Re: semi-local rings

From: Nicolas Ojeda Baer (nojb_at_fibertel.com.ar)
Date: 06/09/04 

Date: 9 Jun 2004 10:26:20 -0700

Like everything, it's obvious once you see it.

Thanks for the help,
nojb.

magidin@math.berkeley.edu (Arturo Magidin) wrote in message news:<ca4r4m$1ctp$1@agate.berkeley.edu>...
> In article <ca2sfv$lvm$1@agate.berkeley.edu>,
> Arturo Magidin <magidin@math.berkeley.edu> wrote:
> >In article <a8adcf56.0406041507.5cdd9253@posting.google.com>,
> >Nicolas Ojeda Baer <nojb@fibertel.com.ar> wrote:
>
> Before Bill Dubuque tells me how to do it easier...
>
> >
> >>I'm trying to prove the following, which I _think_ is true.
> >>
> >>Let R be a semi-local ring (R assumed commutative), that is, a ring
> >>with finitely many maximal ideals M_1, ..., M_r.
> >>
> >>Let T_i : R -> A_{M_i} be the natural map from A to the localization
> >>of A at each maximal ideal. Consider T = (T_1, ..., T_r) the induced
> >>product map A -> \Prod_{i=1}^r A_{M_i}. Then T is an isomorphism.
> >>
> >>I _think_ I've got injectivity worked out, but surjectivity still
> >>misses me.
> >>
> >>Any ideas?
>
> Pick any semi-local non-local domain. The product of the A_{M_i} has
> zero divisors, and so cannot be isomorphic to a domain. A much easier
> observation than the calculations I made; examples of such would be
> the integers localized away from a finite number of primes. E.g., let
> p_1,...,p_n be distinct primes, and let R be the ring of all rationals
> which, when written in lowest terms, have denominator relatively prime
> to p_1*...*p_n. If n>1, the product has zero divisors, but clearly R
> does not.
>
> --
> ======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
> ======================================================================
>
> Arturo Magidin
> magidin@math.berkeley.edu

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