TITLE: Zerodivisors in Finite Commutative Local Rings
QUESTION [6 upvotes]: Let $p$ be a prime and $r$ a positive integer. Let  $\mathbb Z_{p^r}$ the ring of integers modulo $p^r$. Every zero-divisor $z$  in  $\mathbb Z_{p^k}$ can be written in the form $z=ap^k$, where $a$ is a unity. 
1) Let now $(R, \mathfrak m)$ be a finite commutative local ring. If the maximal ideal $\mathfrak m$ is principal and generated by $s$, then every zero divisor $z$ of $R$ can also be written in the form $z=as^k$ where $a$ is a unity or we need to impose some conditions on the ring $R$ in order to conclude that $z=as^k$ where $a$ is a unity?
2) Assume now that $(R, \mathfrak m)$ is a finite commutative local ring. Suppose that the maximal ideal $\mathfrak m$ is generated by $s_1, s_2, \dots, s_l$. What conditions we impose to the ring in order to have that a zero-divisor $z$ can be written in the form $z=a_1s_1^{k_1}+\cdots+a_ls_l^{k_l}$, where $a_i$ are a units, for $i=1, \dots , l$?

REPLY [2 votes]: Considering a commutative local finite ring $R$ with principal maximal ideal:
By Kaplansky's theorem, a finite commutative local ring with a principal maximal ideal is a special principal ideal ring. If its maximal ideal is $aR$, then the rest of its ideals are given by $a^nR$, which will eventually be zero. 
Clearly every element of $R$ is a unit or is nilpotent. Let $a^k r$ be a zero divisor with the power of $a$ as high as possible. (This amounts to determining which $a^nR$ it is in and which it is not in.) By our choice of $k$, $a$ does not divide $r$, hence $r$ is a unit.

Conditions to make the second part occur are probably going to be pretty rare. You are essentially asking that every zero-divisor is in a product of powers of a principal ideal, and that the generators of the maximal ideal are fixed. 
To illustrate, consider the 8-element local ring in this question. Since the maximal ideal has square zero, the product of any two powers of principal ideals is going to be zero, so you can't get anything except single powers of the generators you fixed. If you picked $(x,y)$ as in that example, you can't get $x+y$ as a product of $x$'s and $y$'s with a unit.