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Let $(R, \mathfrak{m})$ be a Gorenstein local ring of characteistic $p>0$. Let $x_1,...,x_d$ be a system of parameters of $R$. Let $I$ be an $\mathfrak{m}$-primary ideal containg $(x_1,...,x_d)$. Then there exists an ideal $J$ containng $(x_1,...,x_d)$ such that $(x_1,...,x_d):J = I$.

Question. Is it true that for all power of $p$, say $q$, we have $$(x_1^q,...,x_d^q):J^{[q]} = I^{[q]}?$$

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up vote 4 down vote accepted

I am pretty sure that you already know a result of Hochster-Huneke, Corollary 4.3 in "Tight Closure, Invariant Theory, and the Briancon-Skoda Theorem", Journal of the AMS, 1990, for regular local rings.

I don't think it is true in general. Let $R = \mathbb{Z}/3[X,Y]_{(X,Y)}/(X^4)$. Let $x,y$ denote the images of $X,Y$ in $R$. Let $m = I = (x,y)$ and $y$ be a system of parameters of $R$. Then $$(y)^{[3]}: m^{[3]} = (y^3): (x^3,y^3) = (y^3) : x^3 = (x,y^3).$$ Since $(x,y^3) \subseteq m \setminus m^2$, the ideal $(x,y^3)$ can not be expressed as the 3rd bracket power of an ideal.

-------- An example when $R$ is reduced
I think this works. One can take $R= \mathbb{Z}/3[X,Y]_{(X,Y)}/ (X^2-Y^2)$, $I = (x,y)$, and $y$. Then we have $$(y^3) : (x^3,y^3) = (y^3) : x^3 = (y^3) : xy^2.$$ Therefore, $y$ is in the colon ideal. In fact, it is $(x,y)$. By the same reason as above it can not be expressed as the 3rd bracket power of an ideal.

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Thank you. Do you have any example with $R$ is reduce. – Pham Hung Quy Oct 23 '13 at 15:24
I added one more example. I wonder if there is a characterization of this property. – Youngsu Oct 23 '13 at 17:48
I think so. I have just see this paper of Wenliang Zhang (, the condition (C) in this paper is much similar to my question. He consider colon property for all ideals. So I think that ring satisfying my question may be regular of $F$-regualar. Do you have any example of non-regular ring for my question? – Pham Hung Quy Oct 24 '13 at 8:32
@ Youngsu: I have proved that the ring satisfying the question must be $F$-regular. – Pham Hung Quy Oct 25 '13 at 8:01
Pham Hung Quy: Hi. That is good to hear. But can you formally write the statement? I'd like to know how one particular ideal controls the regularity of the ring. – Youngsu Oct 25 '13 at 21:25

We prove an characterization of regularity based on the result of Wenliang Zhang ( It should be metion his result

Theorem 1 (W. Zhang). Let $(R, \mathfrak{m})$ be a local ring of characteristic $p>0$. The following are equivalent.

  1. $R$ is regular
  2. $I^{[q]}:J^{[q]} = (I:J)^{[q]}$ for all pair of ideals $I, J$ and for all powers $q$ of $p$.
  3. $I^{[q]}:x^{[q]} = (I:x)^{[q]}$ for all $\mathfrak{m}$-primary ideals $I$, all elements $x\in R$, and for all powers $q$ of $p$.

Definition 2. Let $R$ be a Notherian ring of characteristic $p$, $I$ an ideal. Then

  1. The tight closure of $I$ is $I^{*} = \{ x :\exists c \in R^o, cx^q \in I^{[q]} \,\, \text{for all} \,\, q =p^e \gg 0\}$, where $R^o = R \setminus \cup_{\mathfrak{p}\in min(R)} \mathfrak{p}$.

  2. The Frobenius closure of $I$ is $I^F = \{ x: x^q \in I^{[q]}\,\, \text{for all}\,\, q =p^e \gg 0\}$

Remark 3. It is easy to see that $I \subset I^F \subset I^*$. If $I^{[q]}:x^{[q]} = (I:x)^{[q]}$ for all elements $x\in R$, and for all powers $q$ of $p$, then $I = I^*$ (see Zhang's paper, Lemma 3.1).

Lemma 4. Let $I$ and $J$ is Frobenius closed i.e. $I = I^F$ and $J=J^F$. Then $(I \cap J)^{[q]} = I^{[q]} \cap J^{[q]}$ for all $q$.

proof. easy.

Result 5. Let $(R, \mathfrak{m})$ be a local ring of characteristic $p>0$. The following are equivalent.

  1. $R$ is regular
  2. (a) $R$ is Gorenstein; (b) $\mathfrak{a}^{[q]}:x^{[q]} = (\mathfrak{a}:x)^{[q]}$ for all parameter ideals $\mathfrak{a}$, all elements $x\in R$, and for all powers $q$ of $p$.

Proof. $(1) \to (2)$ is clear.

$(2) \to (1)$, By Remark 3 we have every parameter ideal of $R$ is tight closed. Moreover $R$ is Gorenstein so every ideal of $R$ is tight closed and hence Frobenius closed. It is enough to prove (3) of Zhang's theorem. Let $I$ be an $\mathfrak{m}$-primary ideal and $x$ an element. Choose a parameter ideal $\mathfrak{a}$ contained in $I$. Since $R$ is Gorenstein we have an ideal $\mathfrak{b} = (b_1,...,b_t)$ such that $I = \mathfrak{a}:\mathfrak{b} = \cap_{i=1}^t(\mathfrak{a}:b_i)$. By Lemma 4 we have $$I^{[q]} = \cap_{i=1}^t(\mathfrak{a}:b_i)^{[q]} = \cap_{i=1}^t(\mathfrak{a}^{[q]}:b_i^q).$$ Therefore $$I^{[q]}:x^q = \big(\cap_{i=1}^t(\mathfrak{a}^{[q]}:b_i^q) \big):x^q = \cap_{i=1}^t(\mathfrak{a}^{[q]}:(xb_i)^q) = \cap_{i=1}^t(\mathfrak{a}:(xb_i))^{[q]}.$$ The last equation follows from our asumption. So by Lemma 4 we have $$I^{[q]}:x^q = \big(\cap_{i=1}^t(\mathfrak{a}:(xb_i)) \big)^{[q]} = (I:x)^{[q]}.$$ The proof is complete.

Remark 6. I do not know the charateriztion of rings satisfying the property 2(b) of Result 5.

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