I've been doing some practice problems for an upcoming intro level abstract algebra class and came across this problem (in Herstein 3rd edition, Ch 4.5 #26 & #28). I had some help from my professor with reasoning through the first part but am unsure if I got it or not.
i) Let $R$ be a commutative ring in which $a^2 = 0$ only if $a = 0$. Show that if $q(x) \in R[x]$ is a zero-divisor in $R[x]$, then, if $q(x) = a_{n}x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ there is an element $b \neq 0$ in $R$ such that $ba_0 = ba_1 = ... = ba_n = 0$.
ii) Show that the element b must exist even if the condition "$a^2 = 0$ only if $a = 0$" does not hold
My figuring thus far: Let $h(x) \in R[x]$ such that $h(x)q(x) = 0 = q(x)h(x)$ , where $h(x) = c_mx^m+c_{m-1}x^{m-1} + ... + c_1x+c_0$. We know that the product of these polynomials is zero, so we know that every coefficient of the product must be zero: $q(x)h(x) = a_nc_mx^{m+n} + (a_nc_{m-1} + a_{n-1}c_m)x^{n+m-1} + ... + a_0c_0 = 0 \rightarrow a_n$ and $c_m$ are zero divisors in $R$ (as we will have written $h(x)$ and $q(x)$ such that their leading coefficients are nonzero). We also know that $a_nc_{m-1} + a_{n-1}c_m = 0 \rightarrow c_m(a_nc_{m-1}+a_{n-1}c_m) = 0 \rightarrow 0 = (c_ma_n)c_{m-1} = -c_m^2a_{n-1}$. Following this line of reasoning it seems that $b= c_m^k$ for some appropriate k (I think k=n+1?).
Does this seem correct? In my mind this is an appropriate proof for i) and ii) but I suspect something more nuanced is going on in ii) that I'm missing. EDIT: I think in part two the issue is that I can't say $c_m^2a_{n-1} = 0$ because I haven't shown that $a_{n-1} \neq j^2$ for some $j \in R$, but I'm still not really sure.
Thanks for any and all input.
(Meta: First post. Was this appropriate? If not let me know how I can improve)
