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Jun
7
comment Shortest irreducible polynomials over $\Bbb F_p$ of degree $n$
With your new Maple script you should be able to fill in the rest of your table for the primes above $7$ without much effort. I would be a bit surprised if the currently missing entries do not all turn out to be $3$.
Jun
7
comment Shortest irreducible polynomials over $\Bbb F_p$ of degree $n$
Then you can fill in the rest of your table for $p \leq 7$. Since the 2 and 3 entries swamp everything else, and should continue to do so as $n$ grows, make a table for each of the first few primes $p$ to separately indicate the $n$ up to $500$ or $1000$ where $\lambda(p,n)$ takes on different values greater than $3$. Trivially, of course, $\lambda(2,n)$ will never have even values.
Jun
7
comment Shortest irreducible polynomials over $\Bbb F_p$ of degree $n$
@JimBelk, okay. It would be good if someone else could run through the calculations I made to verify numerically what I had found.
Jun
7
comment How do we prove Dirichlet L-series converges when $\chi$ is non-trivial character and $s>0$?
This should be discussed in almost any analytic number theory book. Have you read any such texts?
Jun
7
comment Shortest irreducible polynomials over $\Bbb F_p$ of degree $n$
In particular, you're going to find $\lambda(3,n) \leq 3$ for $n \leq 57$ except at $n=49$ and $n=57$, $\lambda(5,n) \leq 3$ for $n \leq 70$ except at $n=35$ and $n=70$, and $\lambda(7,n) \leq 3$ for $n \leq 163$ except at $n=124$ and $n=163$.
Jun
7
comment Shortest irreducible polynomials over $\Bbb F_p$ of degree $n$
For each $n > 1$ and $k$ running from $1$ to $n-1$, determine if $x^n + ax^k + b$ is irreducible as $a$ and $b$ run over $\mathbf F_p$ using a computer algebra package. Add $1$ if it is irreducible and add $0$ if it is reducible. Then see for which $n$ the sum over all $k$, $a$, and $b$ is $0$. The calculations were carried out months ago, in order to find the first two $n$, for small $p$, when there are no irreducible binomials or trinomials over $\mathbf F_p$. (It included your data $n = 8$ and $n = 13$ for $p = 2$, but I hadn't mentioned it before.)
Jun
6
comment Shortest irreducible polynomials over $\Bbb F_p$ of degree $n$
Some $(p,n)$ beyond the scope of your table such that $\lambda(p,n) > 3$, i.e., there are no monic irreducible binomials or trinomials in $\mathbf F_p[x]$ of degree $n$, are $(p,n) = (3,49)$, $(3,57)$, $(5,35)$, $(5,70)$, $(7,124)$, and $(7,163)$.
Jun
1
comment The group $(1+p\mathbb Z_p)/(1+p^{n}\mathbb Z_p)$
I see a bounty seeking an "easy argument" that the order of the quotient group is $p^{n-1}$. What is not easy about the argument in the first paragraph of my answer? As coset representatives it shows you can use all $1+a_1p+\cdots+a_{n-1}p^{n-1}\bmod p^{n}$, where each $a_i$ runs from $0$ to $p-1$. No exact sequences are necessary.
Jun
1
comment Problem with Differentiation under the Integral Sign
@BetterWorld, review the meaning of improper integrals. If you did not understand my explanation for why $\int_0^\infty \arctan(bx)/x \,dx$ diverges then please speak to a calculus teacher in person. Your treatment is like studying the convergent series $\sum_{n \geq 1} (-1)^{n-1}/n = 1 - 1/2 + 1/3 - 1/4 + \cdots$ by writing it as $S_1 - S_2$, where $S_1 = 1 + 1/3 + \cdots = \sum_{m \geq 1} 1/(2m+1)$ and $S_2 = 1/2 + 1/4 + \cdots = \sum_{m \geq 1} 1/(2m)$. Both $S_1$ and $S_2$ diverge.
Jun
1
comment Problem with Differentiation under the Integral Sign
For $b > 0$ your $J(b)$ is meaningless: the integral does not converge. There's no problem near $x = 0$, where $\arctan(bx)/x \approx b - b^3x^2/3$, but out near $x = \infty$ the value of the integrand behaves like $(\pi/2)/x$, so the integral diverges since $1/x$ is not integrable out towards $\infty$. The original question is about a difference under the integral sign, where convergence is not problematic. What you've done is similar to studying a convergent series by writing it as a difference of two divergent series: a dangerous move unless you are more careful than you have been.
May
28
comment Is there a finite abelian group $G$ such that $\textrm{Aut}(G)$ is abelian but $G$ is not cyclic?
If you google "abelian automorphism group" then you can find pages (on mathoverflow, for instance) with answers that address your question.
May
28
revised Is there a finite abelian group $G$ such that $\textrm{Aut}(G)$ is abelian but $G$ is not cyclic?
edited title
May
28
comment Geometric structure on the set of valuation rings of a field
Have you tried a rational function field like $\mathbf F_p(x)$? (Classically one could consider $\mathbf C(x)$ too, but then only consider valuation rings with quotient field $\mathbf C(x)$ that contain $\mathbf C$.)
May
28
comment Is there a finite abelian group $G$ such that $\textrm{Aut}(G)$ is abelian but $G$ is not cyclic?
In the spirit of the previous comment, would you like it if someone simply answered your question with "yes" or "no"? That's as useful as your title is.
May
28
comment Prove by combinatorial method that $ \frac{(2m)! \cdot (2n)!}{(m)! \cdot (n)! \cdot (m+n)!} $ is an integer
What you describe at the end is not really making full use of the situation, because the separate factors $(2m)!/m!^2$, $(2n)!/n!^2$, and $(2m)!/((m-n)!(m+n)!)$ if $m \geq n$ (or analogous expression if $n \geq m$) are all integers.
May
28
comment A finite group which has a unique subgroup of order $d$ for each $d\mid n$.
How much group theory do you know already? And is this a homework problem?
May
28
revised A finite group which has a unique subgroup of order $d$ for each $d\mid n$.
deleted 3 characters in body
May
27
comment The group $(1+p\mathbb Z_p)/(1+p^{n}\mathbb Z_p)$
Certainly. Otherwise it is false that $|u/v - 1|_p = |u-v|_p$; they differ by the factor $|v|_p$. That equation is symmetric in $u$ and $v$ precisely when they both have absolute value $1$. Without that equality, you can't say the ratio $u/v$ is as close to $1$ as the difference $u-v$ is to $0$, which is the reason that we can think of the multiplicative group $(1+p\mathbf Z_p)/(1 + p^n\mathbf Z_p)$ as a subset of the additive group $\mathbf Z_p/p^n\mathbf Z_p$ for the purpose of counting its size.
May
27
revised What are statements about the natural numbers where induction is impossible or unnecessary to prove?
added 1 character in body
May
27
comment What branch of Mathematics does the study of Algebraic/Transcendental Numbers lie in?
@Trogdor, there is indeed a subject within number theory called "transcendental number theory." It is concerned with transcendence of special values of functions like exponential, elliptic, or modular functions.