# Thread: Ring Functions

1. ## Ring Functions

I had posted this problem before, however the expression didn't appear as expected. The final expression should read f ^2=f
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
The expresiion should read the square of f equals f. What I mean here is that f multiplied by f equals f. Symbolicaly this is f ^2=f

2. Originally Posted by mwanahisab
I had posted this problem before, however the expression didn't appear as expected. The final expression should read f ^2=f
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
The expresiion should read the square of f equals f. What I mean here is that f multiplied by f equals f. Symbolicaly this is f ^2=f

Hmmm... this would mean that for all $\displaystyle x\in\mathbb{R}\,,\,\,f^2(x)=f(x)\iff f(x)\left(f(x)-1\right)=0\Longrightarrow f(x)=0\,\,\,or\,\,\,f(x)=1$ , since $\displaystyle \mathbb{R}$ is an integer domain (in fact a field, of course).

We can thus define $\displaystyle f(x):=\left\{\begin{array}{ll}0&\,if\,\,\,x\leq 0\\1&\,if\,\,\,x>0\end{array}\right.$ , and there we have a non-trivial idempotent.

Tonio