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

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    $\begingroup$ have you tried Using depth lemma on $$0\to R/(p_1 \cap p_2)\to R/p_1 \oplus R/p_2\to R/(p_1+p_2)\to 0 $$ $\endgroup$
    – user 1
    May 2, 2015 at 17:58

2 Answers 2

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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 $\operatorname{depth}(R/p_1\cap p_2)=1$.

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    $\begingroup$ @Selva If you want a slight generalization to primary decompositions then I suggest you Lemma 1.1 from this paper. $\endgroup$
    – user26857
    May 2, 2015 at 18:10
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Hint. Use depth lemma on $$0\to R/(p_1 \cap p_2)\to R/p_1 \oplus R/p_2\to R/(p_1+p_2)\to 0 $$

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  • $\begingroup$ what about using repeatedly? $\endgroup$
    – user 1
    May 2, 2015 at 18:10

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