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.

### 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 [40] lists the wrong year. It has to be 1989 instead of 1997.