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Math Help - 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..

    f \neq 0, 1

    i think if we define the funtions

    f(a)=\left\{\begin{matrix}<br />
a &  a \leq  0\\ <br />
0 & a \geq  0<br />
\end{matrix}\right.<br />

    and

    g(a)=\left\{\begin{matrix}<br />
0 &  a \leq  0\\ <br />
a & a \geq  0<br />
\end{matrix}\right.

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

    i.e (-2,-2)^2 = (4, 4) ... which is on 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 R to R, where R is a ring, then each element f of \mathcal{F}(R) can be seen as a "sequence" (x_r)_{r\in R} where x_r=f(r), and the binary operations you defined over it to obtain a ring are:

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

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

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

    When R has more than two elements, can you find very simple elements of \mathcal{F}(R) other than (0)_{r\in R} and (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 f\in \mathcal{F}(R) such that Im(f)\subseteq\{0,1\} is idempotent.

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