Motivated by this post we give the following definition:

Definition: A pair of unital rings $(R,S)$ is called a consecutive pair of rings if they do not have non trivial idempotent and satisfy the following two conditions:

1)There is a surjective morphism $\phi:R\to S$.

2)If a unital ring $W$ without non trivial idempotent possess a surjective sequence of morphisms $R\to W \to S$ then either $W\simeq R$ or $W \simeq S$.

Let $C([0,1])$ be the ring of all complex continuous functions.

Is the pair $(C([0,1]), \mathbb{C})$ a consecutive pair of rings?

  • $\begingroup$ It is perhaps worth highlighting, for others reading this, that condition (2) does not assume any topology or continuity $\endgroup$ – Yemon Choi Jan 11 '17 at 14:55

The answers to the question prime ideals in C([0,1]) explain why the answer is no, assuming that Yemon Choi was mistaken in his answer in believing that you are only interesting in continuous homomorphims.

In particular it is explained that there are non-maximal prime ideals in the ring $C([0,1])$. Given such a prime $P$, $C([0,1])/P$ will provide a non-trivial $W$.

  • $\begingroup$ Nice! I forgot that the simplest way to ensure no non-trivial idempotents is to be an integral domain... $\endgroup$ – Yemon Choi Jan 11 '17 at 15:04
  • $\begingroup$ @SimonWadsley Thank you very much for your very elegant answer. $\endgroup$ – Ali Taghavi Jan 12 '17 at 9:51

EDIT: the following answer was based on the assumption that $W$ is Banach and the ring morphisms $R\to W\to {\mathbb C}$ were continuous. My apologies for misunderstanding the question.

I think the answer is yes, because $W$ must be isomorphic to $C(K)$ for some closed subset $K\subset [0,1]$. The condition about idempotents implies $K$ is connected. The only connected subsets of $[0,1]$ are closed sub-intervals, and these are either homeo to $[0,1]$ itself or are degenerate singletons.

  • $\begingroup$ R is just a ring not necessarily a c* algebra $\endgroup$ – Ali Taghavi Jan 11 '17 at 14:02
  • $\begingroup$ You mean $W$, right? I assumed you were only interested in continuous algebra homomorphisms, and then every closed ideal in $C(X)$ is given by the functions vanishing on some open subset $U$ $\endgroup$ – Yemon Choi Jan 11 '17 at 14:20
  • $\begingroup$ Is it obvious that the kernel of $C([0 1]) \to W$ is closed? $\endgroup$ – Ali Taghavi Jan 11 '17 at 14:25
  • $\begingroup$ @AliTaghavi Well that is why I was assuming you were working with continuous homomorphisms. I agree that in general it is not obvious if a homomorphism defined on $C([0,1])$ is continuous (the Dales-Esterle counterexamples to Kaplansky's question) but in the present case we may be able to say more. Let me think about this; I retract my earlier comment on your original question. $\endgroup$ – Yemon Choi Jan 11 '17 at 14:54

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