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 ...