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Thread: More rings, and cosets of rings

  1. #1
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    More rings, and cosets of rings

    So now working up to nilpotent groups, there is this problem in my book where its given as a problems to think about , and it goes:

    $R$ is a commutative ring with unity $1$. Prove that if $J$ is defined by $K=\{{1-s|s\in R
    \text{ nilpotent}\}}$, then it is a group under multiplication. Now suppose $R$ is finite with $|R|=n$, and $s$ nilpotent, show that $(1-s)^n=1$.

    How I tried: So for the group part I think the inverse is pretty easy to find which is just $(1-s)(1 + s + s^2+ \cdots +s^{n-1}) = 1-s^n = 1 \rightarrow$ Inverse = $1+s+s^2+\cdots+s^{n-1}$. Im stuck at trying to show that there exists some 1-t, t\in R, then $(1-t)(1-s)=1-((t+s)-st)$, how do I show that $(t+s)-st$ is nilpotent?

    For the second problem I think I can see that $|K| | |R|$ should be true, but I dont know why this is the case, but clearly $K$ is a coset of the nilpotent $s$ so is this why the order of $K$ divides $R$? But why is $|K|$ diving $|R$| show that $(1-s)^n =1$ ? Could anyone show me how to write this formally so it makes sense in a more coherent manner?
    Last edited by gaussrelatz; Nov 26th 2017 at 07:49 AM.
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  2. #2
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    Re: More rings, and cosets of rings

    in a commutative ring, if a and b are nilpotent then their sum a + b is nilpotent

    Proof: use the Binomial Theorem
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  3. #3
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    Re: More rings, and cosets of rings

    Let S be the set of nilpotent elements of R. As Idea points out, S is closed under addition, and it follows easily that S is a subgroup of the additive group of R. Let m be the cardinality of S. Then m divides n and "obviously" m is the cardinality of K. It should now be obvious that $(1-s)^n=1$ for any element s of S.
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  4. #4
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    Re: More rings, and cosets of rings

    Quote Originally Posted by johng View Post
    Let S be the set of nilpotent elements of R. As Idea points out, S is closed under addition, and it follows easily that S is a subgroup of the additive group of R. Let m be the cardinality of S. Then m divides n and "obviously" m is the cardinality of K. It should now be obvious that $(1-s)^n=1$ for any element s of S.
    Ok I get how to show a+b is nilpotent if a and b are nilpotennt but I cant show that (a+b-ab) is also nilpotent. Can you help me alittle more please? Also we have to show closure under multiplication I dont get why you are showing it is for additive case?
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  5. #5
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    Re: More rings, and cosets of rings

    product of nilpotent elements is nilpotent

    actually, the set of nilpotent elements in a commutative ring is an ideal

    r in R and s nilpotent with s^n=0 implies

    (r s)^n= r^ns^n=r^n*0=0
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