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Thread: Finite rings

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

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

    What is the smallest positive integer $\displaystyle n$ such that there does not exist any finite ring such that $\displaystyle U(R)$ contains exactly $\displaystyle n$ 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 $\displaystyle R$ with multiplicative identity, let $\displaystyle U(R)$ denote the set of its units.

    What is the smallest positive integer $\displaystyle n$ such that there does not exist any finite ring such that $\displaystyle U(R)$ contains exactly $\displaystyle n$ 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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  4. #4
    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 $\displaystyle R$ with multiplicative identity, let $\displaystyle U(R)$ denote the set of its units.

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

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

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

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

    hence: $\displaystyle 5=|G|=(2^{n_1}-1)(2^{n_2}-1) \cdots (2^{n_r}-1),$ 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 $\displaystyle R$ such that $\displaystyle |U(R)|=38.$ 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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