Does an ideal $I=(x_1,...,x_n)$ which contains a regular element implie $x_i$ is regular for some $i$?

Question

Suppose $R$ is a commutative ring. $I=(x_1,...,x_n)$ is an ideal. Suppose $I$ contains a $R$-regular element. Does this imply that $x_i$ is $R$-regular for some $i$?

If not, is this true for Noetherian ring or Noetherian local ring?

Thank you in advance.

Answer 1

No, let $R=\mathbb{C}[\![x,y]\!]/(xy)$. This is a commutative Noetherian local ring. Take $I=(x,y)$, which is the unique maximal ideal of $R$. Then both $x$ and $y$ are zero divisors as $xy=0$ in $R$. But the depth of $R$ is $1$, since $\mathbb{C}[\![x,y]\!]$ has depth $2$ and $xy$ is a nonzerodivisor on $\mathbb{C}[\![x,y]\!]$, so $I$ contains a nonzerodivisor of $R$.

More concretely, the associated primes of $R$ are exactly $(x)$ and $(y)$ so anything outside their union is a nonzerodivisor of $R$, e.g. $x+y$.

Answer 2

In the ring of matrix take

$x_{1} = \left( \begin{array}{lcr} 1 & 0\\ 0 & 0 \end{array} \right) $

$x_{2} = \left( \begin{array}{lcr} 0 & 0\\ 0 & 1 \end{array} \right) $

Then $Id = \left( \begin{array}{lcr} 1 & 0\\ 0 & 1 \end{array} \right) \in I$

Id is a regular element of $I$, but $x_{1}$ and $x_{2}$ are not regular elements