SJR
|
Registered User
|
|
|
May 16 |
comment |
Germs at infinity of sequence of integers Sorry, that link is broken. Use this: math.helsinki.fi/logic/people/juliette.kennedy/… Why can't I delete comments any more ??!! |
|
May 16 |
comment |
Germs at infinity of sequence of integers The module can't be free because the germ of the sequence n! is divisible. <\a href="duckduckgo.com/…; is a paper that studies this module (actually ring) in the context of models of arithmetic. |
|
May 13 |
accepted | First order decidability of rings vs Diophantine decidability |
|
May 10 |
comment |
First order decidability of rings vs Diophantine decidability @Laurent: Yes it really is surprising that we get the definability of N without extra constants, and the proof is extremely clever. Nevertheless, my intuition is that the full first-order theory is in general wildly unrelated to the existential theory. |
|
May 8 |
comment |
First order decidability of rings vs Diophantine decidability @David: You could ask the kind of question you are intersted in, using in place of R some countable subfield of the reals with a nice simple presentation, like the real algebraics. This leads to a complicated network of open and partially solved problems. See "Undecidability of Existential Theories of Rings and fields", by Pheidas and Zahidi, which I believe is available online. |
|
May 8 |
comment |
First order decidability of rings vs Diophantine decidability @David: Let's put it this way: If we allow coefficients in R(t) then we are asking about the effective computability of an uncountable set. This should seem a little bit uncomfortable! |
|
May 8 |
comment |
First order decidability of rings vs Diophantine decidability @David: This came up in the comments when the question was first posted. It was established that the coefficients must lie in some subring for which questions of decidability make sense. The OP concurred, and left it open what subring to choose, subject to this requirement. I chose the integers, or just as good, the rationals. In any case the question doesn't make good sense if we allow coefficients in R(t), unless we introduce some exotic definition of decidability, and there are many. But then we are in territory far from the OP's intent. |
|
May 8 |
comment |
First order decidability of rings vs Diophantine decidability @David: The polynomial equations in question are solvable in $\mathbb{R}(t)$ if and only if they are solvable in $\mathbbr{R}$. Why? On the one hand $\mathbb{R}$ is infinite: Give a solution in $\mathbb{R}(t)$ we can pick a real at which all the components are defined and substitute to get a solution in $\mathbb{R}$. Conversely, a solution in $\mathbb{R}$ is already a solution in $\mathbb{R}(t)$. So the rings $\mathbb{R}$ and $\mathbb{R}(t)$ have the same existential theories. But quantifier elimination for real closed fields gives an algorithm for deciding the existential theory of R. |
|
May 8 |
comment |
First order decidability of rings vs Diophantine decidability At least "archimedian" is necessary for Robinson's argument. Whether it is actually required I don't know |
|
May 8 |
revised |
First order decidability of rings vs Diophantine decidability added 12 characters in body |
|
May 8 |
comment |
First order decidability of rings vs Diophantine decidability Whoops! Archimedian is necessary. I've edited. And hello Joel! Greetings from Thailand. Yes, the subring here can be the rationals, or the integers. |
|
May 8 |
answered | First order decidability of rings vs Diophantine decidability |
|
May 6 |
comment |
Zeros of polynomials in discretely ordered rings @Emil: Ok, thanks. I don't have access to the paper, but I suppose $p=0$ reduces to some form of Pell's equation. I hadn't noticed. |
|
May 6 |
comment |
Zeros of polynomials in discretely ordered rings @Emil: To make your proof work, I need some DOR in which $f$ has a zero. To make this happen I must satisfy in some DOR the three equations (1) $p(a, (1+\sum_iu_i^2)(a-1)+1, (1+\sum_iu_i^2+\sum_iv_i^2)(a-1))=0$, (2) $a=\sum_iw_i^2$, and (3) $q(w)=0$. I have no problem with (2) and (3), but given solutions to (2) and (3) in some DOR, how do I know that there are $u$'s and $v$'s that will satisfy (1)? |
|
May 6 |
comment |
Zeros of polynomials in discretely ordered rings @Emil: You "take a root of $f$ in a DOR $R$..." Why is it obvious that if $p(a,x,y):=x^2+2axy+y^2-1$ then the equation
$p(a,(1+\sum_iu_i^2)(a-1)-1,(1+\sum_iu_i^2+\sum_iv_i^2)(a-1))=0$
has a solution in some DOR?
|
|
May 6 |
comment |
First order decidability of rings vs Diophantine decidability The coefficients of the polynomials in $S$ need to have some cannonical, finitary presentation in order for the question to be well-posed. Perhaps you would be content with coefficients in the prime subring? |
|
May 6 |
comment |
Zeros of polynomials in discretely ordered rings @Emil: I just found your response. Thank you very much! I'll need some time to digest this... |
|
Apr 23 |
comment |
Zeros of polynomials in discretely ordered rings @Joel, Emil: My hope is that DOR is so weak that some construction will give a zero in a rank 1 DOR, merely because DOR$+\exists x,y\,f(x,y)=0$ is consistent. It is a longstanding open problem to effectively determine if an arbitrary polynomial has a zero in at least one DOR. It would be nice if we only had to look at rank 1 DOR's. |
|
Apr 23 |
revised |
Zeros of polynomials in discretely ordered rings added 27 characters in body |
|
Apr 23 |
comment |
Zeros of polynomials in discretely ordered rings @Joel: I'm not sure what you mean by "the root". $\bar{x}$ is a tuple of variables. I'll add that. |
|
Apr 23 |
revised |
Zeros of polynomials in discretely ordered rings deleted 24 characters in body; added 4 characters in body; deleted 1 characters in body |
|
Apr 23 |
asked | Zeros of polynomials in discretely ordered rings |
|
Apr 16 |
answered | References on techniques for solving equations with discontinuous functions such as floor and ceiling? |
|
Apr 6 |
awarded | ● Yearling |
|
Feb 12 |
comment |
Two different analytic curves cannot intersect in infinitely many points Peter, are you overlooking the counterexample given in Ramiro's answer? |
|
Feb 12 |
comment |
Two different analytic curves cannot intersect in infinitely many points A related MO question: mathoverflow.net/questions/109705/… |
|
Jan 29 |
revised |
Sets of integers represented by degree zero rational functions added 2 characters in body |
|
Jan 29 |
accepted | Sets of integers represented by degree zero rational functions |
|
Jan 29 |
revised |
Sets of integers represented by degree zero rational functions added 3 characters in body |
|
Jan 29 |
revised |
Sets of integers represented by degree zero rational functions added 31 characters in body |
|
Jan 29 |
answered | Sets of integers represented by degree zero rational functions |
|
Jan 7 |
comment |
Are All Irrational Elementary Numbers Conjectured to Be Normal? @Michael: Maybe it is not too hard to cook up an "elementary" definition of a non-normal irrational (whatever the OP means by "elementary", which he needs to clarify.) |
|
Jan 5 |
comment |
Does every polynomial diophantine equation have solutions modulo p? I'm reading the question as follows: Suppose that $f_1,\ldots,f_n$ are polynomials in $n$ variables with integer coefficients, and that the $f_i$ have a common complex zero. Must there be some $n$-tuple of integers $\bar{x}$ such that the greatest common divisor of the numbers $f_i(\bar{x})$ is greater than 1? Ironbeard, do you accept this formulation? |
|
Dec 12 |
comment |
orderings of the field R((x, y)) What do you mean by "explicitly determine"? Do you have in mind some way of classifying all orderings on $\mathbb{R}((x))(y)$ that doesn't work for $\mathbb{R}((x,y))$? |
