Let $R$ be the ring of all functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ which are continuous outside $(1,1)$ and let $S$ be the ring of all functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ which are continuous outside a bounded open interval containing zero (depended on $f$). Is it true that $R \cong S$?

2$\begingroup$ The sets of idempotents of $R$ and $S$ have nonisomorphic multiplicative structure. $\endgroup$ – Dag Oskar Madsen Sep 12 '13 at 14:55

$\begingroup$ @DagOskarMadsen I am not sure about that $\endgroup$ – user30300 Sep 13 '13 at 13:33

$\begingroup$ @Sally Give me time, I'm writing down a proof. The key is looking at primitive idempotents. $\endgroup$ – Dag Oskar Madsen Sep 13 '13 at 13:45

$\begingroup$ I agree. Note that in both rings the constant $1$ can be characterized multiplicatively as the only square root $r$ of $1$ such that for any nonzero square $q$ the element $rq$ is not a square. So the argument I wrote below actually shows that $R$ and $S$ have nonisomorphic multiplicative structures. $\endgroup$ – Pietro Majer Sep 13 '13 at 17:37
They are not ring isomorphic, because e.g. $R$ has the following property of a ring $X$, and $S$ does not:
There is a non zero element $u \in X$ such that for any invertible $f\in X$ either $uf$ or $uf$ is a square, and for some $g\in X$ neither $ug$ nor $ug$ is a square.

$\begingroup$ Here $u:=\chi_{[1,+\infty)}$ in the case of $X=R$. One may state several analogous properties in terms of notions such as invertible /idempotent /square /square root of $1$. $\endgroup$ – Pietro Majer Sep 12 '13 at 17:12

1$\begingroup$ Yes. But why is this question still open at MO, do you think? $\endgroup$ – Todd Trimble♦ Sep 13 '13 at 16:41
An idempotent $u$ is called primitive if the equation of idempotents $$(1u)x=0$$ has a unique nonzero solution $x=u$. The primitive idempotents in $R$ are $\chi_{\{a\}}$ for $a \in (1,1)$ and $\chi_{(\infty,1]}$ and $\chi_{[1,+\infty)}$. For ease of notation give two last mentioned primitive idempotents index $1$ and $1$ respectively. The primitive idempotents in $S$ are indexed by $a \in \mathbb R$.
A ring isomorphism $\phi \colon R \rightarrow S$ would then define a bijection $\tilde \phi \colon [1,1] \rightarrow \mathbb R$. Let $A=\tilde \phi^{1}(\mathbb Q) \subseteq [1,1]$, where $\mathbb Q$ denotes the rational numbers. Then $\chi_A$ is an idempotent in $R$ with the property that $\chi_A \cdot \chi_{\{a\}} \neq 0$ for all $a \in A$ and $\chi_A \cdot \chi_{\{a\}}= 0$ for all $a \in [1,1] \smallsetminus A$.
The idempotent $\phi(\chi_A) \in S$ must be of the form $\chi_B$ for some $B \subseteq \mathbb R$. We must have $$\chi_B \cdot \chi_{\{\tilde \phi(a)\}}=\phi(\chi_A \cdot \chi_{\{a\}}) \neq 0$$ for all $a \in A$ and $$\chi_B \cdot \chi_{\{\tilde \phi(a)\}}=\phi(\chi_A \cdot \chi_{\{a\}}) = 0$$ for all $a \in [1,1] \smallsetminus A$. Therefore $B$ contains all the rational numbers and no numbers that are not rational. But $\chi_{\mathbb Q}$ is not an element of $S$ and we reach a contradiction. Therefore there is no ring isomorphism $\phi \colon R \rightarrow S$.

$\begingroup$ By the way I could write the first equation as $ux=x$ so that it becomes clearer I'm only using the multiplicative structure. $\endgroup$ – Dag Oskar Madsen Sep 13 '13 at 15:35

$\begingroup$ @pin2 its okay. I will delete my question $\endgroup$ – user30300 Sep 13 '13 at 18:17

$\begingroup$ Definitely worth upvoting but I accepted @pietroMajer's solution since that solution helped me more. $\endgroup$ – user39207 Sep 13 '13 at 19:24