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 a 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.

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may a […]

The reflection principle is a theorem schema in ZFC, meaning that for each formula $\phi(\vec x)$ we can prove in ZFC a version of the principle for $\phi$. In particular, it gives us that if $\phi$ holds (in the universe of sets) then there is some ordinal $\alpha$ such that $V_\alpha\models \phi$. It follows from this that (assuming its consistency) $\math […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

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.