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Math Help - Finite rings

  1. #1
    Senior Member TheAbstractionist's Avatar
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    Finite rings

    Given a ring R with multiplicative identity, let U(R) denote the set of its units.

    What is the smallest positive integer n such that there does not exist any finite ring such that U(R) contains exactly n elements?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    Given a ring R with multiplicative identity, let U(R) denote the set of its units.

    What is the smallest positive integer n such that there does not exist any finite ring such that U(R) contains exactly n elements?
    Emm...I think you natural number, not integer. As there exists no ring with identity such that the set of units has 0 elements, or -1 elements, etc...
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    I did say positive integer. Isn’t a positive integer a natural number?
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    I did say positive integer. Isn’t a positive integer a natural number?
    Touché - I just read what I wanted to read...my bad.
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  5. #5
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    Quote Originally Posted by TheAbstractionist View Post
    Given a ring R with multiplicative identity, let U(R) denote the set of its units.

    What is the smallest positive integer n such that there does not exist any finite ring such that U(R) contains exactly n elements?
    this is a quite popular question is ring theory i guess. there's a solution which uses the structure theory of rings:

    the answer is n=5. you don't need to assume R is finite. it's clear that for n \leq 4 there exist rings with |U(R)|=n. now let R be a ring and U(R)=G. suppose |G|=5. then 2=0 in R, because
    otherwise \{1,-1 \} would be a subgroup of G, which is impossible because |G| is odd. now look at the group algebra S=\mathbb{F}_2[G]. by Maschke's theorem S is semisimple and thus by Wedderburn-Artin

    theorem S \cong M_{k_1}(D_1) \times M_{k_2}(D_2) \times \cdots M_{k_r}(D_r), where each D_i is a division ring containing \mathbb{F}_2. (here M_t(D) means the ring of t \times t matrices with entries from D.) since S is finite, each D_i is

    finite. but we know that a finite division ring is a field. so each D_i is a finite field extension of \mathbb{F}_2, i.e. |D_i|=2^{n_i}=N_i, for some n_i \geq 1.

    also if k_i > 1, for some i, then |U(M_{k_i}(D_i)|=(N_i^{k_i}-1)(N_i^{k_i}-N_i) \cdots (N_i^{k_i}-N_i^{k_i - 1}), which is odd only for k_i=1. thus S \cong D_1 \times D_2 \times \cdots D_r. but then G=U(S) \cong D_1^{\times} \times D_2^{\times} \times \cdots \times D_r^{\times} and

    hence: 5=|G|=(2^{n_1}-1)(2^{n_2}-1) \cdots (2^{n_r}-1), which is clearly impossible. Q.E.D.


    Remark: the above solution uses some powerful theorems in ring theory and so you should expect to get a lot more than just answering your question out of it. so ... see if you can do that!
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  6. #6
    Senior Member TheAbstractionist's Avatar
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    Briliant!

    Over in the AoPS/MathLinks forum at the moment, some people are trying to determine if there exists a finite ring with unity R such that |U(R)|=38. The latest development appears to be that if such a ring exists, it must have characteristic 4. It looks exciting – I wish I knew more ring theory so I could join in and not be left out of the fun.
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