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From: William Elliot on 9 Aug 2010 06:56
Let B be a Boolean ring and a, a point in B.
Is there a prime ideal that exclude a?

Please don't use the Boolean ring representation theorem,
as the question is a key step in a proof of the theorem.
From: A N Niel on 9 Aug 2010 07:47
In article <20100809035423.C81280(a)agora.rdrop.com>, William Elliot
<marsh(a)rdrop.remove.com> wrote:

> Let B be a Boolean ring and a, a point in B.
> Is there a prime ideal that exclude a?
>
> Please don't use the Boolean ring representation theorem,
> as the question is a key step in a proof of the theorem.

Wouldn't that be the same as a prime ideal that does contain the
complement 1-a of a ?
From: David C. Ullrich on 9 Aug 2010 10:01
On Mon, 9 Aug 2010 03:56:56 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>Let B be a Boolean ring and a, a point in B.
>Is there a prime ideal that exclude a?

Clearly not if a = 0. Yes if a <> 0:

Let b = 1 - a. We need only show that b is not invertible;
then b is contained in a maximal (proper) ideal, which
as Neil points out cannot also contain a.

But suppose that bc = 1. Then

bc^2 = c
bc = c
c = 1
b = 1.

>Please don't use the Boolean ring representation theorem,
>as the question is a key step in a proof of the theorem.

From: quasi on 9 Aug 2010 13:01
On Mon, 9 Aug 2010 03:56:56 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>Let B be a Boolean ring and a, a point in B.
>Is there a prime ideal that exclude a?

Yes, provided the element a is nonzero.

By the Boolean property, the set {a} is multiplicatively closed. Also,
since a is nonzero, the ideal 0 is disjoint from S. By Zorn's lemma,
we can extend the ideal 0 to an ideal which is maximal (with respect
to inclusion) in the set ideals not containing a. Such an ideal is
necessarily prime. A clear discussion of these ideas can be found in
the first few pages of Kaplansky's wonderful text "Commutative Rings".

quasi
From: quasi on 9 Aug 2010 13:03
On Mon, 09 Aug 2010 12:01:02 -0500, quasi <quasi(a)null.set> wrote:

>On Mon, 9 Aug 2010 03:56:56 -0700, William Elliot
><marsh(a)rdrop.remove.com> wrote:
>
>>Let B be a Boolean ring and a, a point in B.
>>Is there a prime ideal that exclude a?
>
>Yes, provided the element a is nonzero.
>
>By the Boolean property, the set {a} is multiplicatively closed. Also,
>since a is nonzero, the ideal 0 is disjoint from S.

Correction: disjoint from {a}

>By Zorn's lemma,
>we can extend the ideal 0 to an ideal which is maximal (with respect
>to inclusion) in the set ideals not containing a. Such an ideal is
>necessarily prime. A clear discussion of these ideas can be found in
>the first few pages of Kaplansky's wonderful text "Commutative Rings".

quasi
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