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I should recall the notion of maximal subring of a commutative unitary ring $R$.

Def: A commutative ring $S$ is called a maximal subring of $R$ if $S \subset R$ and if $T \subset R$ constitute a commutative ring with the restricted addition and multiplication of $R$ and also $S\subsetneq T$ then we could deduce that $T=R$.

I am interested in studing this notion in Rings of continuous functions.

We could easily deduce that for $x \neq y \in X$ The set of the form $$M_{x,y}=\Big(f\in C(X): f(x)=f(y) \Big)$$ forms a maximal subring of $C(X)$

From the above summary and notations I could pose my Questions.

Question1: Is there a maximal subring in $C(X)$ other than all $M_{x,y}$'s?

Question2: Is $X$ compact if all maximal subrings of $C(X)$ is of the form $M_{x,y}$?

PS:I suppose that all subrings of a commutative ring $R$ contains the unitary element of $R$.

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  • $\begingroup$ Fix $x\in X$. What about the set $M_x$ of elements $f\ in C(X)$ such that $f(x)=0$? $\endgroup$
    – Alireza Abdollahi
    Commented Aug 28, 2012 at 15:18
  • $\begingroup$ Hello Dear Alireza. If you notice to my postscript which was added at the end of my problems you could find that $M_x$ is not a subring of $C(X)$, because $1\notin M_x$. $\endgroup$
    – Ali Reza
    Commented Aug 28, 2012 at 15:25
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    $\begingroup$ I assume you mean --closed subrings--. By Zorn's lemma argument there are lots of maximal subrings of $\mathbb{C}=C(\{\emptyset\})$. $\endgroup$
    – Ollie
    Commented Aug 28, 2012 at 17:23
  • $\begingroup$ Dear Ollie, My notation of $C(X)$ consider all continuous real valued functions. On the other hand I do not understand How you could apply Zorn's lemma for this situation. Please describe in more details. Thanks a lot.Also How could you deduce that $\mathbb {C}= C(\{\emptyset \})$ contains a proper maximal subring. $\endgroup$
    – Ali Reza
    Commented Aug 28, 2012 at 17:41
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    $\begingroup$ @Ollie, $C(X)=C(\beta X)$ only holds if all continuous functions on $X$ are bounded. $\endgroup$
    – Jochen Wengenroth
    Commented Aug 29, 2012 at 8:25

3 Answers 3

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Hi, Yes, C(X) has very different maximal subrings. First note that the real line, say R, has uncountable many maximal subrings, see for example A. Azarang and O.A.S. Karamzadeh works about the existence of maximal subrings in Fields and Commutative Rings, Hence for any fixed maximal ideal M_x of C(X), C(X) contains many maximal subrings different from the above. Also note that for any free maximal ideal M, C(X) contains a maximal subring which contains R.

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  • $\begingroup$ Thanks dear Alborz. I think you are skilled in the existence of maximal subrings. could you think about the second Question.If you are familiar with the kolmogoroff theorem in characterization of maximal ideals of C(X), when X is compact, You could find that it's a bit similar to it. cheers $\endgroup$
    – Ali Reza
    Commented Nov 11, 2012 at 11:50
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Question 1 has a positive answer for general $X$. The reason for this is that $C(X)$ cannot distinguish between $X$ and its realcompactification. Hence if $X$ is not realcompact we construct $M_{x,y}$ as above with $x$ in $X$ and with $y$ in the realcompactification, but not in $X$. This is a maximal subring which does not have the required form.

A good reference for this material is Weir's book "Hewitt-Nachbin spaces" (and, of course, Gilman and Jerison).

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  • $\begingroup$ Dear jbc. Thank you so much for your nice and simple construction, from your answer for part 1 of my Question, we could deduce that if the condition of part 2 occurs, then $X$ should be real compact. but it also remains to approach to the compactness of it. Best wishes $\endgroup$
    – Ali Reza
    Commented Aug 29, 2012 at 11:09
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You're assuming that $C(X)$ actually separates the points of $X$. Otherwise one could have $M_{x,y}=C(X)$.

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  • $\begingroup$ Dear Josh. Thank you very much for your notation. but it's better to see that in the study of rings of continuous functions, we could suppose that the topological space that all continuous functions defined on it, is completely regular Hausdorff space. then the case which you mentioned as an answer could not occur. As you Know for every topological space $X$ there exists a completely regular Hausdorff space $Y$ so that $C(X)$ and $C(Y)$ are ring isomorphic. $\endgroup$
    – Ali Reza
    Commented Aug 28, 2012 at 20:21

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