Grade of a maximal prime ideal in a Noetherian UFD

Solution 1:

Since $R$ is noetherian, a maximal prime of $(a,b)$ is nothing but a maximal prime in the set $\operatorname{Ass}_R(R/(a,b))$. We can localize and reduce the problem to the following:

Let $(R,\mathfrak m)$ be a local noetherian UFD, and $a,b\in R\setminus\{0\}$ such that $(a,b)\ne R$. If $\mathfrak m\in\operatorname{Ass}_R(R/(a,b))$, then $\operatorname{depth}R\le 2$.

We have an exact sequence of $R$-modules $$0\to R/(a)\cap (b)\to R/(a)\oplus R/(b)\to R/(a,b)\to0.$$ Since $R$ is a UFD, $(a)\cap (b)$ is a principal ideal (generated by $ab/\gcd(a,b)$), and thus the projective dimension of $R/(a)\cap (b)$ is one. Now we get $\operatorname{pd}_R(R/(a,b))\le 2$. Applying the Auslander-Buchsbaum formula we obtain $\operatorname{depth}R=\operatorname{pd}_R(R/(a,b))\le 2$.