Re: Model Theory question.

On 15 Jan 2007 12:57:53 -0800, fred.galvin@xxxxxxxxx wrote:

Robert Sheskey wrote:
In article <1168809454.164484.46880@xxxxxxxxxxxxxxxxxxxxxxxxxxx>, fred.galvin@xxxxxxxxx says...



David C. Ullrich wrote:
Thanks. Comments below...

On 13 Jan 2007 18:36:43 -0800, fred.galvin@xxxxxxxxx wrote:

fred.galvin@xxxxxxxxx wrote:
David C. Ullrich wrote:
A colleague who's shy about posting asked me
the following:

Let's suppose that L is a language including the language of
ring-with-1 theory {+,.,0,1} and possibly some extra constant
symbols.

Suppose that R is a commutative ring with 1 that is a subring of
a commutative ring with (the same) 1 S. Suppose moreover that S
is a metric ring (metric space, operations continuous) and that
R is dense in S. Note that the metric space structure isn't in my
language - it just happens that these extra facts hold. Also
suppose that I have an interpretation for the extra constant
symbols in L (if any) in R.

Finally, let T be the *positive* theory of R with respect to L.
The "logic question" is: Does S satisfy T? I think that the
answer is yes. It seems to be fairly similar to the fact that a
homomorphic image of R satisfies T, together with the continuity
of the operations.

Do you see anything wrong with this? Is the statement written
down somewhere? Do you, generally, have any comments?

When I asked it turned out that the "positive theory" was the
fragment of the theory that uses only universal and existential
quantifiers and the connectives "and" and "or".

Maybe I'm all wet, my logic (and algebra and a lot of other stuff) is
pretty rusty, but it seems to me that it's false. Let R be the ring of
all real algebraic numbers, let p be a transcendental real number, and
let S be the ring generated by R and p. The positive sentence "Ax Ey
y^3 = x" is true in R, but false in S, because p has no cube root in S.
Right?

Seems right to me. Took me a second to see why p had no cube
root in S (you claim _your_ algebra's rusty), but of course
that's clear. If p has a cube root in S then P(p)^3 = p*Q(p)^3
for some non-trivial P, Q in R[x]. Now since p is transcendental
it follows that P(x)^3 = x*Q(x)^3 and this is clearly impossible,
for example by considering degrees.

This contradicts a silly thought I had, which was that for
a sentence like this the answer would be yes if y depended
continuously on x. Of course I was unthinkingly assuming
there that S was complete (in the metric-space sense).

Hence the question: You happen to know the answer if
we also assume that S is a complete metric space?

I wondered about that too. I have no idea.

Aren't there trivial counterexamples, eg, Q as a subring of R? (Here Q is Q
and R is the reals.) A positive sentence satisfied in Q but not in R:

Ax Eyz [(yz = 1) & (x^2 = 2+y)]

Well done! Now, why didn't I think of that?

_My_ excuse is that "Ax x^2 <> 2" is not a positive sentence,
and I was too stupid to even consider the possibility that
it could be translated into one.

At first I thought it was
because I ignored the fact that the question is about rings with 1, so
the language has a symbol for 1. On second thought, the symbol "1" has
nothing much to do with it, and the reason I didn't think of your nice
solution is just that I'm not very clever.

Nope, that's my excuse, dammit. You may think it's your because
you were the first to post it in _this_ thread, but I've been
using it for years. Get your own excuse - I've had enough of this
business of people talking about how dumb they are.

As a midget standing on a
giant's shoulder, I can now give an example of a positive sentence in
the language with +, *, and = which holds in Q but not in R:

Ex Ay Ez Au Ev [(x+z = y^3) & (zv = u)].


************************

David C. Ullrich
.

Relevant Pages

  • Re: Model Theory question.
    ... ring-with-1 theory and possibly some extra constant ... a commutative ring with 1 S. Suppose moreover that S ... R is dense in S. Note that the metric space structure isn't in my ... A positive sentence satisfied in Q but not in R: ...
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  • Re: Model Theory question.
    ... ring-with-1 theory and possibly some extra constant ... a commutative ring with 1 S. Suppose moreover that S ... R is dense in S. Note that the metric space structure isn't in my ... A positive sentence satisfied in Q but not in R: ...
    (sci.logic)
  • Re: Model Theory question.
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    (sci.logic)
  • Re: Model Theory question.
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    (sci.logic)
  • Re: Model Theory question.
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