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.



Can the differentials in a minimal free resolution ever have a "long" row of $0$'s?
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.


