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From: quasi on 9 Aug 2010 13:12
On Mon, 09 Aug 2010 12:03:53 -0500, quasi <quasi(a)null.set> wrote:

>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.

In fact, such an deal, P say, is actually maximal in B. Since P is
prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J
strictly containing P must contain a (by the maximality condition on
P). It follows that J = B, hence P is a maximal ideal of B.

>< 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 13:35
On Mon, 09 Aug 2010 12:12:31 -0500, quasi <quasi(a)null.set> wrote:

>On Mon, 09 Aug 2010 12:03:53 -0500, quasi <quasi(a)null.set> wrote:
>
>>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
Should be: "in the set *of* ideals"
>>>necessarily prime.
>
>In fact, such an deal, P say, is actually maximal in B. Since P is

Such a deal -- supersize P to maximal -- today only!

Of course that should have been

"In fact, such an ideal"

>prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J
>strictly containing P must contain a (by the maximality condition on
>P). It follows that J = B, hence P is a maximal ideal of B.
>
>>< A clear discussion of these ideas can be found in
>>>the first few pages of Kaplansky's wonderful text "Commutative Rings".

Hopefully all typos corrected.

quasi
From: William Elliot on 10 Aug 2010 02:41
On Mon, 9 Aug 2010, David C. Ullrich wrote:
> <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:

Also not if a = 1, which may or may not be in B.
What if 1 not in B?

> 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

1 = bc = bcc = 1c = c

> b = 1.

1 = bc = b1 = b

>> Please don't use the Boolean ring representation theorem,
>> as the question is a key step in a proof of the theorem.
>
>
From: William Elliot on 10 Aug 2010 04:02
On Mon, 9 Aug 2010, quasi wrote:
>>> <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.

Nor the identity.

>>> By the Boolean property, the set {a} is multiplicatively closed.
>>> Also, since a is nonzero, the ideal 0 is 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.

Why? Is this correct?
If rs in I, the ideal, r not in I, s not in I, then
a in (r,I) = Br + Zr + I and a in (s,I) = Bs + Zs + I;
some b,c in B, i,j in I, n,m in {0,1}
with a = br + nr + i = cs + ms + j

Thus a = aa
= bcrs + ncrs + csi + mbrs + nmrs + msi + brj + nrj + ij in I,
which is a no-no.

> In fact, such an deal, P say, is actually maximal in B. Since P is
> prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J
> strictly containing P must contain a (by the maximality condition on
> P). It follows that J = B, hence P is a maximal ideal of B.

If, which the premise did not affirm, 1 in B. What if 1 not in B?

>>> A clear discussion of these ideas can be found in the first
>>> few pages of Kaplansky's wonderful text "Commutative Rings".

Currently beyond my reach.

----
From: William Elliot on 10 Aug 2010 05:48
On Tue, 10 Aug 2010, William Elliot 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.
>
> Nor the identity.
>
>>>> By the Boolean property, the set {a} is multiplicatively closed.
>>>> Also, since a is nonzero, the ideal 0 is 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.
>
> Why? Is this correct?
> If rs in I, the ideal, r not in I, s not in I, then
> a in (r,I) = Br + Zr + I and a in (s,I) = Bs + Zs + I;
> some b,c in B, i,j in I, n,m in {0,1}
> with a = br + nr + i = cs + ms + j
>
> Thus a = aa
> = bcrs + ncrs + csi + mbrs + nmrs + msi + brj + nrj + ij in I,
> which is a no-no.
>
>> In fact, such an deal, P say, is actually maximal in B. Since P is
>> prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J
>> strictly containing P must contain a (by the maximality condition on
>> P). It follows that J = B, hence P is a maximal ideal of B.
>
> If, which the premise did not affirm, 1 in B. What if 1 not in B?
>
A more general proposition can be similarly proved.

If B is a Boolean ring with identity and I a proper prime ideal,
then I is maximal ideal.

If J ideal, I subset J and a in J\I, then
a not in I; a(a + 1) = 0 in I; a + 1 in I;
a + 1 in J; 1 = a + a + 1 in J; J = B.

Can the same be said of Boolean rings without identity?


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