1. ## Commutative ring functions

Could somebody please solve this problem?
Let the commutative ring F(R) be the set of all functions R-R equipped with the operations of pointwise addition and pointwise multiplication: given f, g are members F(R), define functions f+g and fg by
f+g: a→ f(a) + g(a) and fg: →f(a)g(a)
(Notice fg is not their composite)
Pointwise addition and multiplication are the operations on functions occurring in calculus for example.
f(x) g(x)dx =∫f(x) dx +∫g(x) dx
And D(fg)=D(f)g+f(D(g), where D denotes derivative.
The sum f+g in the first integrand is pointwise addition while the product fg in the derivative formula is pointwise multiplication.

Show that the above commutative ring F(R) contains elements f≠0,1 with f2=f

2. Originally Posted by mwanahisab
Could somebody please solve this problem?
Let the commutative ring F(R) be the set of all functions R-R equipped with the operations of pointwise addition and pointwise multiplication: given f, g are members F(R), define functions f+g and fg by
f+g: a→ f(a) + g(a) and fg: →f(a)g(a)
(Notice fg is not their composite)
Pointwise addition and multiplication are the operations on functions occurring in calculus for example.
f(x) g(x)dx =∫f(x) dx +∫g(x) dx
And D(fg)=D(f)g+f(D(g), where D denotes derivative.
The sum f+g in the first integrand is pointwise addition while the product fg in the derivative formula is pointwise multiplication.

Show that the above commutative ring F(R) contains elements f≠0,1 with f2=f
How about $\displaystyle f(x)=\begin{cases}0\quad\text{if}\quad x=0\\1\quad\text{if}\quad x\ne 0\end{cases}$?