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Oct 4, 2018 at 8:47 comment added BrianT Thanks. Is there a kind of geometric interpretation of being Macaulay ? In terms of the zero set of $J$ for instance ?
Oct 4, 2018 at 8:34 comment added abx It needs dimension $n-k$ and CM, as shown by my counter-example.
Oct 4, 2018 at 8:23 comment added BrianT Sorry to insist, but I don't understand what conditions are sufficient and necessary. Is it enough that $\mathbb{C}[u] / J$ has dimension $n-k$, or does it need to have dimension $n-k$ and be CM, given the two points above ?
Oct 4, 2018 at 3:47 comment added abx Yes, sorry I didn't make that explicit.
Oct 3, 2018 at 21:26 comment added BrianT Suppose that it is indeed Cohen-Macaulay. Don’t we need the dimension to be equal (and not at least) $n-k$ ?
Oct 3, 2018 at 16:39 comment added abx I think that $\mathbb{C}[u]/J$ Cohen-Macaulay is sufficient.
Oct 3, 2018 at 16:07 comment added BrianT Thank you for your comment. What would be a sufficient additional condition then ?
Oct 3, 2018 at 14:48 comment added abx No. Take $n=2$, $k=1$, $J=(u_1^2,u_1u_2)$, $I=(u_2)$. Then 1. and 2. are satisfied, but $u_2$ is a zero divisor in $\mathbb{C}[u]/J$.
Oct 3, 2018 at 14:00 history asked BrianT CC BY-SA 4.0