I'm trying to understand the theorem for systems of parameters in Noetherian local rings, which says:
Let $R$ be a Noetherian local ring with maximal ideal $m$. Then there exists an $m$-primary ideal in $R$ generated by $\dim R$ elements.
In the proof that I'm reading is said that the fact follows from these $2$ lemmas:
1) Let $R$ be a Noetherian ring and $a$ an ideal in $R$ of height $r$. Assume there are elements $a_1, ..., a_{s-1} \in a$, such that the height of $(a_1, ..., a_{s-1})$ is $s-1$ for some $1 \leq s \leq r$. Then there exists an element $a_s \in a$ such that the height of $(a_1, ..., a_s)$ is $s$.
2) (Krull's Principal Ideal Theorem) Let $R$ be a Noetherian ring and $a$ an element of $R$ that is neither zero divisor nor a unit. Then every minimal prime divisor $p$ of $(a)$ is of height $1$.
So if I can find an element that is neither a unit nor a zero divisor in any Noetherian local ring, then everything is OK. But I'm not sure that every such ring has such element, i.e. is there a Noetherian local ring consisting only of units and zero divisors ? Thanks in advance !
Edit: OK, I see from the answers that there exist rings consisting only of units and zero divisors. So my question now is how to deal with them? It seems like the two lemmas are not sufficient or I'm missing something. Any ideas?
Of course! Any artinian local ring consist of units and nilpotents. ($\mathbb Z/4\mathbb Z$ is such an example.)
Edit. The two lemmas are not enough to prove the theorem. Let $R=K[X,Y]_{(X,Y)}/(X^2,XY)$. Then $\dim R=1$, and the maximal ideal of $R$, that is, $\mathfrak m=(x,y)$, consists of zerodivisors since $x\mathfrak m=0$. However, there is a principal $\mathfrak m$-primary ideal, namely $\mathfrak m^2=(y^2)$.