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Thread: Stuck with Nilpotents

  1. #1
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    Stuck with Nilpotents

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

    Ok this is what im stuck with:
    Assume that R is commutative. if $\displaystyle a,b \in R$ 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 $\displaystyle a \in R$ is called nilpotent if there exists a positive integer n so that $\displaystyle a^n = 0$"

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

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

    Assuming there exist two integers $\displaystyle p$ and $\displaystyle q$ such that $\displaystyle a^p=b^q=0$, can you find $\displaystyle n$ such that $\displaystyle (a+b)^n=0$ ?
    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 $\displaystyle \mathcal{M}_2(\mathbb{R})$ (the ring of real $\displaystyle 2\times 2$ matrices).
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  3. #3
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    "Assuming there exist two integers $\displaystyle p$ and $\displaystyle q$ such that $\displaystyle a^p=b^q=0$, can you find $\displaystyle n$ such that $\displaystyle (a+b)^n=0$ ?"

    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 $\displaystyle p$ and $\displaystyle q$ such that $\displaystyle a^p=b^q=0$, can you find $\displaystyle n$ such that $\displaystyle (a+b)^n=0$ ?"

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

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

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

    In other words :

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

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

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