117b – Undecidability and Incompleteness – Lecture 4

We showed that is Diophantine. This implies that the prime numbers are Diophantine. Therefore, there is a polynomial in several variables with integer coefficients whose range, when intersected with the natural numbers, coincides with the set of primes. Amusingly, no nonconstant polynomial with integer coefficients can only take prime values.

We proved the bounded quantifier lemma, the last technical component of the proof of the undecidability of Hilbert’s tenth problem. It implies that the class of relations definable by a formula in the structure coincides with the class of relations (i.e., the Diophantine ones).

To complete the proof of the undecidability of the tenth problem, we will show that any c.e. relation is Diophantine. For this, recall that a set is c.e. iff it is the domain of a Turing machine. We proceeded to code the behavior of Turing machines by means of a Diophantic representation. This involves coding configurations of Turing machines and it remains to show how to code the way one configuration changes into another one.

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4 Responses to 117b – Undecidability and Incompleteness – Lecture 4

I was a bit confused about this in class, but it’s probably just because I was majorly sleep-deprived: “Therefore, there is a polynomial in several variables with integer coefficients whose range on the natural numbers coincides with the prime numbers. Amusingly, no nonconstant polynomial with integer coefficients can only take prime values.”

The distinction, then, is that there IS a polynomial whose range is all the primes AND a bunch of negatives, but there is NO polynomial whose range is primes only? Is that correct?

Yes, exactly. I think I may have actually written something incorrect on the board, so I will briefly review it on Tuesday. We have that the primes are Diophantine. From this, we saw how to build a polynomial P such that, if R is its range, then the intersection of R with the naturals is the set of primes. But any such R necessarily takes negative values (and 0?) as well. I rephrased the entry a little to make this clear.

If anyone’s interested, I remembered reading about this polynomial (though I didn’t know the historical context in which it was derived) a while back in the book The Music of the Primes. According to Amazon’s search-inside-this-book function, it’s on page 200, which I have downloaded from Amazon and is now stored on my web page. It’s good to see these things actually written out!

Thanks!
The same polynomial is shown in page 331 of the handout “Hilbert’s tenth problem. Diophantine equations: positive aspects of a negative solutions”, it is due to J. P. Jones. The polynomial we obtain if we simply apply the definitions shown in lecture is different than this one.
It is curious that this polynomial is written as the product of two polynomials, and yet its (positive) values are precisely the prime numbers.

This is Theorem 39 in the paper (see Theorem 4.(i) for a user-friendly preview). But the fact that $(2^\kappa)^+\to(\kappa^+)^2_\kappa$ is older (1946) and due to Erdős, see here: Paul Erdős. Some set-theoretical properties of graphs, Univ. Nac. Tucumán. Revista A. 3 (1942), 363-367 MR0009444 (5,151d). (Anyway, it is probably easier to read a more modern pre […]

One of the best places to track these things down is The mathematical coloring book, by Alexander Soifer, Springer 2009. Chapter 35 is on "Monochromatic arithmetic progressions", and section 35.4, "Paul Erdős’s Favorite Conjecture", is on the problem you ask about. As far as I can tell, the question is sometimes called the Erdős-Turán con […]

Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory. How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$? Really, even pointers on how to say anything […]

This question is partly motivated by Timothy Chow's recent question on the division paradox. Say that a set $X$ admits a paradoxical partition if and only if there is an equivalence relation $\sim$ on $X$ such that $|X|

A solution can be obtained as suggested by Keith Conrad in the comments, via Chebotarëv's theorem. Details can be found in $\S3.4$ of Coloring the $n$-Smooth Numbers with $n$ Colors Andrés Eduardo Caicedo, Thomas A. C. Chartier, Péter Pál Pach The Electronic Journal of Combinatorics 28 (1) (2021), #P1.34, 79 pp. DOI: https://doi.org/10.37236/8492 Many t […]

No, this is not possible. Dave L. Renfro wrote an excellent historical Essay on nowhere analytic $C^\infty$ functions in two parts (with numerous references). See here: 1 (dated May 9, 2002 6:18 PM), and 2 (dated May 19, 2002 8:29 PM). As indicated in part 1, in Zygmunt Zahorski. Sur l'ensemble des points singuliers d'une fonction d'une variab […]

This is a difficult question in general. Ideally, to show that $f$ is analytic at the origin, you show that in a suitable neighborhood of $0$, the error of the $n$-th Taylor polynomial approaches $0$ as $n\to\infty$. For example, for $f(x)=\sin(x)$, any derivative of $f(x)$ is one of $\sin(x)$, $\cos(x)$, $-\sin(x)$, or $-\cos(x)$, and the error given by the […]

To complement Yann's answer: This is a nice problem, the possible length of Borel hierarchies in different spaces or without assuming the axiom of choice. It has been studied in detail by Arnie Miller. See Arnold W. Miller. On the length of Borel hierarchies, Ann. Math. Logic, 16 (3), (1979), 233–267. MR0548475 (80m:04003), Arnold W. Miller. Long Borel […]

This is a good question, because a priori $\mathsf{PA}$ lacks the flexibility of $\mathsf{ZFC}$ that allows us to deal with consistency problems semantically (by building models) and, anyway, the obvious model of most subtheories of $\mathsf{PA}$ is just the standard model. The way this is done in the context of $\mathsf{ZFC}$ is using the reflection theorem […]

Yes, of course. An example is the statement that all Goodstein sequences terminate. The point is that this sentence is not only independent of $\mathsf{PA}$, but in fact of the theory resulting from adding to $\mathsf{PA}$ all $\Pi^0_1$ statements true in the standard model of arithmetic. Note that $\mathrm{Con}(\mathsf{ZFC})$ is an example of such a $\Pi^0_ […]

I was a bit confused about this in class, but it’s probably just because I was majorly sleep-deprived: “Therefore, there is a polynomial in several variables with integer coefficients whose range on the natural numbers coincides with the prime numbers. Amusingly, no nonconstant polynomial with integer coefficients can only take prime values.”

The distinction, then, is that there IS a polynomial whose range is all the primes AND a bunch of negatives, but there is NO polynomial whose range is primes only? Is that correct?

Yes, exactly. I think I may have actually written something incorrect on the board, so I will briefly review it on Tuesday. We have that the primes are Diophantine. From this, we saw how to build a polynomial P such that, if R is its range, then the intersection of R with the naturals is the set of primes. But any such R necessarily takes negative values (and 0?) as well. I rephrased the entry a little to make this clear.

If anyone’s interested, I remembered reading about this polynomial (though I didn’t know the historical context in which it was derived) a while back in the book The Music of the Primes. According to Amazon’s search-inside-this-book function, it’s on page 200, which I have downloaded from Amazon and is now stored on my web page. It’s good to see these things actually written out!

Thanks!

The same polynomial is shown in page 331 of the handout “Hilbert’s tenth problem. Diophantine equations: positive aspects of a negative solutions”, it is due to J. P. Jones. The polynomial we obtain if we simply apply the definitions shown in lecture is different than this one.

It is curious that this polynomial is written as the product of two polynomials, and yet its (positive) values are precisely the prime numbers.