On the depths of symbolic powers of the Stanley-Reisner ideal of a bow-tie complex

Question

Consider the polynomial ring $S=k[x_1,...,x_5]$.
Consider the Stanley-Reisner ideal $I$ (i.e. the face ideal) of the simplicial complex which is a bow-tie $\Delta:=\left<x_1x_2x_3,x_3x_4x_5\right>$.
So $I=\left<x_1x_4,x_1x_5,x_2x_4,x_2x_5\right>$.
Let $I^{(m)}$ denote the $m$-th symbolic power of $I$ .

Is it true that depth $S/I^{(m)}\ge 2, \forall m\ge 1$ ?

Answer 1

We will use the following isomorphism

Let $R[x]$ be a polynomial ring over a ring R. If $J$ is an $R$-ideal, then $R[x]/JR[x] \cong R/J [x]$.

Notice that the ideal $I$ has a primary decomposition $$ I = (x_1,x_2) \cap (x_4,x_5), $$ and the generators of $I$ do not involve the variable $x_3$.

Now, to show that depth $S/I^{(m)} \ge 2$, as $x_3$ is a nonzerodivior (use the quoted statement above), if suffices to show that depth $S'/J^{(m)} \ge 1$ in $S' = k[x_1,x_2,x_4,x_5]$ with $J = (x_1x_4,x_1x_5,x_2x_4,x_2x_5)S'$. As the associated primes of $J^{(m)}$ are $(x_1,x_2)S'$ and $(x_4,x_5)S'$, an element such as $x_1 + x_4$ is a nonzerodivisor on $S'/J^{(m)}$.