Is a normal series gotten?

A regular sequence is an $n$ - layer collection $\{r_1, \cdots, r_n\} \subset R$ of components of a ring $R$ such that for any kind of $2 \leq i \leq n$, $r_i$ is not an absolutely no divisor of the quotient ring $$ \frac R {\langle r_1, r_2, \cdots, r_{i-1} \rangle}.$$

Does the order of the $r_i$ is issue? That is, is any kind of permutation of a normal series normal?

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2019-12-02 02:50:09
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Answers: 2

In basic, no. If $R$ is a neighborhood noetherian ring, after that of course, though.

A counterexample is if $R = k[x,y,z]/(x-1)z$ over an area $k$, and also the components $x, (x-1)y$. This is a normal series despite the fact that $(x-1)y$ is a zerodivisor (so the exchanged series is not normal).

It holds true under neighborhood theories by the Krull junction theory ; see

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

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2019-12-03 04:26:29
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Not as a whole. A typical instance is $R=k[x,y,z]$, where $k$ is an area. After that, $x,y(1-x),z(1-x)$ is normal yet $y(1-x),z(1-x),x$ is not.

On the silver lining, if $R$ is Noetherian, neighborhood after that every permutation of a normal series is normal.

Actually extra holds true. We can expand this idea to components over rings analogously. After that, if $M$ is a finitely created component over a Noetherian, neighborhood ring, after that every permutation of a normal series is normal.

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2019-12-03 04:26:18
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