This fact might be either trivial, wrong, or well known.


Let $R$ be a commutative ring. Let $u\_1,\dots,u\_{s-1},u\_s\in R$ and $m,M\in R$. Let us assume that $m,M$ satisfy
$$(m-u\_1) \dots (m-u\_{s-1})=0,$$
$$(M-m)(M-u\_s)=0.$$


**Question.** Is it true that $M$ satisfies:
$$(M-u\_1)\dots (M-u\_{s-1})(M-u\_{s})=0?$$


Obviously if $R$ has no zero divisors, then the answer is positive. It is evidently positive for $s=2$. I think it is still true for $s=3$ by a straightforward computation.



An identity in an arbitrary commutative ring
This is actually trivial. Write $M-u\_k=(M-m)+(m-u\_k)$; this gives $(M-u\_1)\dots (M-u\_{s-1})\in R(M-m)$. Multiplying by $(M-u\_s)$ gives $(M-u\_1)\dots (M-u\_{s})=0$.


