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A regular sequence is an $n$-fold collection $\{r_1, \cdots, r_n\} \subset R$ of elements of a ring $R$ such that for any $2 \leq i \leq n$, $r_i$ is not a zero divisor of the quotient ring $$ \frac R {\langle r_1, r_2, \cdots, r_{i-1} \rangle}.$$

Does the order of the $r_i$'s matter? That is, is any permutation of a regular sequence regular?

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up vote 5 down vote accepted

Not in general. A standard example is $R=k[x,y,z]$, where $k$ is a field. Then, $x,y(1-x),z(1-x)$ is regular but $y(1-x),z(1-x),x$ is not.

On the bright side, if $R$ is Noetherian, local then every permutation of a regular sequence is regular.

In fact more is true. We can extend this notion to modules over rings analogously. Then, if $M$ is a finitely generated module over a Noetherian, local ring, then every permutation of a regular sequence is regular.

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You beat me by one minute! – Akhil Mathew Nov 29 '10 at 23:46
@Akhil: and with the same example :) – Timothy Wagner Nov 29 '10 at 23:52

In general, no. If $R$ is a local noetherian ring, then yes, though.

A counterexample is if $R = k[x,y,z]/(x-1)z$ over a field $k$, and the elements $x, (x-1)y$. This is a regular sequence even though $(x-1)y$ is a zerodivisor (so the swapped sequence is not regular).

It is true under local hypotheses by the Krull intersection theorem; see

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If $R$ is a local non-Noetherian integral domain the answer is still no! See M. Hochster - Pathological maximal R-sequences in quasilocal domains, Portugaliae Mathematica V. 38, Fasc. 3-4 (1979), 33-36. – user26857 Jun 17 '12 at 0:17

Here is a general result for when any permutations of elements of a regular sequence forms a regular sequence:

Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module. If $x_1,...,x_n$ be an $M$-sequence s.t. $x_i \in J(A)$ for $1 \leq i \leq n$, where $J(A)$ is the Jacobson radical of $A$, then any permutation of $x_1,...,x_n$ becomes an $M$-sequence.

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