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This question is related to
Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements from $k[X_1,X_2,X_3,X_4]$?

Let $R=k[X_1,X_2,X_3,X_4]$ and $I= (X_1,X_2) \cap (X_3,X_4)$. I know that $4≥\operatorname{ara}(I)≥\operatorname{ht} I=2$. Is $\operatorname{ara}(I)≥3$? Is $\operatorname{ara}(I)≥4$? (Here $\operatorname{ara}$ means arithmetic rank, that is, the least number of elements of $R$ required to generate an ideal which has the same radical as $I$.)

Thanks for any hint.

Community
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user 1
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  • Are you following any book or lecture note on local cohomology? I would try to use Meyer-Vietories sequence. – Youngsu Mar 28 '14 at 16:50
  • i read 6 chapter of "Local Cohomology-An Algebraic Introduction with Geometric Applications-By M. P. Brodmann -online publish 2010" – user 1 Mar 28 '14 at 18:45

1 Answers1

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This ideal is well-known to have arithmetic rank $3$, and is not a set-theoretic complete intersection.

To see that $\text{ara}(I) \le 3$, notice that $I = (X_1X_3, X_1X_4, X_2X_3, X_2X_4) = \sqrt{(X_1X_3, X_2X_4, X_1X_4 + X_2X_3)}$, since $(X_1X_4)^2 = X_1X_4(X_1X_4 + X_2X_3) - (X_1X_3)(X_2X_4)$ (and similarly for $(X_2X_3)^2$).

To see that $\text{ara}(I) \ge 3$, note that $H^i_I(R) = 0$ for $i > \text{ara}(I)$, where $H^i_I(\_)$ is the $i^\text{th}$ local cohomology with support in $I$. It thus suffices to show $H^3_I(R) \ne 0$, which can be done by Mayer-Vietoris. If $p = (X_1, X_2)$, $q = (X_3, X_4)$, then $I = p \cap q$, and $p + q = m = (X_1, X_2, X_3, X_4)$. Now $H^4_p(R) = H^4_q(R) = 0$ (since $p, q$ are complete intersections) and $H^4_m(R) \ne 0$ (since $R$ is Cohen-Macaulay). By Mayer-Vietoris, there is an exact sequence

$$H^3_I(R) \to H^4_m(R) \to H^4_p(R) \oplus H^4_q(R) = 0$$

so $H^3_I(R)$ surjects onto a nonzero module, hence is nonzero.

user26857
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zcn
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  • @ucf: It suffices to check that the (images of the) $4$ given generators are linearly independent in the $k$-vector space $I \otimes_R R/m \cong I/mI$. This approach was explained in my answer to your linked question, and another approach was also given by user26857. Please consider accepting one of the answers there, or clarifying your confusion more – zcn Mar 29 '14 at 19:27