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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.
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.