.

    (a+I)(b+I)=ab+I=ba + I=(b + I)(a + I) where the middle step is because the ring is commutative.

    Same for the b) part. If R has unity call it 1. Then if 1\in I \Rightarrow I=R so it has unity since R has unity. Otherwise 1+I \in R/I and (a+I)(1+I)=a+I

    For 5 just write out the definition of nilpotent and do the ideal test. It is straightforward. b) you just modded out by all the nilpotent elements, how could there be any left in R/N?

    6) is a pretty standard proof in pretty much all algebra books. Let (p) be a nonzero prime ideal. Suppose I = (m) (using PID) be an ideal containing (p). Show either I=(p) or I=(R). p\in (m) so p=rm, r\in R. (p) is prime, so either r\in (p) or m\in (p).

    Case 1 : m\in (p) \Rightarrow (p)=(m)= I

    Case 2 : r\in (p) \Rightarrow r=ps but then from above
    p=rm=psm \Rightarrow sm=1 \Rightarrow m is a unit, so I=(m)=R.
    QED
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  5. #5
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    Thanks. And like I said, I know what a maximal ideal is for 3, but I don't know how to show one exists. Does anyone know?
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  6. #6
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    How do I show for #3 that a maximal ideal exists and that it is unique? I don't see how I can show that this is the case. Do I have to give a specific example of what the ideal is?
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