Exercise III.5.10 in Grillet's Abstract Algebra 


*Let $R$ be commutative, with characteristic either $0$ or greater than $m$. Show that a root $r$ of $A \in R[X]$ has multiplicity $m$ if and only if $A^{(k)}(r)=0$ for all $k<m$ and $A^{m}(r) \neq 0$. Show that the hypothesis about the characteristic of $R$ cannot be omitted from this result.


By definition of multiplicity $A(x)=(x-r)^{m}B(x)$, where $B(r) \neq 0$. It's easy to see, that $A^{k}(r)=0$ for all $k<m$ and $A^{m}(r)=m!B(r)$. But why $m!B(r) \neq 0$? Shouldn't $R$ necessarily be a domain?
 A: I think you need the characteristic to be prime, not necessarily that $R$ be a domain.
If the characteristic is prime, then note that:
Proposition. Let $R$ be a commutative ring, $r\in R$, $a,b\in\mathbb{Z}$. If $ar=br=0$, then $\gcd(a,b)r = 0$. In particular, if $R$ has prime characteristic $p\gt 0$, and $ar=0$ for some $0\lt a\lt p$, then $r=0$.
Proof. Write $\gcd(a,b) = ax+by$ with $x,y\in\mathbb{Z}$. Then $\gcd(a,b)r = (ax+by)r = x(ar)+y(br) = 0+0=0$.
If $0\lt a\lt p$, then $\gcd(a,p)=1$. Since $ar=pr=0$, then $r = \gcd(a,p)r = 0$. $\Box$
If the characteristic is prime, then we know $\gcd(m!,p)=1$, so the result would follow as well (in that direction, in any case).
If the characteristic $d\gt 0$ is not prime, then there exists $m$, $1\lt m\lt d$, such that $\gcd(m!,d)=k\gt 1$. Write $d=kn$, and let $A(x) = x^m(x+n)$. Then $x=0$ is a root of multiplicity $m$, and $A^{(m)}(0) = m!(n) = 0$, because $d=kn|m!(n)$.
In characteristic $0$ you will have problems if there are elements $a\in R$, $a\neq 0$,  such that $na=0$ for some $n\gt 0$. (Note that characteristic $0$ just means that the only positive integer such that $na=0$ for all $a$ is $n=0$). If such an $a$ and $n$ exist, then you can take $A(x) = x^n(x+a)$ to get the same problem.
