0

Suppose $R$ is a regular local ring of dimension 3 and $I=(a,b)$ is of height one. Is it always true that $I_p$ is not principal for all height two primes $p$? Is there an example?

I suspect that this may be false in general. But somehow the ring being regular local may make things interesting.

dongrugose
  • 189
  • Let ${\frak m}=(x_1, x_2, x_3) , I=(x_1x_2,x_1x_3)$ and $p=(x_1 , x_2)$. I am not sure, but it seems that $I_p=(x_1)_p$ and so it is principal. – Mohammad Bagheri Sep 14 '19 at 21:59
  • 2
    Regular local rings are UFDs, so $I=(a_1d,b_1d)$ where $d$ is the gcd of $a,b$. If $I$ is height one, $d$ is not a unit. If $I$ is two generated, but not one generated, then both $a_1,b_1$ are non-units and then any minimal prime containing $a_1,b_1$ has height 2 and localization at any one these prime makes $I$ non-principal. – Mohan Sep 14 '19 at 23:16
  • @Mohan, thats works! thanks! – dongrugose Sep 15 '19 at 02:57

0 Answers0