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There's an exercise in Herstein marked as very hard

Let $R$ be a commutative ring. If $q(x) \in R[x]$ be a zero divisor in $R[x]$, then if $\displaystyle q(x) = \sum_{0 \leq i \leq k} a_i x^i$, prove there's $b \in R$ such that for all $0 \leq i \leq k$, we have $ba_i = 0_R$

Now how I am supposed to solve this ?

I know that I'm supposed to prove how much I have worked by myself before asking for help, but I tried for more than two hours but still can't even solve the case when $\text{deg}[q] = 1$

katana_0
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  • Well $deg(q)=1$: $0=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$. Even if the other factor has larger degree, we only need to use that the linear part kills the linear $q(x)=ax+b$. The cases, $c=0$, $d=0$ are done. Then $ad=-bc$. If $bc=0$ we are done, take $c$ as the killing element. Otherwise take $d$. –  Jul 25 '18 at 13:29
  • @nextpuzzle AAAAARGH WHY I'M SO STUPID.... I feel like wallbashing my head seeing your proof for the linear case. – katana_0 Jul 25 '18 at 13:31
  • Didn't Hardy once asked why a holomorphic function bounded on the circle is bounded in the interior. What matters is the work that you accumulate in the long run. –  Jul 25 '18 at 13:32
  • As stated, the question allows for the trivial answer $b=0_R$. – Servaes Jul 25 '18 at 13:33
  • @nextpuzzle it's been awhile since I have studied complex analysis, so what is the reason that a holomorphic function bounded on the circle must be bounded in the interior? Is it just the property of continuity? – Prism Jul 25 '18 at 22:42

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