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数论 MSE 2 票 0 回答 31 浏览 未读

Primes as $X-Y$ with $\gcd(X,Y)=1$ and $\operatorname{rad}(XY)$ equal to the product of all primes below $p$ — is this known?

Matthew Miller
Title: Primes as $X-Y$ with $\gcd(X,Y)=1$ and $\operatorname{rad}(XY)$ equal to the product of all primes below $p$ — is this known? While experimenting with multiplicative decompositions of primes, I arrived at the following question. I would like to know whether it is already in the...
代数几何 MSE 0 票 0 回答 26 浏览 未读

reduction of a conjugate point in $X_0(p)$

Camilo Gallardo
Consider a non-cuspidal point $x\in X_0(p)(K)$ representing an elliptic curve $E/K$ over a quadratic field $K$, and suppose that $E$ has potentially multiplicative reduction over a prime $\mathfrak{q}$ of $K$ over the rational prime $q$. Let $0,\infty$ be the cusps of $X_0(p)$. Using the...
代数几何 MSE 2 票 1 回答 34 浏览 未读

Defining property of morphisms of algebraic spaces out of an étale local on source and target property of scheme morphisms. Stacks Project vs Olsson

Elías Guisado Villalgordo
This is about the umpteenth discrepancy of definitions in algebraic geometry. It's about recipes to define properties of morphisms of algebraic spaces out of properties of morphisms of schemes. When the morphism of algebraic spaces in question is representable the recipe is clear [SP, 025V;...
代数几何 MSE -1 票 0 回答 34 浏览 未读

Irreducible topological space

Lars
We say that a topological space $X$ is irreducible if it can not be written as the union of two proper closed subsets. My question is: Do we consider this definition when the topological space is defined in terms of open subsets or closed subsets? When working with the zariski topology, the...
代数数论 MSE -1 票 0 回答 38 浏览 未读

How can the Albert-Brauer-Hasse-Noether theorem be interpreted topologically via sheaf cohomology? Seeking precise duality dictionary and references

Damien Leandro
I am looking for a topological or geometric interpretation of the Albert–Brauer–Hasse–Noether (ABHN) theorem, which establishes the local-global principle for central simple algebras over a global field $K$. The classical exact sequence is given by: $$0\rightarrow \text{Br}(K)\rightarrow...
解析数论 MSE 0 票 0 回答 19 浏览 未读

Exponential sum associated to Maass cups forms of level $N$

Hossain
We consider the $L$-function associated with a nonzero Maass cusp form $f$ of weight $0$, level $N$, and Laplace eigenvalue $1/4+r^2$. Let $t(n)$ be the normalized Fourier coefficient corresponding to the Maass cusp form $f$. I need the estimate of $$\sum_{n\le T}t(n)e^{2\pi i n x},$$ where $x...
伽罗瓦理论 MSE 4 票 3 回答 126 浏览 未读

Galois group of $x^6+3$ over $\mathbb{F}_5$

khashayar
One question from a past qualifying exam is to find the Galois group of $x^6+3$ over the finite field $\mathbb{F}_5$. With some creativity, we can write: $$x^6+3=x^6+8=(x^2)^3+2^3=(x^2+2)(x^4-2x^2+4)=(x^2+2)(x^4+4x^2+4-16x^2)\\ =(x^2+2)((x^2+2)^2-(4x)^2)=(x^2+2)(x^2-2x+2)(x^2+2x+2).$$ Since...
伽罗瓦理论 MSE 0 票 1 回答 29 浏览 未读

Finding the Galois group of over $\mathbb{Q}$ using the Galois groups over finite fields

khashayar
One question in a past qualifying exam asked us to find the Galois group of $x^6+3$ over $\mathbb{F}_5$, $\mathbb{F}_7$, and $\mathbb{Q}$. Each part of this question has been answered individually on this website: over $\mathbb{F}_5$, $\mathbb{F}_7$, and $\mathbb{Q}$. I asked this question again...
代数几何 MSE 0 票 0 回答 43 浏览 未读

Why does a nowhere vanishing section of $\omega_{E/S}$ induce an isomorphism $\mathcal O_E \cong \Omega^1_{E/S}$?

Mehshav
I am reading Arithmetic Moduli of Elliptic Curves by Katz and Mazur, and I have a question about the beginning of Chapter 2, §2. Let $f:E\to S$ be an elliptic curve. Since the sheaf of relative diffrentials $\Omega^1_{E/S}$ is an invertible sheaf on $E$, one defines...
数论 MSE 5 票 1 回答 209 浏览 未读

A continued fraction for the reciprocal of Gauss's constant

Pedja
I found the following infinite continued fraction: $$ \frac{1}{G} = \frac1{(2\pi)^{-3/2}\,\Gamma^2(1/4)} = \cfrac{4}{1+\cfrac{5}{1+\cfrac{6}{1+\cfrac{9}{1+\cfrac{8}{1+\cfrac{13}{\ddots}}}}}} $$ where $G$ denotes Gauss's constant, and the partial numerators are defined by interweaving sequences...
数论 MSE 0 票 0 回答 59 浏览 未读

Why does repeatedly prepending a fixed bit-block converge the Collatz step-count difference to the block's own length?

MAEDA AKIHIRO
I've been experimenting with a self-similar construction for the Collatz map (the $n \to n/2$ / $n \to 3n+1$ function) and found a pattern I can partially — but not fully — explain. I'd appreciate a sanity check and any pointers to relevant literature. Construction. Fix an odd integer $x$ with...
代数几何 MSE 2 票 0 回答 65 浏览 未读

Can the resultant ideal $\mathrm{Res}(f, g)$ of two homogeneous polynomials be defined in terms of the projective vanishing locus $V_+(f, g)$?

Jakob Werner
Let $A$ be a commutative ring and $f, g \in A[S, T]$ be two homogeneous polynomials in two variables of homogeneous degrees $d$, resp. $e$. Their resultant $\newcommand{\Res}{\mathrm{Res}}\Res(f, g)$ is defined to be the determinant of the linear map of free modules of rank $d + e$ $$ (f, g)...
代数几何 MSE 0 票 0 回答 59 浏览 未读

About the Jacobian conjecture counterexample and the determinant

mick
So recently the Jacobian conjecture has been disproven. The counterexample had a determinant of $-2$. See for instance : https://www.newscientist.com/article/2580374-ais-solution-to-87-year-old-riddle-takes-mathematicians-by-surprise/ or Wikipedia. Now I wonder if this polynomial can lead to an...
伽罗瓦理论 MSE 0 票 1 回答 53 浏览 未读

Does the size of the automorphism group divide the separable degree

khashayar
Let $E/K$ be a finite field extension, $G=\operatorname{Aut}_K(E)$ be the group of automorphisms fixing elements of $K$, and $E^G$ be the fixed field of $E$ under $G$. We know $E/E^G/K$ and so $$[E:K]=[E:E^G][E^G:K]=|G|\,[E^G:K].$$ Additionally, $[E:K]=[E:K]_s[E:K]_i,$ where $[E:K]_s$ denote the...
伽罗瓦理论 MSE 2 票 4 回答 227 浏览 未读

Is every field Galois over its prime subfield?

khashayar
A field extension $F/K$ is Galois if $F^{\operatorname{Aut}_K(F)}=K$ in Hungerford's Algebra, and it is Galois if it is normal and separable in Lang's Algebra. For the finite extension, these two definitions are the same. For infinite algebraic extensions or for transcendental extensions, which...
代数几何 MSE 1 票 0 回答 108 浏览 未读

Counterexamples to Jacobian Conjecture not surjective

Dave Rusin
I notice that the recently-publicized counterexample(s) to the Jacobian Conjecture are not surjective. Is that necessarily the case? That is, (Q) If $F:\mathbb C^n \to\mathbb C^n$ is algebraic and everywhere locally injective, and also surjective, must it be injective? Maybe the relevant setting...
代数几何 MSE 1 票 0 回答 32 浏览 未读

Few Questions about Contraction of Exceptional Curve $E$ on a Smooth Surface

user267839
Let $X,Y$ be two algebraic surfaces (=smooth, proper $2$-dim schemes over fixed base field $k$) and let $E \subset X$ exceptional curve, ie $E \cong \Bbb P^1$ with self intersection $E^2=-1$. By Castelnuovo's contraction theorem $E$ can be contracted to a smooth point of a smooth surface leaving...
伽罗瓦理论 MSE 1 票 1 回答 38 浏览 未读

Field extension over a fixed field has smaller or equal degree than the size of the automorphism group

khashayar
Let $F/K$ be a finite field extension, $G=\text{Aut}_K(F)$ be the group of automorphisms of $F$ that fix elements of $K$, and $F^G$ be the fixed field of $G$. We then have $$[F:F^G]\le |G|.$$ This is proven in Hungerford Chapter V, Lemma 2.9. Hungerford used this lemma to prove "$F^G=K$ iff...
数论 MSE -2 票 0 回答 42 浏览 未读

Why does the Euclidean algorithm outperform prime factorization for finding the GCD of large integers?

Cat Mock
While creating quantitative aptitude problems for management entrance exam preparation, I noticed that the Euclidean algorithm is almost always preferred over prime factorization for computing the greatest common divisor.
数论 MSE 0 票 0 回答 60 浏览 未读

Solving system of congruences involving powers

Yathi
I am in the middle of a problem which needs showing that the following system of congruences has finite number of solutions. I have verified up to some extent through sage that this has only two solutions for $(q, r)$ (with $q<r$) namely $(11, 17)$ and $(23, 103)$. The system of congruences is...