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Let $R$ be a commutative ring with 1.

a) Suppose $R$ has no nonzero nilpotent elements (that is, $a^n=0$ implies $a=0$). If $f(X)=a_0+a_1X+\cdots+a_nX^n$ in $R\left[X\right]$ is a zero-divisor, prove that there is an element $b\ne0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$.

b) Do part a) dropping the assumption that $R$ has no nonzero nilpotent elements.

There is a question in the picture about rings... I can't do it. Can you help me please?

There are some solutions in this link but i did not understand. This question was marked as duplicate. Do a) and b) have difference answer? In a) R has no nonzero nilpotents, but in b) dropping the assumption that R has no nonzero nilpotent elements it says do it .. i thought a) and b) have different solutions.

If in a) if R has no nonzero nilpotent elements, will you use this property in proof of a) ? and in b) there is no such a condition, so what i can do ? Could you clarify a) and b) ? thank you for your helping ...

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Read here without nilpotents assumption:… – DonAntonio Dec 24 '13 at 12:39

Some hints. Suppose $g=\sum_{i=0}^{d}b_iX^i\in R[X]$ is nonzero and $fg=0$

  1. Show that we may assume $b_0\neq 0$.
  2. Write down the equations that express that the coefficient of $1, X, X^2, X^3$ in $fg$ are zero.
  3. Determine how $b_0$ may be used to create elements that annihilate $a_0,a_1, a_2, a_3$.
  4. Now take a guess at what nonzero element $b$ might do the job.
  5. Prove your guess.
  6. Using what you established above, deduce part (b) of the question by "stopping before it is too late."
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