Quotient rings of $C(X)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:29:52Z http://mathoverflow.net/feeds/question/104770 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104770/quotient-rings-of-cx Quotient rings of $C(X)$ AliReza Olfati 2012-08-15T15:17:59Z 2012-11-28T20:06:15Z <p>Let $X$ be a Tychonoff topological space. Consider the ring $C(X)$ of all continuous real valued functions on $X$. For what conditions on an ideal $I$ of $C(X)$, we could deduce that the quotient ring is isomorphic to a ring of the form $C(Y)$.i.e. </p> <p><em><strong>Question</em></strong> : An ideal $I$ has the algebraic property $\mathcal{P}$ if and only if The exists a Tychonoff topological space $Y$ so that: $$\frac{C(X)}{I} \cong C(Y)$$</p> http://mathoverflow.net/questions/104770/quotient-rings-of-cx/104779#104779 Answer by Pietro Majer for Quotient rings of $C(X)$ Pietro Majer 2012-08-15T17:18:17Z 2012-08-15T17:18:17Z <p>To expand Qiaochu Yuan's comment. If $X$ is a Hausdorff compact space, any closed ideal $I$ of the Banach algebra $C(X)$ is the kernel of the restriction map $C(X)\rightarrow C(Y)$ for some closed subset $Y$ of $X$, namely $Y:=\cap_{f\in I}f^{-1}(0)$ (that is, all functions that vanish on $Y$. This follows from the Stone-Weierstrass theorem applied to the quotient space $X/Y$, obtained collapsing $Y$ to a point). The above restriction map is surjective by the Tietze extension theorem. On the quotient, we have the ordered and isometric algebra isomorphism $C(X)/I \rightarrow C(Y)$. </p> http://mathoverflow.net/questions/104770/quotient-rings-of-cx/104794#104794 Answer by Igor Khavkine for Quotient rings of $C(X)$ Igor Khavkine 2012-08-15T21:25:31Z 2012-08-15T21:25:31Z <p>Repeating Yemon Choi's cue, while being a bit more specific, problem 10D from Gillman and Jerison should get you started. It gives some necessary conditions (like $I$ must be a $z$-ideal) and relates the potential space $Y$ to a closed subset of the Hewitt realcompactification $\upsilon X$ of $X$. After a quick glance, I'm not sure how far you have to go to get sufficient conditions. Let me know if you have trouble getting the relevant information from the book.</p> http://mathoverflow.net/questions/104770/quotient-rings-of-cx/112080#112080 Answer by jbc for Quotient rings of $C(X)$ jbc 2012-11-11T15:28:33Z 2012-11-11T15:28:33Z <p>Since you ask for a duality between quotients of $C(X)$ and the closed subsets of $X$, the following does not answer your question directly. Nevertheless, I hope it might be of interest. I am a tireless proponent of the thesis that if you want to extend the duality between compact spaces and algebra of continuous functions thereon, then the appropriate context is the space $C^b(X)$ of bounded, continuous functions, not with the norm but with the strict topoogy. This was introduced by Buck for functions on locally compact spaces and then extended, using different methods, to completely regular ones. Many of the results for Gelfand-Naimark theory for compact spaces can be carried over in the natural way, in particular, precisely the duality between closed subsets of $X$ and quotient algebras with respect to ideals---one need only demand that the ideal be closed in the above topology. This theory is presented in the book "Saks spaces and applications to functional analysis".</p> http://mathoverflow.net/questions/104770/quotient-rings-of-cx/114810#114810 Answer by Ali for Quotient rings of $C(X)$ Ali 2012-11-28T20:06:15Z 2012-11-28T20:06:15Z <p>Hi Alireza, An ideal $I$ of $C(X)$ is an annihilator ideal if $int\bigcap Z[I]\subset intZ(f)$, $f\in C(X)$, then $f\in I$. Now let $I$ be an annihilator ideal of $C(X)$, $X$ a completely regular space and $Y=\bigcap Z[I]$. Then define $\phi$ of $C(X)$ on to $C(Y)$, which $\phi(f)=f|_{Y}$. Then it is easy to see that $\phi$ is a ring homomorphism. By above definition we have, $ker\phi={f: Y\subset Z(f)}=I$. So $\frac{C(X)}{I} ≅C(Y)$.</p>