Grade of an ideal equal to that of all its associated primes? By a well-known theorem in commutative algebra, if $I$ is an ideal of a commutative ring $R$ then $\operatorname{grade}(I)$ is the minimum of $\operatorname{grade}(P_i)$, where $P_i$ are the associated primes of $I$. Hence, at least one of $P_i$ has grade equal to $\operatorname{grade}(I)$. My question is: > Are there examples showing that not all $P\in\operatorname{Ass}(R/I)$ have necessarily grade equal to $\operatorname{grade}(I)$? Thanks in advance to any helper or solver.

There are very simple examples: $R=K[x,y,z]$ and $I=(xy,xz)$. In fact, $I=(x)\cap(y,z)$. Obviously, the height of $I$ is one, and thus its grade is also one ($R$ is Cohen-Macaulay). The associated primes of $I$ are $(x)$ (which has grade one) and $(y,z)$ (which has grade two).

**Added later** : An ideal $I$ with the property $\operatorname{grade}(I)=\operatorname{grade}(P)$ for all $P\in\operatorname{Ass}(R/I)$ is called _grade-unmixed_. An important class of such examples is given by the _perfect_ ideals, that is, ideals $I$ with the property $\operatorname{grade}(I)=\operatorname{pd}_R(R/I)$.