Re: I-adic completions

On 09.10.2006 14:00, Jannick Asmus wrote:
On 07.10.2006 10:22, Jannick Asmus wrote:
On 07.10.2006 04:37, canadan wrote:
Let A be a noetherian ring and I contained in J ideals. If A is
J-adically complete then A is I-adically complete. Why?

I believe that it has to do with the Artin-Rees theorem.
If I and J are arbitrary ideals with I contained in J, the assertion
does not seem to be true to me. Is it possible that J denotes the
nilradical of A?

... or, more generally, that J is contained in the nilradical of A?

However, the assertion as you stated is not true.

Counterexample: Let A=k[[X]], J=<X> and I=<X(X-1)>.
The sequence {X^n} is Cauchy in (A,I), since X^n-X^(n-1) converges to 0
in (A,I). The sequence converges to 0 in (A,J). If (A,I) was complete,
then {X^n} would converge to 0 in (A,I), too, since (A,I) is Hausdorff
and the identity map (A,I) -> (A,J) is continuous. But no X^n is
contained in any I^k which means in other words that {X^n} does not
converge to 0. Contradiction!

Withdrawn - since I=J because of the fact that X-1 is a unit in A. :-(
.

Relevant Pages

  • Re: I-adic completions
    ... On 07.10.2006 10:22, Jannick Asmus wrote: ... I believe that it has to do with the Artin-Rees theorem. ... the assertion as you stated is not true. ... Contradiction! ...
    (sci.math)
  • Re: I-adic completions
    ... On 07.10.2006 04:37, canadan wrote: ... I believe that it has to do with the Artin-Rees theorem. ... If I and J are arbitrary ideals with I contained in J, the assertion ... Is it possible that J denotes the ...
    (sci.math)
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