Find $\operatorname{depth}(R/p_1\cap p_2)$

Question

Let $R=K[x_1,\dots,x_n]$ be a polynomial ring over a field, $K$. Let $I$ be a square free monomial ideal of $R$. Let $p_1 ,p_2$ be minimal prime ideals of $I$ generated by subsets of $\{x_1,\dots,x_n\}$ and $I =p_1\cap p_2 $.

What is $\operatorname{depth}(R/p_1\cap p_2)$?

Answer 1

By the Lemma 2.1 from this paper you get $$\operatorname{depth}(R/p_1\cap p_2)=\min(\operatorname{depth}R/p_2,1+\operatorname{depth}R/(p_1+p_2)),$$ which is now only a matter of knowing how many variables generates $p_2$ and $p_1+p_2$. (Of course, I've assumed that $p_1\nsubseteq p_2$ and viceversa.)

If, for instance, $p_1+p_2=(x_1,\dots,x_n)$ you always get $\operatorname{depth}(R/p_1\cap p_2)=1$.

Answer 2

Hint. Use depth lemma on $$0\to R/(p_1 \cap p_2)\to R/p_1 \oplus R/p_2\to R/(p_1+p_2)\to 0 $$