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?

Printable View

- Nov 22nd 2009, 06:21 PMjohnt4335Commutative 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?

- Nov 24th 2009, 06:51 AMux0
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? - Nov 24th 2009, 08:08 AMclic-clac
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? - Nov 24th 2009, 09:40 AMjohnt4335
Ohh okay, so you mean something along the lines of the matrix I.

- Nov 24th 2009, 09:54 AMclic-clac
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.