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Math Help - Stuck with Nilpotents

  1. #1
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    Stuck with Nilpotents

    "Suppose that R is a ring. Then an element  a \in R is called nilpotent if there exists a positive integer n so that a^n = 0"

    Ok this is what im stuck with:
    Assume that R is commutative. if  a,b \in R are nilpotent, show that a + b is nilpotent.

    I also need a help finding a couterexample, to show that the same result does not hold in non-commutative rings.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by Lipticboven View Post
    "Suppose that R is a ring. Then an element  a \in R is called nilpotent if there exists a positive integer n so that a^n = 0"

    Ok this is what im stuck with:
    Assume that R is commutative. if  a,b \in R are nilpotent, show that a + b is nilpotent.
    As R is a commutative ring one can use the binomial theorem :

    (a+b)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}

    Assuming there exist two integers p and q such that a^p=b^q=0, can you find n such that (a+b)^n=0 ?
    I also need a help finding a couterexample, to show that the same result does not hold in non-commutative rings.
    You can look for this counterexample in \mathcal{M}_2(\mathbb{R}) (the ring of real 2\times 2 matrices).
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  3. #3
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    "Assuming there exist two integers p and q such that a^p=b^q=0, can you find n such that (a+b)^n=0 ?"

    Is it when n = max(p,q)?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Lipticboven View Post
    "Assuming there exist two integers p and q such that a^p=b^q=0, can you find n such that (a+b)^n=0 ?"

    Is it when n = max(p,q)?
    No : for p=q=2 one has (a+b)^{\max(p,q)}=(a+b)^2=2ab which may or may not equal 0, depending on how a and b were chosen.

    A sufficient condition which gives us (a+b)^n = 0 is

    for all k such that 0\leq k\leq n, either a^k=0, either b^{n-k}=0.

    In other words :

    for all k such that 0\leq k\leq n, either k\geq p, either n-k\geq q.

    (you don't have to find the lowest integer n such that (a+b)^n=0, you have to find an integer n such that (a+b)^n=0)

    Quote Originally Posted by flyingsquirrel View Post
    You can look for this counterexample in \mathcal{M}_2(\mathbb{R}) (the ring of real 2\times 2 matrices).
    I'd like to add that strictly triangular matrices are nilpotent. They look like \begin{bmatrix} 0 & \alpha \\ 0 & 0\end{bmatrix} or \begin{bmatrix} 0 & 0 \\ \beta & 0\end{bmatrix}.
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