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From: William Elliot on 4 Aug 2010 05:30
On Tue, 3 Aug 2010, Hagen wrote:

>>>> Let R be a Boolean ring and a in R.
>>>> Is there a prime ideal excluding a?
>>
>> I'm still trying to fix a flawed proof
>> for which a hopefully yes answer to that
>> question would complete the proof.
>
> For a Boolean ring with unity this is true:
> it is known that such a ring is isomorphic to a
> subring S of (Z_2)^I for some set I, and such that
> p_i (S) = Z_2 for every i \in I,
> where p_i: (Z_2)^I --> Z_2 is the natural projection
> to the i-th component.

I is the collection of homomorphisms of the Boolean ring into Z_2.

The Boolean ring representation theorem is not to be used
because proof of that is the flawed proof I'm trying to fix.

I see now another approach to proving this theorem using the
theorem that a Boolean group is embeds into a product of Z_2's.

> Hence the ideal I_i we defined in an earlier post
> always is a prime ideal of S.

> No if a<>0 is some element of S, there exists
> i \in I such that a(i)<>0. Hence a is not in I_i.
>
From: Chip Eastham on 4 Aug 2010 10:54
On Aug 3, 2:32 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Aug 3, 8:14 am, Chip Eastham <hardm...(a)gmail.com> wrote:
>
>
>
> > On Aug 3, 5:57 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
>
> > > On Mon, 2 Aug 2010, Arturo Magidin wrote:
> > > >> If a is not in the intersection of all maximal ideals
> > > >> and I is an ideal maximal within ideals not containing a,
> > > >> then I is a maximal ideal.
>
> > > >> Steps for a proof.
> > > >> There's some maximal ideal J with a not in J.
> > > >> J subset I; I = J is maximal.
>
> > > > There is no reason whatsoever why two ideals that are maximal among
> > > > those that do not contain I need to be related to one another by
> > > > inclusion.
>
> > > Ok, I see the problem, confusing maximal with maximum.
> > > How does one approach this problem?
>
> > Sometimes it helps to have standard names for things.
>
> > All (commutative) rings have maximal ideals.
>
> You are missing "with 1". Rings without 1 may lack maximal ideals.
>
> For example, every abelian group can be given the structure of a ring
> by defining multiplication by ab=0 for all a and b; with this
> definition, a subset is an ideal if and only if it is a subgroup. So
> if you take an abelian group that has no maximal subgroups (e.g., the
> rationals under addition) then you get a ring that has no maximal
> ideals. There are less silly examples, of course.
>
> You need the unity to guarantee existence of maximal ideals, and if
> you have unity then you do not need commutativity.
>
> --
> Arturo Magidin

Thanks, Arturo, for sorting out the "with 1"
condition. Nice example!

regards, chip
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