FOM: Boolean rings; moving on toward interesting f.o.m. issues
Stephen G Simpson
simpson at math.psu.edu
Sat Mar 14 17:46:37 EST 1998
Vaughan Pratt 12 Mar 1998 15:24:19 writes:
> Please prioritize them so I can answer the most pressing ones
> first, it would be silly if I ran out of time after answering the
> least pressing ones.
Please answer as many as you have time for, starting at the end and
working backwards. I'm hoping that we'll eventually get to some
common ground that can serve as a basis for rational discussion
between f.o.m. researchers and category theorists. If you don't have
time to participate in establishing such a basis, I'm not sure how to
proceed.
To my mind, the most pressing issue right now is whether you still
insist that Boolean algebras are isomorphic to Boolean rings, when
most people understand that they aren't.
Could it be that you didn't understand where the quantifiers were
supposed to go? Mossowski 13 Mar 1998 15:36:30 apparently didn't. He
interpreted the question as asking whether it is not the case that
> forall Boolean algebras A. forall Boolean rings R. A is isomorphic
> to R
But what we are actually talking about is whether particular Boolean
algebras can be isomorphic to particular Boolean rings. The truth is
that, as most people know, they can't. In Mossakowski's notation,
what I am saying is:
For all Boolean algebras A, for all Boolean rings R, A is not
ismorphic to R.
This is of course well known. The proof of this is: (1) According to
the usual conventions, the signatures of Boolean algebras and Boolean
rings are different, so for this reason alone they aren't
isomorphic. (2) Even if we follow the elegant but somewhat
non-standard notational convention used by Jech and others (+,.,- for
join, meet, complement), then although the signatures become the same,
we still don't have isomorphism, because for instance 1+1=0 in a
Boolean ring, 1+1=1 in a Boolean algebra.
Do you agree with this?
Perhaps you are professing confusion because I have not stated which
"bases" I intend. You raised this issue in your posting of 12 Mar
1998 16:00:41:
> Since the language of Boolean rings has as many bases as the language
> of Boolean algebras, this is clearly a basis-dependent question: U is
> a basic operation just when it is in the chosen basis.
Since you insist on raising this normally well-understood issue, let
me answer by saying that the standard bases are implicit in the
question: {join,meet,complement,top,bottom} for Boolean algebras,
{plus,minus,times,0,1} for rings. Is this clear enough? It should
be, because these bases were in the definitions of Boolean algebras
and Boolean rings that you yourself quoted from Sikorski, in your
posting of 10 Mar 1998 20:39:59. (Of course I admit that the question
may have a different answer if we use non-standard bases, e.g. if we
saddle Boolean algebras with the Boolean ring basis, or vice versa.)
Another source of confusion may be your categorical perspective. I
say this because according to Mossokowski it is reasonable to
interpret the question in terms of isomorphism of categories. But I
didn't mention categories, and I intended no such interpretation.
While I recognize that *the category of all Boolean rings* is
isomorphic to *the category of all Boolean algebras*, I also recognize
that there are significant differences between Boolean algebras and
Boolean rings. You need to recognize this too. If you can't
recognize things like this, then you won't be able to rationally
discuss f.o.m. issues with f.o.m. professionals.
[ Incidentally, isomorphism of categories has its counterpart in
f.o.m. research: interpretations or translations of theories. For
instance, one could say that the theory of Boolean algebras and the
theory of Boolean rings are biinterpretable in a certain sense. This
is a very flexible and suggestive approach. Some important
f.o.m. results, which were first proved this way by
f.o.m. researchers, were later reformulated by category theorists in
categorical terms, with concomitant loss of clarity and motivation.
An example is Cohen's work on the continuum hypothesis, later
reformulated by Lawvere and Tierney. ]
Yet another source of categorical confusion is that you and
Mossokowski may think that "isomorphism" can only refer to isomorphism
within some particular category, and I haven't specified the category.
In actual fact, categories are irrelevant to the question we are
discussing. The notion of isomorphism that is used here is the
absolute one, which is standard in most mathematics textbooks (except
possibly category theory books):
Two structures (A,Phi) and (B,Psi) are isomorphic if
dom(Phi)=dom(Psi)=s, i.e. s is the common signature, and if there
is a 1-1 onto mapping i:A->B such that for all n-ary relation
symbols R in s and all a1,...,an in A, Phi(R)(a1,...,an) iff
Psi(R)(i(a1),...,i(an)), and for all n-ary operation symbols o in
s, i(Phi(o)(a1,...,an))=Psi(o)(i(a1),...,i(an)).
(quoted from my posting of 12 Mar 1998 15:53:52). Here of course A
and B are sets, the domains or universes of the structures in
question. Note that this very clear, very standard, very
well-understood mathematical notion does not involve categories, not
even implicitly. This is a key point. If we are to have a basis for
rational discussion, you and the other category theorists will need to
step outside your categorical mindset.
In the case of a Boolean algebra A and a Boolean ring R, what I am
saying is that there is no mapping i:A->R as above. OK? Note that
there are no categories here. Is that OK?
To summarize: Boolean algebras are not isomorphic to Boolean rings.
Once I have forced Pratt and the other category theorists to admit
crucial technical distinctions of this kind, then perhaps we can move
on to more interesting foundational questions: foundational
motivation, general intellectual interest, why algebraic logic is not
the same as f.o.m., etc.
-- Steve
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