# Properties of d-sequence

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 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?

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 $a \in X^{m-1}(X \cap I)$.