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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

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    what topology do you have on $\mathbb{N}$? – 2017-01-27
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    Discrete topology – 2017-01-27
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    So don't you just mean that $f,g$ are arbitrary functions from N to R? – 2017-01-27
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    @heptagon, $f,g$ are any arbitrary continuous functions from $(\mathbb N, \tau_{dis})\rightarrow (\mathbb R,\tau_u)$ such that $f(1)=g(1)$. – 2017-01-27
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    I proved $R$ is not local ring. I just need to prove it is indecomposable. – 2017-01-27
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    @Leonardo I don't understand what you are trying to do with your statement about locality and indecomposability. – 2017-01-27
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    @Leonardo I can't get around the following. If you use the *discrete* topology on $\mathbb N$, then *every* function from $\mathbb N\to\mathbb R$ is continuous. Then take a function $f$ which is zero everywhere except at $1$ where it is $1$. Then $f$ is idempotent and $(f,f)$ is a nontrivial idempotent of your ring, and the ring is not indecomposable. But I also could have done something silly, and I wait for that to be pointed out (or to have it suddenly come to me.) – 2017-01-27
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    @rschwieb If $R$ is indecomposable and not local , then $R$ cannot be a clean , since an indecomposable clean ring is local ring. Yes, $R$ is decomposable , so this theorem is useless. But, I noticed if I can prove $R$ is not a $pm$-ring then $R$ is not clean.( I found this in some paper about clean ring. A $pm$-ring is a ring in which every prime ideal is contained in a unique maximal ideal , for sure, $R$ is commutative ring with unity . Also, every clean ring is a $pm$-ring ) The strange thing – 2017-01-27

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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 $f(1)=g(1)$.

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    exactly, I failed to explain that in a good way. Thanks – 2017-01-29
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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.

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    This is really unclear. I think you probably have not written what you meant. – 2017-01-28
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    I edited the answer – 2017-01-28
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    by this definition, if $f(2)=1$, this would cause $u_1(x)=0$, so it would not be a unit. – 2017-01-28