# Are these rings of functions isomorphic?

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 $R \cong S$?

• The sets of idempotents of $R$ and $S$ have non-isomorphic multiplicative structure. – Dag Oskar Madsen Sep 12 '13 at 14:55
• @DagOskarMadsen I am not sure about that – user30300 Sep 13 '13 at 13:33
• @Sally Give me time, I'm writing down a proof. The key is looking at primitive idempotents. – Dag Oskar Madsen Sep 13 '13 at 13:45
• 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 non-isomorphic multiplicative structures. – 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.

• 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 $1$. – Pietro Majer Sep 12 '13 at 17:12
• Yes. But why is this question still open at MO, do you think? – Todd Trimble Sep 13 '13 at 16:41

An idempotent $u$ is called primitive if the equation of idempotents $$(1-u)x=0$$ has a unique non-zero 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 $\phi \colon R \rightarrow S$.

• By the way I could write the first equation as $ux=x$ so that it becomes clearer I'm only using the multiplicative structure. – Dag Oskar Madsen Sep 13 '13 at 15:35
• @pin2 its okay. I will delete my question – user30300 Sep 13 '13 at 18:17
• Definitely worth upvoting but I accepted @pietroMajer's solution since that solution helped me more. – user39207 Sep 13 '13 at 19:24