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We know the fact that if $R$ is a regular local ring then $R_{P}$ the localization of $R$ at $P$, $P\in\mathrm{Spec}(R)$ is a regular local ring.

So, I wonder the converse is true or not? My counter-example is taking $R=\mathbb Z$.

-If $P=(0)$ then $R_{P}$ is a field so $R_{P}$ is a regular local ring

-If $P=(p)$, $p$ is a prime number then $\dim R_{p}=1=v(PR_{P})$ (generate by $\frac{p}{1}$) so it is also a regular local ring.

However, $\mathbb Z$ is not a regular local ring. Can you check that to me please?

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  • $\begingroup$ I know it is not true. I just wanna know my counter example is true or false. $\endgroup$
    – Soulostar
    Jan 9, 2017 at 9:58
  • $\begingroup$ Because I do not know if the localization of R at P is a regular local ring with every P in Spec(R) then is R a regular local ring? It is not true and I try to find the example of that. $\endgroup$
    – Soulostar
    Jan 9, 2017 at 10:02
  • $\begingroup$ Oh I see your meaning. You mean that $R$ is regular iff $R_{P}$ is regular local ring with every prime ideal $P \in Spec(R)$. But it is just quite similar to my question. The theorem (Serre 1955) is: Suppose that $(R;m)$ is a regular local ring. Then $R_{P}$ is again a regular local ring for every a prime ideal $P \in $R$. I wonder the converse of this theorem is true or not? so we just find some rings which is not local then the converse is not true. $\endgroup$
    – Soulostar
    Jan 9, 2017 at 10:26
  • $\begingroup$ No, it is if every $P\in Spec(R)$ the localization of $R$ at P is regular local ring then is R regular local ring? $\endgroup$
    – Soulostar
    Jan 9, 2017 at 10:29
  • $\begingroup$ But how about your meaning? Can you give me the example? @user26857 $\endgroup$
    – Soulostar
    Jan 9, 2017 at 10:45

1 Answer 1

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Let $I$ be a prime ideal of $R_P$, then $I^c$ is a prime ideal of $\mathbb{Z}$, so $I^c=q\mathbb{Z}$ for some prime number $q$.

We have $I=I^{ce}=(q\mathbb{Z})^e=(\frac{q}{1})$. However, $q\mathbb{Z}$ is maximal so $I$ is also maximal. So an arbitrary nonzero prime ideal of $R_P$ is maximal, which means $\dim R_P=1=v(PR_P)$

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  • $\begingroup$ Yes, I see that is what I mention. :D $\endgroup$
    – Soulostar
    Jan 9, 2017 at 8:28
  • $\begingroup$ O...k. So what do need to check? $\endgroup$
    – T C
    Jan 9, 2017 at 8:28
  • $\begingroup$ I just wanna know it is true or false :). $\endgroup$
    – Soulostar
    Jan 9, 2017 at 8:33
  • $\begingroup$ Well, it's true then $\endgroup$
    – T C
    Jan 9, 2017 at 8:37

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