Motivated [by this post](https://mathoverflow.net/questions/259251/a-consecutive-resolution-of-continum-algebras-to-a-simple-continum-algebra) 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? 
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Is the pair $(C([0 \;1]),\mathbb{C})$ a consecutive pair?
The answers to the question [prime ideals in C([0,1])](https://mathoverflow.net/questions/35793/prime-ideals-in-c0-1) 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 $W$.


