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Math Help - Ring Theory

  1. #1
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    Ring Theory

    Hi everyone
    I am having trouble with this problem. It is number 6 in section 8.2 of Dummit and Foote if you have it.

    R is an integral domain and suppose that every prime ideal is principal.
    a) (I have done this) Assume that the set of ideals of R that are not principal is nonempty and prove that this set has a maximal element under inclusion.

    b) (Where I am stuck) Let I be an ideal which is maximal with respect to being nonprincipal, and let a, b in R with

    ab in I, but a not in I and b not in I (so I is not a prime ideal)

    Let Ia be the ideal generated by (I,a) and Ib be the ideal generated by (I,b). Define J={r in R| rIa is a contained in I}

    Prove that Ia=(v) and J=(w) (generated by v and w, respectively) are principal ideals in R with I properly contained in Ib contained in J and IaJ=(vw) contained in I.

    ---------
    First with Ia, I cannot figure out how to express Ia as being generated by v. I also tried to suppose that it is not principal, but haven't gotten anywhere. Any ideas?
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  2. #2
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    Quote Originally Posted by robeuler View Post
    Hi everyone
    I am having trouble with this problem. It is number 6 in section 8.2 of Dummit and Foote if you have it.

    R is an integral domain and suppose that every prime ideal is principal.
    a) (I have done this) Assume that the set of ideals of R that are not principal is nonempty and prove that this set has a maximal element under inclusion.

    b) (Where I am stuck) Let I be an ideal which is maximal with respect to being nonprincipal, and let a, b in R with

    ab in I, but a not in I and b not in I (so I is not a prime ideal)

    Let Ia be the ideal generated by (I,a) and Ib be the ideal generated by (I,b). Define J={r in R| rIa is a contained in I}

    Prove that Ia=(v) and J=(w) (generated by v and w, respectively) are principal ideals in R with I properly contained in Ib contained in J and IaJ=(vw) contained in I.

    ---------
    First with Ia, I cannot figure out how to express Ia as being generated by v. I also tried to suppose that it is not principal, but haven't gotten anywhere. Any ideas?
    i guess your ring is commutative. let I_a be the ideal generated by I and a. then I_a=I+Ra. obviously I \subseteq I_a and a \in I_a - I. hence: I \subsetneq I_a. similarly I \subsetneq I_b, because b \in I_b - I and

    I \subseteq I+Rb=I_b. now let r \in I, then ra \in I and thus: rI_a=rI +Rra \subseteq I. thus by definition: r \in J. hence: I \subseteq J. we also have: bI_a=bI + Rba \subseteq I, because ab \in I. thus b \in J.

    hence I_b=I+Rb \subseteq J. so: I \subsetneq I_b \subseteq J. now since I is maximal among non-principal ideals, every ideal which properly contains I must be principal. in particular I_a and J are principal.

    so I_a=Rv, \ J=Rw, for some v,w \in R. clearly I_aJ=Rvw. finally JI_a \subseteq I is a trivial result of the definition of J. \ \ \ \Box.
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  3. #3
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    It always seems so clear after you put it like that.

    Here is the last one I am struggling with, if you have the patience.

    Let G be a finite group {g1,g2,...,gn} and assume R is a commutative ring. If r is any element of the augmentation ideal of RG then r(g1+g2+...+gn)=0.

    The hint is to use the fact that the augmentation ideal in group rings is generated by {g-1|g in G}.

    I tried to look at the augmentation map directly, but i can't figure out a general case regarding any element of the augmentation ideal. It doesn't help that my notion of the group ring is weak... but that is slowly improving.

    thanks again
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  4. #4
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    Quote Originally Posted by robeuler View Post
    It always seems so clear after you put it like that.

    Here is the last one I am struggling with, if you have the patience.

    Let G be a finite group {g1,g2,...,gn} and assume R is a commutative ring. If r is any element of the augmentation ideal of RG then r(g1+g2+...+gn)=0.

    The hint is to use the fact that the augmentation ideal in group rings is generated by {g-1|g in G}.

    I tried to look at the augmentation map directly, but i can't figure out a general case regarding any element of the augmentation ideal. It doesn't help that my notion of the group ring is weak... but that is slowly improving.

    thanks again
    this is a nice, although quite easy, problem! first we need a simple lemma:

    Lemma: let G=\{g_1, \cdots , g_n \} be a group, R a commutative ring, and r=\sum_{i=1}^nr_ig_i \in RG. suppose that \sum_{i=1}^n r_i = 0. then r\sum_{i=1}^n g_i=0.

    Proof: clearly for any 1 \leq k \leq n we have: G=\{g_1, \cdots , g_n \}=\{g_kg_1, \cdots , g_kg_n \}. thus: r\sum_{i=1}^n g_i=(\sum_{i=1}^n r_ig_i)(\sum_{i=1}^n g_i)=\sum_{k=1}^n r_k\sum_{i=1}^n g_kg_i=\sum_{i=1}^n(\sum_{k=1}^nr_k)g_i = 0. \ \ \Box

    now let g_1=1_G and r \in RG be in the augmentation ideal. then r=\sum_{i=2}^nr_i(g_i - 1)=(-r_2-r_3 - \cdots - r_n)g_1 + r_2g_2 + \cdots r_ng_n. thus r\sum_{i=1}^n g_i=0, by the Lemma. Q.E.D.
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  5. #5
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    Thanks again!

    I should have mentioned, I actually proved the lemma in part a of the exercise.

    And it is a nice problem I agree. Dummit and Foote has a very nice collection of problems and examples.
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