nt.number theory – perfect fields in positive characteristic

Let $k$ be an infinite perfect field in positive characteristic $p$, i.e. every element of $k$ is a $p$th power. I am interested in properties of finite fields that can be extended to $k$. For example:

  1. Let $L$ be a finite Galois extension of $k$. Is $mathrm{Gal}(L/k)$ always cyclic?
  2. Let $L$ and $L’$ be two finite Galois extensions of $k$. Are $L$ and $L’$ always $k$-isomorphic fields?

The only example of such a field $k$ I have encountered is $mathbb{F}_{q}(x,x^{p^{-1}},x^{p^{-2}},x^{p^{-3}},ldots)$, where $q=p^d$ for some prime $p$, and the only finite Galois extensions I see are those of the form $mathbb{F}_{q^e}(x,x^{p^{-1}},x^{p^{-2}},x^{p^{-3}},ldots)$, which makes me expect a positive answer to both questions.

Basically, I want to know whether Galois theory over infinite perfect fields in positive characteristic is as ”easy” as over finite fields. In addition to an answer to the questions above, any properties supporting/contradicting this vague statement are apreciated.

nt.number theory – Explicit construction of division algebras of degree 3 over Q

In his book Introduction to arithmetic groups, Dave Witte Morris implicitly gives a construction of central division algebras of degree 3 over $mathbb{Q}$ in Proposition 6.7.4. More precisely, let $L/mathbb{Q}$ be a cubic Galois extension and $sigma$ a generator of its Galois group.If $p in mathbb{Z}^+$ and $p neq tsigma(t)sigma^2(t)$ for all $t in L$, then
$$ D=left{ begin{pmatrix}
x & y & z\
psigma(z) & sigma(x) & sigma(y)\
psigma^2(y) & psigma^2(z) & sigma^2(x)
end{pmatrix} :(x,y,z)in L^3 right}

is a division algebra.

On page 145, just before Proposition 6.8.8, Morris claims that it is knows that every division algebra of degree 3 arises in this manner. This should follow from the fact that every central division algebra of degree 3 is cyclic. I could not find this explicit construction in my references (e.g. Pierce – Associative Algebras, though maybe I missed something) and I would like to know if there is a reference or a quick way to see that this exhausts all central division algebras of degree 3 over $mathbb{Q}$.

nt.number theory – Weak Goldbach conjecture with distinct primes for odd integers between 4×10^18 and 10^27

This is related to the conjecture that all odd integers greater than 17 can be written as the sum of 3 distinct primes.

Schinzel showed that the Goldbach conjecture implied this in 1959 and as the Goldbach conjecture has been verified up to 4×10^18 by Oliveria e Silva, Herzog and Pardi, this conjecture holds up to there as well.
Vinogradov’s proof that all sufficiently large odd integers are the sum of three primes implies that as the number of representations of a sufficiently large odd integer as the sum of three primes is large enough, it must be the sum of 3 distinct primes.
Harald Helfgott’s proof of the weak Goldbach conjecture also implies that all odd integers greater than 10^27 can be represented as the sum of 3 distinct primes (assuming I’ve understood it correctly).

So the interval between 4×10^18 and 10^27 is the remaining interval on which to verify whether all odd integers greater than 17 can be written as the sum of 3 distinct primes.

In Helfgott’s proof, he uses the fact that the Goldbach conjecture is verified up to 4×10^18 along with a prime ladder to show that the odd integers in this interval are the sum of 3 primes. But that prime ladder is a list of primes from 3 to beyond 10^27 such that consecutive primes have difference at least 6 and at most 4×10^18. Thus for odd n between 4×10^18 and 10^27, there is always a p in the ladder such that n-p is equal to an even number less than 4×10^18 and is therefore the sum of 2 primes, q and r, so n = p + q + r.

As the verification of the Goldbach conjecture up to 4×10^18 showed that each even integer was the sum of 2 distinct primes, it would be necessary to create a similar prime ladder such that consecutive primes had difference at least 8 (as n-p=6 could only be written as 3 primes as n = p + 3 + 3 which is not a representation with distinct primes) and ensuring that odd integers between 4×10^18 and 8×10^18 are the sum of 3 distinct primes (as p could potentially equal q or r for n in this interval).

How difficult is it to make such a ladder and verify that primes between 4×10^18 and 8×10^18 are the sum of three distinct primes? For the second part I imagine finding a prime, s, just below 4×10^18 – 8 and a prime, t, just above 4×10^18 would solve it as by adding any even number less than s to s you could reach any odd number up to just below 8×10^18 and by adding even numbers less than t to t you could reach the other odd numbers less than 8×10^18.

nt.number theory – Powers of automorphic Eisenstein series

Let $G$ be a reductive group defined over $mathbb{Q}$. Let $P$ be a standard parabolic subgroup of $G$ with Levi decomposition
$$P = MN.$$
We denote by $R_{disc,M}$ the discrete spectrum of $M$. Let $I_{M}^G(I_{N(mathbb{A})}otimes R_{disc,M})$ be the induced representation. For $Phiin I_{M}^G(I_{N(mathbb{A})}otimes R_{disc,M})$ we define an Eisenstein serie as follows
$$E(g,Phi,s) = sum_{gammain P(mathbb{Q})setminus G(mathbb{Q})}Phi(gamma g)e^{<rho_P,H_P(gamma g)>},$$
where $rho_P$ is the Weyl vector and $H_P$ is the map
H_P:;P&to Hom(Char(P),mathbb{R}),\
g&mapsto (chiin Char(P)mapsto log(|chi(g)|)),

with $Char(P)$ the group of rational characters of $P$. Let
$$mathbf{E}(g) = E(g,Phi,s)^2,$$
is there any way to understand $mathbf{E}(g)$ as an Eisenstein serie of $G$? I mean, is there any section $phiin I_{M}^G(I_{N(mathbb{A})}otimes R_{disc,M})$ such that
$$mathbf{E}(g) = sum_{gammain P'(mathbb{Q})setminus G(mathbb{Q})}Phi(gamma g)e^{<rho_{P’},H_{P’}(gamma g)>},$$
for some parabolic $P’$?.

nt.number theory – Natural numbers in the form $lfloorfrac{a^3+b^3}2+frac{c^3+d^3}6rfloor$

Let $mathbb N={0,1,2,ldots}$. Several years ago I proved that
$${aw^3+bx^3+cy^3+dz^3: w,x,y,zinmathbb N}not=mathbb N$$
for any positive integers $a,b,c,d$ (cf. http://maths.nju.edu.cn/~zwsun/179b.pdf ).

I’m curious whether there are positive integers $m$ and $n$ such that
$$left{leftlfloorfrac{a^3+b^3}m+frac{c^3+d^3}nrightrfloor: a,b,c,dinmathbb Nright}=mathbb N.$$
My computation suggests that $(m,n)=(2,6)$ might meet my purpose. Moreover, I have formulated the following conjecture.

Conjecture. Each $ninmathbb N$ can be written as the integral part of
$(a^3+b^3)/2+(c^3+d^3)/6$ with $a,b,c,dinmathbb N$, $agemax{b,1}$ and $cgemax{d,1}$.

I have verified this for all $n=0,ldots,60000$. For example, $219$ has a unique required representation:
$$219=leftlfloor frac{4^3+0^3}2 +frac{10^3+5^3}6rightrfloor.$$
For the number of ways to write $ninmathbb N$ in the given form, one may consult http://oeis.org/A343326.

QUESTION. Is the above conjecture true? How to prove it?

Your comments are welcome!

nt.number theory – Is there a planar rational point set within which the distance of any two points is an irrational number?

i.e. could we find a subset $Xsubset mathbb{R}^2$ such that $Xsubset mathbb{Q}^2$ and that for any $x,yin X$ the distance $|x-y|$ is an irrational number?

I’m considering the following assertion of which I’m not sure :

Given finite rational points $p_1,p_2,dots,p_n$ , and an open ball $D$ on the plane, there is a rational point $xin D$ such that $|x-p_i|in mathbb{R}backslash mathbb{Q}$ for $i=1,2,dots,n$.

But this assertion accounts to be the following seemingly number-theoretic problem:

Given $n$ pairs $(a_i,b_i) (i=1,2,dots,n)$ of positive integers such that $a_i^2+b_i^2$ is not square of any integer. Could we find an integer $Ngeq 2$ such that the integral pairs $(Na_i+1,Nb_i)$ still satisfy the previous property(i.e. $(Na_i+1)^2+(Nb_i)^2$ is not square of any integer).

BTW the distribution of the Pythagorean triples might help.

nt.number theory – Fibonacci with seeds, modulo $n$

We can completely classify modulo which $n$ there is a surjective sequence. Indeed, I claim that $text{fib}_{n, x_0, x_1}$ is surjective for some seed values $x_0,x_1$ iff the usual Fibonacci sequence $F_k$ is surjective modulo $n$. As stated on OEIS, this happens precisely when $n$ is of one of the forms $5^k,2cdot 5^k,4cdot 5^k,3^jcdot 5^k,6cdot 5^k,7cdot 5^k,14cdot 5^k$.

One implication is obvious. For the other, assume $text{fib}_{n, x_0, x_1}$ is surjective modulo $n$. In particular we have $text{fib}_{n, x_0, x_1}(k)equiv 0pmod n$. Shifting the index we may assume $k=0$, i.e. $x_0=0$. But then we have $text{fib}_{n, x_0, x_1}(k)equiv x_1F_kpmod n$. This sequence is surjective modulo $n$ iff $F_k$ is and $x_1$ is coprime to $n$.

Old answer:

Not necessarily. Let $F_k$ be the regular Fibonacci sequence. Then we have $text{fib}_{n, x_0, x_1}(k)=x_0F_{k+1}+(x_1-x_0)F_kmod n$. Letting $pi(n)$ be the $n$th Pisano period, this implies that $text{fib}_{n, x_0, x_1}$ is periodic with period dividing $pi(n)$. There are plenty of numbers for which $pi(n)<n$, for instance all numbers modulo which $x^2-x-1$ has a root (as by Binet’s formula and Euler’s theorem we then have $pi(n)midvarphi(n)<n$). For such $n$ the sequence cannot be surjective.

nt.number theory – Classification of vector spaces with a quadratic form and an order n automorphism

Introductory general nonsense (for motivation: feel free to skip): Let $G$ be a finite group and $k$ be a field of characteristic $0$. Consider the set $mathcal{S}$ of isomorphism classes of finite dimensional vector spaces $V$ over $k$ endowed with (a) a nondegenerate quadratic form $qcolon Vto k$, and (b) a linear action $G to mathit{GL}(V)$, which are compatible in the sense that $G$ preserves $q$ (viꝫ. $q(gcdot v)) = q(v)$ for $gin G$). Note that we can take direct sums and tensor products of such data, giving $mathcal{S}$ a semiring (“ring without subtraction”) structure; we can also form the Grothendieck group $mathcal{R}$ of $mathcal{S}$, which is a ring.

Classifying (a) alone and (b) alone is well studied: (a) gives the Grothendieck-Witt ring of $k$, and (b) gives the representation ring of $G$ over $k$. I’m curious about what can be said about both data simultaneously (and compatibly). We have obvious ring homomorphisms from $mathcal{R}$ to the Grothendieck-Witt ring of $k$ and to the representation ring of $G$ over $k$, but I think $mathcal{R}$ (generally) isn’t a fiber product of them, and I suppose there isn’t much we can say at this level of generality (though I’d be happy to be wrong!).

I might still point out that if $V = V_1 oplus V_2$ is a decomposition of $V$ as representations of $G$ and there is no irreducible factor common in $V_1$ and the dual $V_2^vee$ of $V_2$, then necessarily $V_1$ and $V_2$ are orthogonal for $q$ (proof: apply Schur’s lemma to the $G$-invariant linear map $V_1 to V_2^vee$ obtained from $q$). So we are reduced to classifying elements of $mathcal{R}$ (or $mathcal{S}$) whose underlying representation is of the form $U^r$ for $U$ an irreducible self-dual representation of $G$, or of the form $(Uoplus U^vee)^r$ for an irreducible non-self-dual representation $U$.

Anyway, let me concentrate on the important special case where $k=mathbb{Q}$ and $G=mathbb{Z}/nmathbb{Z}$. The irreducible representations of $mathbb{Z}/nmathbb{Z}$ over $mathbb{Q}$ are of the form $U_d$ (self-dual) for $d$ dividing $n$ where $U_d$ splits over $mathbb{C}$ as sum of one-dimensional representations on which a chosen generator acts through each of the primitive $d$-th roots of unity. So I ask:

Actual question: Given $d,n,rgeq 1$ be integers such that $d$ divides $n$, let $U_d$ be the irreductible representation of $mathbb{Z}/nmathbb{Z}$ over $mathbb{Q}$ such that the generators act with characteristic polynomial given by the $d$-th cyclotomic polynomial. Can we classify quadratic forms on $(U_d)^r$ which are invariant under the action of $mathbb{Z}/nmathbb{Z}$ (i.e., describe the corresponding elements of the (known) Grothendieck-Witt ring of $mathbb{Q}$)? Or equivalently, in the other direction, given a quadratic form $(V,q)$ over $mathbb{Q}$ (through its image in the G-W ring), can classify $mathbb{Z}/nmathbb{Z}$-actions (over $V$, linear, preserving $q$) which make $V$ isomorphic to $(U_d)^r$?

I don’t even know the answer when $q$ is the standard Euclidean form (viꝫ. $mathbb{Q}^m$ with the quadratic form $x_1^2 + cdots + x_m^2$): for which $d,n,r$ is there a $mathbb{Z}/nmathbb{Z}$-invariant quadratic form on $(U_d)^r$ that is isomorphic to this?

Note: there is a $mathbb{Z}/nmathbb{Z}$-invariant standard Euclidean structure on $mathbb{Q}^n = bigoplus_{d|n} U_d$ with cyclic permutation of the coordinates, which induces a quadratic form on each of the $U_d$ so that this direct sum is orthogonal (the class of this form in the G-W ring can be computed by the Möbius inversion formula; because there is a scaling involved, it depends on $n$, not just $d$). It might be tempting to think that all $mathbb{Z}/nmathbb{Z}$-invariant quadratic forms on $U_d$ are obtained in this way: if my Witt ring calculations are correct, this is not the case: $U_{30}$ does not get a standard Euclidean structure that way; but there is a standard Euclidean structure on $U_{30}$, namely, take the Coxeter element of the Weyl group of $E_8$ as acting on $mathbb{Q}^8$ with its standard Euclidean structure, which is then $U_{30}$ as a representation of $mathbb{Z}/30mathbb{Z}$.

nt.number theory – satisfying Euclid’s twin primes conjecture allows us to give a solution to the strong version of Goldbach’s

enter image description hereThe Fundamental Theorem of Arithmetic proves that the infinite (hereinafter ∞) natural numbers are primes or compounds and are the producer of prime numbers; satisfying the two versions of Goldbach’s conjecture will allow us to prove that even ∞ natural numbers are the sum of 2 prime numbers, ∞ odd numbers are the sum of only 3 prime numbers; satisfying Euclid’s twin primes conjecture allows us to give a solution to the strong version of Goldbach’s conjecture, “all even ∞ are the sum of two primes” by not looking for the impossible combinatorics between two primes whose quantity and value we do not know , but analyzing the distances between two consecutive prime numbers spaced only by the multiple numbers.