Let $x\_1,\ldots,x\_n$ is a sequence in a Noetherian local ring $R$. We say $x\_1,\ldots,x\_n$ is a $d$-sequence if


1) $x\_i\notin (x\_1,\ldots,\hat{x\_i},\ldots,x\_n),$


2) for all $k\geq i+1$ and all $i\geq 0,$ $(x\_0=0)$ $$((x\_0,\ldots,x\_i):x\_{i+1}x\_k)=((x\_0,\ldots,x\_i):x\_k).$$


In this paper [here](https://deepblue.lib.umich.edu/bitstream/handle/2027.42/23781/0000019.pdf?sequence=1) of Huneke, he showed that (Theorem 2.1) if $x\_1,\ldots,x\_n$ is a $d$-sequence modulo an ideal $I$ of $R$ then $$(x\_1,\ldots,x\_n)^m\cap I=(x\_1,\ldots,x\_n)^{m-1}I$$ for all $m\geq 1.$


As far as I have understood the proof of the theorem, I think the minimality property of $d$-sequence (i.e. property "1") is not required to prove the above result.


Is there anything wrong in my understanding of the proof?



Properties of d-sequence
Huneke only states an inclusion. But you are right, one gets a slightly stronger statement, with weaker assumptions. Namely, for any ring $R$, and any sequence $x\_1,\dots,x\_n$ a of elements of $R$, consider the following two properties (both implied by property $2)$ of $d$-sequences) :


$(a)$ For any $j \leq n$, one has $((x\_i)\_{i < j} : x\_j^2) = ((x\_i)\_{i < j} : x\_j)$.


$(b)$ For any $j \leq k \leq n$, one has $((x\_i)\_{i < j} : x\_j) \subseteq ((x\_i)\_{i < j} : x\_k)$.


Now let $X = (x\_1,\dots,x\_n)$ be an ideal generated by a sequence which has properties $(a)$&$(b)$ in $R/I$, for some ideal $I$. Then one has
$$
X^m \cap I = X^{m-1}(X \cap I)
$$
for any $m \geq 1$.


**Sketch of proof:**
Induction on $n + m$.
The case $n=1$ is handled as does Huneke, using $(a)$, and the case $m =1$ is obvious. For $n,m \geq 2$, let $a$ be an element of $X^m \cap I$, which can thus be written as $a = b + x\_1 c$ with $b \in (x\_2,\dots,x\_n)^m$ and $c \in X^{m-1}$. 


Now apply the induction hypothesis to $X'=(x\_2,\dots,x\_n)$ and $I' = (I,x\_1)$. Since $b$ is in $X'^m \cap I'$, this yields 
$$
b \in X'^{m-1}(X' \cap I') \subseteq X^{m-1}(X \cap I) + x\_1 X'^{m-1}.
$$
One can thus write $a = b' + x\_1 c'$ with $b'$ in $X^{m-1}(X \cap I)$ and $c'$ in $X^{m-1}$. Now $c'$ is in $X \cap (I:x\_1)$. By checking that Huneke's proof of $X \cap (I:x\_1) \subseteq I$ in proposition $2.1$ still holds only assuming $(a)$ and $(b)$, we get $c' \in X^{m-1} \cap I$. 


By induction, we get $c' \in X^{m-2}(X \cap I)$, and thus $a \in X^{m-1}(X \cap I)$.


