# Tag Info

## Hot answers tagged nt.number-theory

17

I think that there is indeed some possibility to lower the bound, and this is something I've looked at seriously a few times. I spent a semester (in 2019) with the Computational Number Theory Group here at BYU trying to do it, without success. Let me outline some of the difficulties we found. Issue 1. The main technique used in establishing the current ...

9

For (3), the answer is no, because there is no irreducible element in $\overline{\mathbb{Z}}$: for example, every element is a square. Similarly for (1), the answer is no because there are infinite ascending chains of ideals: for example, $(x) \subset (x^{1/2}) \subset (x^{1/4}) \subset \cdots$, where $x$ is not a unit (e.g. $x=2$) and $\{x^{1/2^n}\}$ is a ...

8

I claim that for even $n\in \{0,2,4,\ldots, 4\cdot 5^n-2\}$ each remainder of $F_n$ modulo $5^n$ is realized at most twice (thus exactly twice), and the same for odd $n\in \{1,3,5,\ldots, 4\cdot 5^n-1\}$. Denoting $u=(1+\sqrt{5})/2$, $v=(1-\sqrt{5})/2$ we have Binet formula $F_n=(u^n-v^n)/(u-v)$. If $k,m$ are even, then $$F_k-F_m=\frac{u^k-u^{-k}-u^m+u^{-m}}{... 8 Let s\in\mathbb{C} be any point with \Re s>0. The Dirichlet series of \eta(s) converges, hence L(s) converges if and only if L^*(s) converges. For \sigma_0>1/2, it is also known that L(s) converges in the half-plane \Re s>\sigma_0 if and only if \zeta(s) has no zero in that half-plane. Combining the previous two statements, it ... 7 The conjecture is true. That is, if the integral cubic polynomial$$P(Z)=a_3 Z^3+a_2 Z^2+a_1 Z+1$$is irreducible in \mathbb{Z}[Z] (hence also in \mathbb{Q}[Z] by Gauss's lemma), then there are only finitely many positive integer solutions of the equation$$(xz+1)(yz+1)=P(z).$$1. First we consider the case when x\mid a_3 or y\mid a_3. By symmetry, ... 6 I think we can describe P a bit more, using Dirichlet’s unit theorem. Since \overline{\mathbf{Z}} = \varinjlim_{[K:\mathbf{Q}] < \infty} \mathcal{O}_K, the same is true for the units. Now Dirichlet tells us that taking logs of the archimedean valuations v \mid \infty of K gives us an embedding:$$\mathcal{O}_K^\times/\mu_\infty(K) \hookrightarrow \...

5

The ring $\overline{\mathbb{Z}}$ is called the ring of algebraic integers. You can find information about prime ideals, e.g., in https://math.stackexchange.com/questions/156231/non-zero-prime-ideals-in-the-ring-of-all-algebraic-integers.

5

It is possible to find equations for $\pi : X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi$. Yang shows that there exist two modular units $F,G$ on $X_1(... 5 The answer of RP_ is correct, so my main contribution is cleaning it up, providing more detail, and uniformising the different cases. Definition. Let$\Omega$be the set of prime numbers. For subsets$S, T \subseteq \Omega$, write$S \sim T$if the symmetric difference$S \mathbin\triangle T = (S \setminus T) \cup (T \setminus S)$is finite; note that this ... 4 Here's a sketch of an answer. I think the answer is that you can get three types of sets: (i) finite sets, (ii) co-finite sets, and (iii) sets of the form $$S_f = \{ p : f(x) ~ \textrm{has a root in \mathbb{F}_p}\} - S_0$$ for some polynomial$f \in \mathbb{Z}[X]$, and a finite set of primes$S_0$. By the Chebotarev density theorem, the sets$S_f$have a ... 4 I don't have a solution but since we're facing a narrow margins situation I'll answer. I followed the approach I guessed Gerhard Paseman's comment was talking about but there was an unforeseen complication. If$\gcd(2^x-1,3^y-1)=1$we have $$2^x-1=p^2$$ $$3^y-1=2q^2$$ for integers$p,q$. We can factor the first one in$\mathbb{Z}[i]$$$(-i)^x(1+i)^{2x}=(p+i)(... 4 (1) No, as Brunault explaned, we may get an infinite series of ideals. In more general setting, if A is an integral domain then any infinitely ramified (Indeed, an integral extension B of A is "infinitely ramified" if there is a tower A\subsetneq A_1\subsetneq A_2\subsetneq \cdots such that A_{i+1}/A_i are ramified) integral extension of ... 4 In view of the identity (n!)^2/((n-1)!)^2=n^2, the set S_p generates the subgroup Q_p<\mathbb F_p^\times of quadratic residues; thus, if S_p is a subgroup, then in fact S_p=Q_p. Clearly, a necessary and sufficient condition for this to happen is that the sequence of factorials \{1!,\dotsc,(p-1)!\} hits at least one element out of each pair (a,... 4 By Section 4 of P. Dusart [Math. Comp. 68(1999), 411--415], for any x\ge3275 there is a prime p such that$$x\le p\le x\left(1+\frac1{2\log^2x}\right)\le x\left(1+\frac1{2\log^2 3275}\right)<1.01x.$$For any integer n\ge463, we have p_n>3275 and hence$$p_{n+1}<1.01p_n<p_n+\frac12p_n<p_n+p_{n-1}.$$3 By Ichino's triple product formula, of which Watson's formula is a special case, this integral will be given by the expected ratio of L-values (depending only on the automorphic reps generated by the \phi_i) multiplied by a product of local correction terms (which depend on the \phi_i themselves). Since your level is 1, the local integrals at the finite ... 3 The set J(C)_{\Theta}[n] has the structure of a smooth irreducible algebraic curve, and the restriction of J(C)\xrightarrow{\times n } J(C) to C defines a morphism J(C)_{\Theta}[n]\rightarrow C which is a finite etale cover with Galois group J(C)[n]. I don't think it is good to think of J(C)_{\Theta}[n] as generalized torsion points. I think it's ... 3 Using the notation from my answer to the previous question, we have$$D(n+1)−(D(n+2)+1)\not\equiv 0\pmod{p}$$if and only if g_0(n+2) = 2 g_0(n+1) - p and g_1(n+2) \equiv 2 g_1(n+1) + 1\pmod{p}, in which case$$D(n+2) \equiv -(n+2 + \frac{2 g_1(n+1) + 1}{2 g_0(n+1)c})\equiv D(n+1) - 1 - \frac{1}{2 g_0(n+1)c} \pmod{p}.$$Similarly,$$D(n+2)−(D(n+3)+1)\not\... 2 This is more of a long comment than a proper answer. The question appears to be related to generalised eigenvalue problems. Specfically, let$X=diag(x_1,...,x_n)$, with$x_k\in\{0,1\}$. Then you want to solve$Cv=Xv$, with$v\neq 0$, which is a kind of "inverse" gen. eigenvalue problem, the gen. eigenvalue is 1, but you don't know$X$. It's ... 1 To define this correspondence you don't want to write down either side explicitly, but rather do it abstractly. Then you can calculate what it is in explicit terms if desired. The first steps are (1) Send each cusp in$\overline{M}_n$to its formal neighborhood in$\overline{M}_n$, and then to its punctured formal neighborhood in$M_n$. (2) Observe that ... 1 Q1 is too vague, so let me answer Q2 and Q3. Assume that a prime$p$and a primitive root$g$modulo$p$are fixed. Let$g_0(n) := g^n\bmod p$,$g_1(n) := \frac{g^n - g_0(n)}{p}\bmod p$, and$c := g_1(p-1)$. Assuming that$p$is not Wieferich prime base$g$, we have$c\ne 0$. From these definitions we have $$g^n \equiv g_0(n) + g_1(n)p \pmod{p^2}$$ and$\$g^{...

Only top voted, non community-wiki answers of a minimum length are eligible