Is a regular sequence ordered? 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?

 A: 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.
A: 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
http://amathew.wordpress.com/2010/10/30/the-notion-of-a-regular-sequence/
A: 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.
