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From: David C. Ullrich on 10 Aug 2010 08:35
On Mon, 9 Aug 2010 23:41:39 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>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,

??? The proof below does not use the fact that a <> 1.
In fact _any_ maximal (proper) ideal is an example of
a prime ideal excluding 1.

>which may or may not be in B.

I thought that the definition of "Boolean ring" included
the existence of an identity.

>What if 1 not in B?

That's not really a meaningful way to put it.
You meant "what if B has no multiplicative identity?".

In that case the whole argument breaks down;
the fact that any (proper) ideal in a ring is
contained in some maximal (proper) ideal
uses the existence of a multiplicative identity.

>> 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: quasi on 10 Aug 2010 12:50
On Tue, 10 Aug 2010 01:02:12 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>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.

The element a _can_ be the identity. The only exclusion is a = 0.

>>>> 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.

Yes, that's essentially the argument.

>> 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.

I used the Wikipedia definition

<http://en.wikipedia.org/wiki/Boolean_ring>

which says that a Boolean ring is a ring with identity.

>What if 1 not in B?

Then the prime ideal P may or may not be maximal.

>>>> 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.

Well, when your reach extends further, reach for it -- I think you
would really appreciate his style of writing, as well as his cool
choice of exercises.

quasi
From: quasi on 10 Aug 2010 13:09
On Tue, 10 Aug 2010 02:48:07 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>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?

Good question, but I suspect not. I don't have time right now to try
to find a counterexample, but why not see if you can find one?

quasi
From: quasi on 10 Aug 2010 13:58
On Tue, 10 Aug 2010 12:09:41 -0500, quasi <quasi(a)null.set> wrote:

>On Tue, 10 Aug 2010 02:48:07 -0700, William Elliot
><marsh(a)rdrop.remove.com> wrote:
>
>>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?
>
>Good question, but I suspect not. I don't have time right now to try
>to find a counterexample, but why not see if you can find one?

Ok, here's a counterexample (I think) ...

Let A be the boolean ring with identity defined as the quotient

A = Z_2[x,y] / (x^2-x, y^2-y)

and let B = (x,y), the ideal of A generated by x,y.

Then B is a Boolean ring without identity, and in the ring B, the zero
ideal is prime but not maximal.

quasi
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