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 $i=1, \dots , l$?

  • $\begingroup$ Added some more information based on your revised second question. $\endgroup$ – rschwieb Nov 29 '12 at 17:56

In (1), the ring $R$ is assumed finite, and so is an Artin local ring with principal maximal ideal (the principal maximal ideal by assumption). It follows that $s^k=0$ for some $k\geq 1$ (the maximal ideal of an Artin local ring is nilpotent). The proper ideals of $R$ are then $(0),(s),\ldots,(s^{k-1}),R$, and any non-zero element of $R$ can be written as $us^j$ for $u\in R^\times$ a unit and a unique integer $j$ with $0\leq j\leq k-1$. Clearly such an element is a zero divisor if and only if $j\geq 1$.

To see why these facts hold (granting the fact that the maximal ideal of an Artin local ring is nilpotent, and assuming that $k\geq 1$ is the smallest integer such that $(s^k)=0$), let $I$ be a non-zero, proper ideal of $R$. Then $I\subseteq (s)$, and there is an integer $j$, $1\leq j\leq k-1$ such that $I\subseteq (s^j)$ but $I$ is not contained in $(s^{j+1})$ (since $I\neq 0$). Choose $r\in I$ such that $r\notin(s^{j+1})$. Then $r\in(s^j)$, so we can write $r=s^jr^\prime$ for some $r^\prime$. If $r^\prime\in(s)$ then $r\in (s^{j+1})$, so $r^\prime\notin(s)$, that is, $r^\prime$ is a unit. So $(s^j)=(r^\prime)\subseteq I$.

If any inclusion in the chain $R\supseteq(s)\supseteq(s^2)\supseteq\cdots(s^{k-1})\supseteq 0$ is not strict, i.e. if $(s^j)=(s^{j+1})$ for some $j$ with $0\leq j\leq k-1$, then $s(s^j)=(s^j)$ implies $(s^j)=0$ by Nakayama, a contradiction (since $R\neq 0$ and $k$ is the smallest integer $\geq 1$ with $(s^k)=0$).

Now if $r\in R\setminus\{0\}$, we have $r=(s^j)$ for a unique $j$ between $0$ and $k-1$. This means we can write $r=s^ju$ where $u\notin(s)$ (because $u\in(s)$ implies $r\in(s^{j+1})\subseteq(s^j)$, so $(s^j)=(s^{j+1})$, a contradiction as above), that is, $u\in R^\times$.

What's important in the above is not that $R$ is finite. I only used that to deduce that $R$ is Artinian. Everything I said is true for an arbitrary Artin local ring with principal maximal ideal.

I don't really understand your second question, because you say that $R$ has principal maximal ideal but then you list several generators.

  • $\begingroup$ Dear keenan thank you. I have edited the second question. $\endgroup$ – zacarias Nov 29 '12 at 16:06
  • $\begingroup$ Dear @zacarias, You're welcome. I'm afraid I don't know the answer to your second question. $\endgroup$ – Keenan Kidwell Nov 29 '12 at 16:11

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.


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