Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
$0=(m-u_1)\dots(m-u_{s-1})(M-u_s)=(M-u_1+m-M)\dots(M-u_{s-1}+m-M)(M-u_s)=(M-u_1)‌​\dots(M-u_{s-1})(M-u_s)+\text{terms containing }(m-M)(M-u_s)$. –  Sasha Anan'in Mar 3 at 18:17
add comment

2 Answers 2

up vote 8 down vote accepted

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 $(M-u_1)\dots (M-u_{s})=0$.

share|improve this answer
add comment

Yes.

I claim that

(1) $\prod\limits_{i=1}^s \left(M-u_i\right) = \prod\limits_{i=1}^{t} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s} \left(M-u_i\right) $ for every $t \in \left\lbrace 0,1,...,s-1\right\rbrace$.

In fact, you can prove (1) by induction over $t$. The base case $t = 0$ is tautological. In the induction step, you need to show that

$\prod\limits_{i=1}^{t} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s} \left(M-u_i\right) = \prod\limits_{i=1}^{t-1} \left(m-u_i\right) \cdot \prod\limits_{i=t}^{s} \left(M-u_i\right)$

for every $t \in \left\lbrace 1,2,...,s-1\right\rbrace$. Clearly, this equation rewrites as

$\left(m-u_t\right)\left(M-u_s\right)A = \left(M-u_t\right)\left(M-u_s\right)A$,

where $A = \prod\limits_{i=1}^{t-1} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s-1} \left(M-u_i\right)$.

Thus, the induction will be complete once we show that

$\left(m-u_t\right)\left(M-u_s\right) = \left(M-u_t\right)\left(M-u_s\right)$.

But this is clear since

$\left(m-u_t\right)\left(M-u_s\right) - \left(M-u_t\right)\left(M-u_s\right) = -\underbrace{\left(M-m\right)\left(M-u_s\right)}_{=0} = 0$.

So we have proven (1) by induction. Applying (1) to $t=s-1$, we obtain

$\prod\limits_{i=1}^s \left(M-u_i\right) = \underbrace{\prod\limits_{i=1}^{s-1} \left(m-u_i\right)}_{=\left(m-u_1\right)\left(m-u_2\right)...\left(m-u_{s-1}\right)=0} \cdot \prod\limits_{i=s}^{s} \left(M-u_i\right) = 0$,

qed.

share|improve this answer
    
OK, this is the same as abx did but in more basic terms. –  darij grinberg Mar 3 at 18:25
    
I am amazed that you could write that kind of answer in less than 15 minutes! –  TaQ Mar 3 at 19:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.