Correction to: Depth functions of symbolic powers of homogeneous ideals

Correction

1 Correction to: Invent. math.  https://doi.org/10.1007/s00222-019-00897-y

The original proof of Theorem 3.3 is incorrectly claims that $$\overline{I^{(t)}} = \bigcap _{F \in {\mathcal {F}}(I)} \overline{I_F^t}$$. We have found a counter-example to this claim. The proof remains correct if we replace $$\overline{I^{(t)}}$$ by $$\bigcap _{F \in {\mathcal {F}}(I)} \overline{I_F^t}$$ for all $$t \ge 1$$; see the corrected proof below. The correction concerns only this proof and does not affect any result of the paper.

Theorem 3.3

Let I be a monomial ideal in R such that $$I^{(t)}$$ is integrally closed for $$t \gg 0$$. Then $${{\,\mathrm{depth}\,}}R/I^{(t)}$$ is a convergent function with
\begin{aligned} \lim _{t \rightarrow \infty } {{\,\mathrm{depth}\,}}R/I^{(t)} = \dim R - \dim F_s(I), \end{aligned}
which is also the minimum of $${{\,\mathrm{depth}\,}}R/I^{(t)}$$ among all integrally closed symbolic powers $$I^{(t)}$$.

Proof

Let m be the minimum of $${{\,\mathrm{depth}\,}}R/I^{(t)}$$ among all integrally closed symbolic powers $$I^{(t)}$$. Choose an integrally closed symbolic power $$I^{(s)}$$ such that $${{\,\mathrm{depth}\,}}R/I^{(s)} = m$$. By Theorem 2.6(ii), there exists an integer a such that $${{\,\mathrm{depth}\,}}R/I^{(s)}\ge {{\,\mathrm{depth}\,}}R/I^{(t)}$$ for $$t \ge as^2$$. This implies $${{\,\mathrm{depth}\,}}R/I^{(t)} = m$$ for all integrally closed symbolic powers $$I^{(t)}$$ with $$t \ge as^2$$. Since $$I^{(t)}$$ is integrally closed for $$t \gg 0$$, we get $${{\,\mathrm{depth}\,}}R/I^{(t)} = m$$ for $$t \gg 0$$.

Let $$I_t = \bigcap _{F \in {\mathcal {F}}(I)} \overline{I_F^t}.$$ We will show that $$m = \min _{t \ge 1} {{\,\mathrm{depth}\,}}R/I_t$$. Since $$I_t = I^{(t)}$$ for $$t \gg 0$$ by Proposition 2.2,
\begin{aligned} m \ge \min _{t \ge 1} {{\,\mathrm{depth}\,}}R/I_t. \end{aligned}
By Proposition 2.3, we have
\begin{aligned} {{\,\mathrm{depth}\,}}R/I_s \ge {{\,\mathrm{depth}\,}}R/\big (I_s)^{(t)} \end{aligned}
for all $$s, t \ge 1$$. For $$t \gg 0$$, $$I^{(st)}$$ is integrally closed and so is $$I_F^{st}$$ for all $$F \in {\mathcal {F}}(I)$$ by Proposition 2.2. This implies $$I_F^{st} \subseteq \big (\overline{I_F^s}\big )^t \subseteq \overline{I_F^{st}} = I_F^{st}.$$ Hence, $$\big (\overline{I_F^s}\big )^t = I_F^{st}$$. So we get
\begin{aligned} (I_s)^{(t)} = \bigcap _{F \in {\mathcal {F}}(I)} I_F^{st} = I^{(st)}. \end{aligned}
Therefore,
\begin{aligned} {{\,\mathrm{depth}\,}}R/I_s \ge {{\,\mathrm{depth}\,}}R/ I^{(st)} \ge m \end{aligned}
for all $$s \ge 1$$. Now, we can conclude that
\begin{aligned} m = \min _{t \ge 1} {{\,\mathrm{depth}\,}}R/I_t. \end{aligned}
It remains to show that $$\min _{t \ge 1} {{\,\mathrm{depth}\,}}R/I_t = \dim R - \dim F_s(I).$$ For that, we need the following auxiliary observation (cf. [41, Proposition 2.5]).
Let $${\mathcal {F}}$$ denote the filtration of the ideals $$I_t$$, $$t \ge 0$$. Let $$R({\mathcal {F}}) = \bigoplus _{t \ge 0}I_ty^t$$. Then $$R({\mathcal {F}})$$ is an algebra generated by monomials in $$k[x_1,...,x_n,y]$$. We have
\begin{aligned} R({\mathcal {F}}) = \bigcap _{F \in {\mathcal {F}}(I)} \bigoplus _{t \ge 0}\overline{I_F^{t}}y^t. \end{aligned}
For each $$F \in {\mathcal {F}}(I)$$, the algebra $$\bigoplus _{t \ge 0}\overline{I_F^{t}}y^t$$ is the normalization of the finitely generated algebra $$\bigoplus _{t \ge 0}I_F^{t}y^t$$. Hence, $$\bigoplus _{t \ge 0}\overline{I_F^{t}}y^t$$ is a finitely generated algebra. The monomials of $$\bigoplus _{t \ge 0}\overline{I_F^{t}}y^t$$ form a finitely generated semigroup. Since the semigroup of the monomials of $$R({\mathcal {F}})$$ is the intersections of these semigroups, it is also finitely generated [14, Corollary 1.2]. From this, it follows that $$R({\mathcal {F}})$$ is a finitely generated algebra. Moreover, as an intersection of normal rings, $$R({\mathcal {F}})$$ is a normal ring. By [20, Theorem 1], this implies that $$R({\mathcal {F}})$$ is Cohen–Macaulay.
Let $$G({\mathcal {F}}) = \bigoplus _{t \ge 0}I_t/I_{t+1}$$. Then $$G({\mathcal {F}})$$ is a factor ring of $$R({\mathcal {F}})$$ by the ideal $$\bigoplus _{t \ge 0}I_{t+1}y^t$$. Hence, $$G({\mathcal {F}})$$ is a finitely generated algebra. By [4, Theorem 4.5.6(b)], we have $$\dim G({\mathcal {F}}) = \dim R$$. By the proof of the necessary part of [40, Theorem 1.1], the Cohen–Macaulayness of $$R({\mathcal {F}})$$ implies that of $$G({\mathcal {F}})$$. By [5, Theorem 9.23], these facts imply
\begin{aligned} \min _{t \ge 1} {{\,\mathrm{depth}\,}}R/I_t&= {{\,\mathrm{grade}\,}}{\mathfrak {m}}G({\mathcal {F}}) = {{\,\mathrm{ht}\,}}{\mathfrak {m}}G({\mathcal {F}}) \\&= \dim G({\mathcal {F}}) - \dim G({\mathcal {F}})/{\mathfrak {m}}G({\mathcal {F}})\\&= \dim R - \dim G({\mathcal {F}})/{\mathfrak {m}}G({\mathcal {F}}). \end{aligned}
We have $$G({\mathcal {F}})/{\mathfrak {m}}G({\mathcal {F}}) = \bigoplus _{t \ge 0}I_t/({\mathfrak {m}}I_t+I_{t+1})$$. Since $$F_s(I) = \bigoplus _{t \ge 0} I^{(t)}/{\mathfrak {m}}I^{(t)}$$, $$I^{(t+1)} \subseteq {\mathfrak {m}}I^{(t)}$$ [10, Proposition 9] and $$I_t = I^{(t)}$$ for $$t \gg 0$$, the graded algebras $$G({\mathcal {F}})/{\mathfrak {m}}G({\mathcal {F}})$$ and $$F_s(I)$$ share the same Hilbert quasi-polynomial [4, Theorem 4.4.3]. From this, it follows that $$\dim G({\mathcal {F}})/{\mathfrak {m}}G({\mathcal {F}}) = \dim F_s(I)$$. Therefore,
\begin{aligned} \min _{t \ge 1} {{\,\mathrm{depth}\,}}R/I_t = \dim R -\dim F_s(I). \end{aligned}
$$\square$$

Moreover, the reference  lists the wrong year. It has to be 1989 instead of 1997. 