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Let $(R,m)$ be a local Noetherian ring and $\hat R$ be the $m$-adic completion of $R.$ Let $I$ be parameter ideal (generated by system of parameters) of $\hat R.$

How to prove the following.

There exists a parameter ideal $J$ of $R$ such that $J\hat R=I.$

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  • $\begingroup$ The obvious candidate is $I \cap R$ and it works. $\endgroup$ – MooS Nov 22 '17 at 15:49
  • $\begingroup$ @MooS Could you please explain little more? Is it true for any faithful extension? $\endgroup$ – Cusp Nov 22 '17 at 16:17
  • $\begingroup$ It is easy to show that $I \cap R$ is a parameter ideal. It is also easy to see that this is your only choice. If $J\hat R=I$, we have $I \cap R=(J\hat R) \cap R=J$, where the latter holds because the completion is faithfully flat. So the main part of the proof is to show that $(I \cap R)\hat R = I$, which does not hold for arbitrar faithfully flat maps. It does not hold for inclusions of fields into non-fields for instance. $\endgroup$ – MooS Nov 23 '17 at 8:38

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