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Let $i,n,m$ be positive integers. For every nonnegative integer $k<i+1$ , let $a_k$ be elements of a ring $A$ that satisfies :


$1)$ The ring $A$ is isomorphic to ${\Bbb R}^{n}\times{\Bbb C}^{m}$.

$2)$ For every nonzero element $q$ of $A$ we have $q^2 \neq 0$.


Consider the polynomial $B_i(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_i x^i$.

Conjecture : Iff $B_i(x)=0$ has a solution $x$ that is an element of the ring $A$ then $a_i$ is not a zero-divisor.

How to prove this ?

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What is the ring structure that you put on $A$? Is it actually possible to have zero divisors in a ring isomorphic to $\mathbb{R}^n\times\mathbb{C}^m$ ? –  Abramo Dec 5 '13 at 23:49
    
What about $i=0$ and $a_0=0$ ($B(x)=0$) or $B(x)=x$? –  Berci Dec 5 '13 at 23:50
1  
This is pretty catastrophically false... –  universalset Dec 5 '13 at 23:52
    
I edited the OP. –  mick Dec 8 '13 at 22:37
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