Is $R=\{ (f,g)\in C(\mathbb N) \times C(\mathbb N): f(1)=g(1)\}$ a clean ring? Suppose that $f,g:\mathbb N\rightarrow \mathbb R$ be two continuous functions in the ring of continuous function over $\mathbb N=\{1,2,3,...\}$ (i.e.$f,g \in C(\mathbb N)$)
Let $$R=\{ (f,g)\in C(\mathbb N)\times C(\mathbb N) : f(1)=g(1)\} $$
Is $R$ a clean ring?

Clean ring: means any element in the ring can be written as a sum of unit and idempotent .
One of the theorem may be good here is :
Any local ring is equivalent to an indecomposable clean ring 
I feel $R$ is not clean ring 
BTW, $R$ is indecomposable ring means $R$ is not isomorphic to a direct sum of nontrivial rings
 A: You apparently intend your operations to be coordinatewise. Imbuing $\mathbb N$ with the discrete topology makes all functions continuous, so we are actually not restricted there.
Then the units of your ring are of the form $(u,v)$ where $u(x)\neq 0\neq v(x)$ for any $x$, and $u(1)=v(1)$. The idempotents are of the form $(e,f)$ where $e(x),f(x)\in \{0,1\}$ for all $x$. 
It's clear how you "fix" $f$ and $g$ if they happen to be zero at $x$: you just subtract $1$ on that position, and that will make it nonzero. So to that end:
$$e_1(x)=\left\{\begin{array}{rcl}
 1&  &f(x)=0 \\ 
 0&  &f(x)=1 
\end{array}\right.$$
and 
$$e_2(x)=\left\{\begin{array}{rcl}
 1&  &g(x)=0 \\ 
 0&  &g(x)=1 
\end{array}\right.$$
$e_1(1)=e_2(1)$ since $f(1)=g(1)$.
A: after many tries , I found $R$ is a clean ring, since we can choose an idempotent element $(e_1,e_2)$ corresponding any element in $(f,g)$ as follows :
For any element $(f,g)\in R$, we must have $(f,g)=(u_1,u_2)+(e_1,e_2)$, where $(u_1,u_2)$ is unit and $(e_1,e_2)$ is an idempotent.
By choosing $$e_1=\left\{\begin{array}{rcl}
 1&  ,&x\in\mathbb N \setminus  \{1\} , f(x)=0, or \ f(1)=0 \\ 
 0&,  &x=1 
\end{array}\right.$$
Similarly for $e_2$. So, $(e_1,e_2)$ is an idempotent and $(u_1,u_2)=(f-e_1,g-e_2)$ is a unit. 
