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Assume just for sake of simplicity that $R = k[x_1 , \dots , x_n]$ is a standard graded polynomial ring over a field. If one considers the ideal $$I = \left({x}_{1}{x}_{3},{x}_{2}^{2},{x}_{2}{x}_{3},{x}_{3}^{2}\right)$$ and computes the minimal free resolution, the very last differential takes the form $$\begin{pmatrix} x_3 \\ 0 \\ -x_2 \\ x_1 \\ \end{pmatrix}.$$ This means that the very back differential has a "row" of $0$'s. My question is the following: is it possible to have a differential appearing in a minimal free resolution (of a cyclic module, say) with an entire row being equal to $0$ (and the row has length at least $2$).

Computationally, I have found many "near-miss" examples, where the differentials have a very long row and a single nonzero entry. It seems that I can never quite get a long row of $0$'s. Also, in the Cohen-Macaulay case this is definitely impossible since a row of $0$'s would correspond to a column of $0$'s in the minimal free resolution of the canonical module.

Any example (or explanation of why this cannot happen) would be greatly appreciated.

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Since I can not comment on the other's posts as commenting needs at least 50 reputation, I'm forced to release an answer. It seems that the second power, $I^2$, of your ideal $I$ has a zero row of length 4. Compute it via Macaulay2 and look at the matrix of the very last differential.

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  • $\begingroup$ This is certainly an answer to my question. The row of zeros gets even longer as one increases the powers, too, so it looks like it becomes arbitrarily long. $\endgroup$
    – Rellek
    Jun 22, 2021 at 1:16

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