So I read some different versions of Krull's Principal Ideal Theorem and the Theorem 144 from Kaplanskys book "Commutative Algebra", which states that If there are prime ideals $P \subsetneq Q$ in a Noetherian ring and there is another prime ideal in between those, then there are infinitely many prime ideals in between (a direct consequence of Prime Avoidance).
Now I was wondering if the following, stronger formulation works:
Let $R$ be a local Noetherian ring with $\dim R=2$. Then every non-invertible element in $R$ is contained in some prime ideal of height $1$ (exactly!).
So I know that this is the case if $R$ does not contain any zero-divisors, but I struggle in the case if there are some. So I was neither able to figure out a proof nor give a counterexample, but I really would love to know an answer, as it would also show that every prime ideal of height $>1$ can be written as an union of (infinitely) many prime ideals of height $1$.
Yes. Let $M$ be the maximal ideal of $R$. Since $\dim R=2$, we have $\operatorname{ht}M=2$; let $$P_0<P_1<M$$ be any strictly ascending chain of primes of maximal length. Now, fix any non-unit $a\in R$. If $a\in P_1$, then we are done, since $\operatorname{ht}P_1=1$. Thus suppose $a\notin P_1$, and consider the domain $R\big/P_0$. Since $a\notin P_1$, also $a\notin P_0$, so the coset $a+P_0$ is non-zero in $R\big/P_0$; it is also certainly a non-unit, since it lies in the maximal ideal $M\big/P_0$. Thus we may find $Q\big/P_0$ a minimal prime over $a+P_0$, for some $Q\in\operatorname{Spec}R$ with $a\in Q$ and $Q>P_0$. Note that $\operatorname{ht}Q\geqslant 1$; we claim also that $\operatorname{ht}Q\leqslant 1$, after which we will be done. To see this, note that $Q\big/P_0$ is a minimal prime over a principal ideal, so that $\operatorname{ht}Q\big/P_0\leqslant 1$ by Krull's principal ideal theorem. In particular, since $\operatorname{ht}M\big/P_0=2$, we have $Q\neq M$. Thus, if we had $\operatorname{ht}Q>1$, say with a chain of prime ideals $Q_0<Q_1<Q$, then $$Q_0<Q_1<Q<M$$ would be a strictly ascending chain of primes of $R$, contradicting that $\operatorname{ht}M=2$. So we are done.
As you probably know, this is not the case if you drop the hypothesis that $R$ is local. Consider the ring $R=\mathbb{Q}\times\mathbb{Q}[x,y]$. The prime ideals of $R$ are precisely $\mathbb{Q}\times P$, where $P\in\operatorname{Spec}\mathbb{Q}[x,y]$, and $0\times\mathbb{Q}[x,y]$. In particular, since $\dim\mathbb{Q}[x,y]=2$, we have that $\dim R=2$, since $$\mathbb{Q}\times 0<\mathbb{Q}\times\langle x\rangle<\mathbb{Q}\times\langle x,y\rangle$$ is a ascending chain of primes of maximal length. On the other hand, consider the non-unit $a=(0,1)\in R$. The only prime ideal of $R$ containing $a$ is $0\times \mathbb{Q}[x,y]$, which has height zero, so this gives the desired counterexample.