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From: Hagen on 3 Aug 2010 04:39
> On Tue, 3 Aug 2010, Hagen wrote:
> >>>>> On Aug 1, 4:29 am, William Elliot <ma..
> >>
> >>> You mean R consists of those elements of (Z_2)^N
> >> having finite support?
> >> Yes.
> >>
> >> In general, if R is a subring of a product of
> rings,
> >> prod{ Rj | j in I }, then is I_j = { r | r(j) = 0
> }
> >> a proper prime ideal?
> >
> > No. You have to assume that the ring Rj is an
> > integral domain. In this case I_j either is a
> proper
> > prime ideal or equals R.
> >
> All that's needed is for Rj to not have any zero
> divisors.
>
> >>>> Let R1 be the extension of R with an
> (multiplicative) identity 1,
> >>>> added. R1 is the set of functions of (Z_2)^N
> with finite or cofinite
> >>>> support and pointwise addition and
> multiplication.
> >>>
> >>> 'the extension' ? Is it unique in some sense?
> >>
> >> I would hope so; as the smallest subring of
> (Z_2)^N
> >> containing R and 1.
> >>
> > Then R1 = {1+r . r \in R} \cup R
> > Hence for an element of R1 either has finite
> support
> > or has infinitely many components equal to 1.
>
> Or equivalently, cofinite support.

Ok, that's a better characterization of R1.
I did not see this the last time.

> > Thus the element r with r(2k)=0 and r(2k+1)=1 is
> > not in R1.
> >
> Agreed, R1 /= (Z_2)^N.
>
> >>> Maximal ideals are always prime - not just in
> Boolean rings.
> >>>
> >> If I maximal ideal, ab in I, a not in I, then b in
> >> (a,I)
> >> . . = { ra + na + i | r in ring, n in Z, i in I }.
> >>
> >> Ok, that's easy to see for communative rings with
> >> identity and Boolean rings with or without
> idenity.
> >> What if the ring doesn't have identity or isn't
> communative?
> >
> > We are talking about a commutative ring with unity
> > here - namely about R1.
>
> In other other than Boolean rings as you mention, are
> there
> any classes of rings beyond communative rings with
> identity,
> for which maximal rings are prime?

Maximal ideals are prime in every ring with unity
(commutative or not).

This is not true in rings without 1: take for example
the additive group R=Z/4Z and define a multiplication
on R by a*b = 0 for all a,b \in R. Then (R,+,*) is a
commutative ring without unity.
The subgroup 2Z/4Z is a maximal ideal in R but of
course R has no proper prime ideals.

> >> 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.
>

H
From: Hagen on 3 Aug 2010 04:46
> >> 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.
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.

H
From: Chip Eastham on 3 Aug 2010 09:14
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.

If a (commutative) ring has a maximum ideal (necessarily
unique), it is called a "local" ring.

The complement of the maximum ideal in a local ring
(with identity) consists of all the invertible ring
elements.

regards, chip
From: Arturo Magidin on 3 Aug 2010 14:32
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
From: Arturo Magidin on 3 Aug 2010 14:43
On Aug 3, 4:57 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?

Chip Eastham has already pointed out that the proposition:

"If R is a commutative ring, a is an element of R, and a is not in
the intersection of all maximal ideals, then every ideal of R that is
maximal among the ideals that do not contain a is a maximal ideal of
R"

is false (example given: Take R=Z, where the intersection of all
maximal ideals is trivial; a = 2; then I=4Z is an ideal that is
maximal among those that do not contain a, but is not a maximal ideal
of R).

So the first step to approach the problem would be to realize that it
is false.

It is true that there exists a maximal ideal of R that contains a if
and only if a is not in the intersection of all maximal ideals (though
that is trivial when seen in that light). The two notions will
certainly coincide in the case where a is a unit, and possibly in
other situations, though it seems difficult to characterize the
latter.

--
Arturo Magidin
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