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Why order is an important matter in the following case of regular sequence?

The following lines are mentioned in https://en.wikipedia.org/wiki/Regular_sequence

$x, y(1-x), z(1-x)$ is a regular sequence in the polynomial ring $ \mathbb{C}[x, y, z]$, while $y(1-x), z(1-x), x$ is not a regular sequence.

Why is that ?

The regular sequence means the sequence of elements of $ \mathbb{C}[x, y, z]$ which are independent.

Edit: $ \mathbb{C}[x, y, z]/y(1-x)$ contains the zero-divisor $z(1-x)$. So the second sequence do not form regular sequence

user26857
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MAS
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    Did you attempt to verify the definition located on that page? – KReiser Mar 17 '20 at 21:37
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    Did you try checking if some elements of the second sequence are zero divisors when you mod out by previous elements of the sequence? – Samir Canning Mar 17 '20 at 21:37
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    Are you able to show that $z(1-x)$ is a zero-divisor in $\Bbb{C}[x,y,z]/(y(1-x))$? (It's fairly easy to do so.) – Eric Towers Mar 17 '20 at 21:41
  • @EricTowers, thank you. But how to argue by independence argument as above – MAS Mar 17 '20 at 21:42
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    I got downvote but the comments were really helpful to understand the question because I am doing self-study to understand the current topics in my recent questions in this great platform. I don't mind really with negative vote but want to thank people who downvoted but a left a comment at least. – MAS Mar 17 '20 at 22:02
  • It should not surprise you that the question (in it's initial form) was downvoted: it had no context, and the obvious response was to check the definition (as evidenced by the first three comments). Yes, the question is less bad now, but downvoters do not always come back to check on the questions they downvote. In the future, you should consider attempting to make your post better before publishing it - this will help you avoid such downvotes. – KReiser Mar 17 '20 at 22:37
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    @KReiser, Thank you, but I put all information and link in the question. I am very sure that all sources are available in my question. So I will always say "improving question by editing or suggestion correction before downvoting is the best way to uniformly distribute the openness to all of this great platform". As I am learning by self-study without having any professional class on these topics, it is very natural to seek simple helps from others. I don't really mind any downvote but I highly appreciate any comment on my question – MAS Mar 17 '20 at 22:54
  • Here is a link to what you posted the first time: link. You can see that there is no context in this version of the post, nor any obvious effort from you, which makes that version a poor question. Poor questions tend to get downvotes! Yes, you have worked on these issues now after making 4 edits, but this effort is really something you should be capable of making before you post - you've been here long enough that you should understand this. – KReiser Mar 17 '20 at 23:01
  • @KReiser, thank you, we do mistake when we learn. I think the above whole discussion is an important part of learning and teaching. Because everything can no be "smooth" – MAS Mar 18 '20 at 08:56

1 Answers1

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You don't give a definition of independence. Neither does the summary you are citing. Of course, the summary says "In commutative algebra, a regular sequence is a sequence of elements of a commutative ring which are as independent as possible, in a precise sense." You seem to have discarded the last phrase. In what "precise sense"?

The actual definition on that page, after a warm-up defining a module regular sequence defines a regular sequence.

For a commutative ring $R$ [...] An $R$-regular sequence is called simply a regular sequence. That is, $r_1$, ..., $r_d$ is a regular sequence if $r_1$ is a non-zero-divisor in $R$, $r_2$ is a non-zero-divisor in the ring $R/(r_1)$, and so on.

This is the precise sense meant by the summary. You shouldn't expect the summary to tell you what you need to know; it's a summary. If you want to know definitions, read the definitions.

But how does this precise definition lead to the vague statement in the summary? Suppose $r_i$ is not a zero divisor in $R$ (so not trivially a zero divisor). If $r_i$ is also not a zero divisor in $R/(r_1, \dots, r_{i-1})$, then $r_i$ is independent of the "span" of $(r_1, \dots, r_{i-1})$ in $R$. Alternatively, if $r_i$ is also a zero divisor in $R/(r_1, \dots, r_{i-1})$, then $r_i$ is in the "span" of $(r_1, \dots, r_{i-1})$.

In both cases, I use "span" in quotation marks because "span" is a property of vectors in a vector space and we are not working in a vector space. If we continue the analogy to a vector space, We have a vector space $V$, and a sequence of vectors $v_1$, ..., $v_n$. We are constructing the vector space quotients $V/\mathrm{span}(v_1)$, $V/\mathrm{span}(v_1, v_2)$, ..., $V/\mathrm{span}(v_1, ..., v_n)$. Each quotient is different if each $v_i \not\in \mathrm{span}(v_1, \dots, v_{i-1})$, that is each $v_i$ is independent of the subspace spanned by $v_1$ through $v_{i-1}$. Notice that if $v_1, \dots v_i$ are not independent (by definition) there is a nonzero linear combination of them that produces zero; this is our analog to being a zero-divisor in the quotient. However, a ring is not a vector space.

We are working in a ring and a collection of elements in a ring generate an ideal. So a regular sequence is regular because each term of the sequence is outside the ideal generated by the previous elements of the sequence. We detect this by detecting that some $v_i$ either is or is not a zero-divisor in $R/(v_1, \dots v_{i-1})$, so either is or is not independent, in the sense we have described, of the previous sequence members.

Eric Towers
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