Advertisement

Correction to: Depth functions of symbolic powers of homogeneous ideals

  • Hop Dang Nguyen
  • Ngo Viet TrungEmail author
Correction
  • 100 Downloads

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

Notes

Acknowledgements

The original paper and this correction are supported by Vietnam National Foundation for Science and Technology Development under Grant Number 101.04-2019.313. The authors thank Arvind Kumar for pointing out the mistake of the original proof.

Copyright information

© Springer-Verlag GmbH Germany, part of Springer Nature 2019

Authors and Affiliations

  1. 1.Institute of MathematicsVietnam Academy of Science and TechnologyHanoiVietnam
  2. 2.International Centre for Research and Postgraduate Training, Institute of MathematicsVietnam Academy of Science and TechnologyHanoiVietnam

Personalised recommendations