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?
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?
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?
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.