0

Let $R$ be a commutative Noetherian $k$-algebra which is an integral domain and let $I$ be an ideal of $R$, $I:=\langle a_1,\ldots,a_n \rangle$ with $a_1,\ldots,a_n \in R$ a regular sequence.

Assume that $b_1,\ldots,b_m \in R$ also generate $I$, namely: $I=\langle b_1,\ldots,b_m \rangle$.

When $b_1,\ldots,b_m$ is a regular sequence? Probably not always?

Perhaps the following questions are relevant: i and ii.

Any hints and comments are welcome!

user237522
  • 6,467
  • The question is vague. It is better to ask the case you are most interested in. The answer to your second question is "No, not always." Also no to the first one unless $m = n$. – Youngsu Feb 10 '20 at 04:21
  • @Youngsu, thank you for your comment. I am interested in polynomial rings over fields, $k[x_1,\ldots,x_n]$, where $n \geq 2$ ($k[x]$ is a PID, so there is no much to ask). Please, could you elaborate on "Also no to the first one unless $m=n$"? Thanks. – user237522 Feb 10 '20 at 05:01
  • https://math.stackexchange.com/questions/707098/existence-of-a-finite-length-maximal-regular-sequence – user237522 Feb 10 '20 at 05:09
  • @Youngsu, please, in particular, in $k[x,y]$, if we know that $I=\langle u,v \rangle= \langle a,b,c,d \rangle$, with $u,v$ regular, can we pick two of ${a,b,c,d}$ and get a regular sequence? (which will not necessarily generate all $I$). – user237522 Feb 10 '20 at 05:14
  • If $I=(x,y)$ and $a=(xy+1)x$, $b=(xy+1)y$, $c=xy$, $d=0$, then no two of these form a regular sequence. – user26857 Feb 11 '20 at 06:41
  • @user26857, nice, thanks. What if $abcd \neq 0$? Or what if every two of ${a,b,c,d}$ are algebraically independent over $k$? – user237522 Feb 11 '20 at 11:59
  • Similarly one can construct an example with none of the generators equal to zero. – user26857 Feb 11 '20 at 12:30
  • The idea is to choose these generators such that the gcd of any two of them is not 1. This is what I did in the example above. – user26857 Feb 11 '20 at 14:50
  • @user26857, thank you for the clarification. – user237522 Feb 11 '20 at 17:21

0 Answers0