# Commutative Ring

• Nov 22nd 2009, 06:21 PM
johnt4335
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?
• Nov 24th 2009, 06:51 AM
ux0
Quote:

Originally Posted by johnt4335
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}
a & a \leq 0\\
0 & a \geq 0
\end{matrix}\right.
$

and

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

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

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

• Nov 24th 2009, 08:08 AM
clic-clac
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?
• Nov 24th 2009, 09:40 AM
johnt4335
Ohh okay, so you mean something along the lines of the matrix I.
• Nov 24th 2009, 09:54 AM
clic-clac
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.