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Thread: Commutative Ring

  1. #1
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    Commutative Ring

    How is it that a commutative ring F(R) defined by the operations of pointwise addition and pointwise multiplication contains elements f != 0, 1 that satisfy f^2 = f?
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  2. #2
    ux0
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    Quote Originally Posted by johnt4335 View Post
    f != 0, 1 that satisfy f^2 = f?
    I think you meant to say..

    $\displaystyle f \neq 0, 1$

    i think if we define the funtions

    $\displaystyle f(a)=\left\{\begin{matrix}
    a & a \leq 0\\
    0 & a \geq 0
    \end{matrix}\right.
    $

    and

    $\displaystyle g(a)=\left\{\begin{matrix}
    0 & a \leq 0\\
    a & a \geq 0
    \end{matrix}\right.$

    then we can prove that there are elements that work for $\displaystyle f^2 = f$...

    i.e $\displaystyle (-2,-2)^2 = (4, 4)$ ... which is on $\displaystyle g(a)$


    Could someone verify this please?
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  3. #3
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    Hi

    If by F(R) you mean the set of all maps from $\displaystyle R$ to $\displaystyle R,$ where $\displaystyle R$ is a ring, then each element $\displaystyle f$ of $\displaystyle \mathcal{F}(R)$ can be seen as a "sequence" $\displaystyle (x_r)_{r\in R}$ where $\displaystyle x_r=f(r)$, and the binary operations you defined over it to obtain a ring are:

    addition: $\displaystyle (x_r)_{r\in R}+(y_r)_{r\in R}=(x_r+y_r)_{r\in R}$ ,

    multiplication: $\displaystyle (x_r)_{r\in R}\times(y_r)_{r\in R}=(x_r\times y_r)_{r\in R}$ ,

    with as zero $\displaystyle (0)_{r\in R}$ and identity element $\displaystyle (1)_{r\in R},$ which are obviously idempotent.

    When $\displaystyle R$ has more than two elements, can you find very simple elements of $\displaystyle \mathcal{F}(R)$ other than $\displaystyle (0)_{r\in R}$ and $\displaystyle (1)_{r\in R}$ (but similar...) that are idempotent?
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  4. #4
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    Ohh okay, so you mean something along the lines of the matrix I.
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  5. #5
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    I guess; what I mean is any $\displaystyle f\in \mathcal{F}(R)$ such that $\displaystyle Im(f)\subseteq\{0,1\}$ is idempotent.

    So if $\displaystyle R$ is finite and has $\displaystyle n$ elements, there are at least $\displaystyle 2^n$ elements in $\displaystyle \mathcal{F}(R)$ that are idempotent.
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