@MaryStar you said a is a root of x^(p^n)-x-1 right? what is the degree of that? its degree is p^n. that means F(a)/F has degree p^n, so F(a) is isomorphic to F^(p^n) as an F-vector space, so it has p^(p^n) elements.
@arctictern Ah ok... I want to apply the following theorem for $E=\mathbb{F}_p(a)$ and $F=\mathbb{F}_{p^n}$. Since m divides n, it must hold that $n\mid p^n$, or not? Does this hold?
@MaryStar in your old problem you said a was a root of an irreducible polynomial $x^{p^n}-x-1$. as I explained to you, this means $\Bbb F_p(a)\cong \Bbb F_{p^{\large p^n}}$. So why are you talking about $\Bbb F_{p^n}$?
maybe just start at the very beginning and say what your original problem is.
and in any case $Gal(F_p(a)/F_{p^n})$ doesn't make sense unless $n$ is itself a power of $p$. this makes me skeptical you are telling us the original problem.
Let $p$ be a prime, $n\in \mathbb{N}$ and $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ irreducible. We have that $a\in \overline{\mathbb{F}_p}$ is a root of $f$. We have that $\mathbb{F}_p(a)$ is a finite extension of $\mathbb{F}_p$.
We have the following: - $\mathbb{F}_p(a)$ contains all the roots of $f$. - for each $b\in \mathbb{F}_{p^n}$ we have that $a+b$ is a root of $f$. - $\mathbb{F}_{p^n}\leq \mathbb{F}_p(a)$ and $n=p^i$ for some $i\in \{0, 1, \ldots , n\}$.
I want to show that $\text{Gal}(\mathbb{F}_p(a)/\mathbb{F}_{p^n})$ is cyclic and let $\tau$ be a generator. I want to compute also th…
If $L/K$ is an extension of finite fields, then ${\rm Gal}(L/K)$ will be generated by $x^{|K|}$.
In your case ${\rm Gal}(\Bbb F_p(a)/\Bbb F_{p^n})$ will be generated by $\tau(x)=x^{p^n}$.
really, the details seem mostly irrelevant. Just assume you have any extension $L/K$ of finite fields. Show it's Galois, and its Galois group is cyclically generated by $\tau(x)=x^{|K|}$.
If $K$ is a finite field, $f(x)\in K[x]$ an irreducible polynomial and $a$ a root of $f(x)$, then $K(a)$ does happen to be the splitting field of $f$, but not by definition. (If $K$ is not a finite field for instance, it may not be a splitting field.)
Suppose I have the piecewise function $f(x) = \begin{cases}1, & 0<x<1 \\ 0 & x>1 \end{cases}$
and I wanted to extend it to an even function over the entire real line. How would I do that?
I know that normally for a function $\phi$ defined on the half line, the even extension would normally be ...
I don't know any other name of $\langle a, b | [a,b]^2 = 1\rangle$ off the top of my head. It's the fundamental group of the 2-torus with the two meridianal generators and the two longitudinal generators tubed off respectively, but that's probably unhelpful.
Maybe I lied though. What I wanted to do is mark the pair of meridians and pair of longitudes in $\Sigma_2$ and add tubes $S^1 \times I$ connecting the two meridians/longitudes resp. in each pair. But of course that introduces more generators (longtidinal curves in the tubes I glued) so you have to cap that off by a disk.
Sorry, I was in a hurry. None of this is helpful anyway so whatever
@Null for every $n\in\Bbb N$ you can construct a matrix $A$ such that $A,A^2,A^3,...,A^{n-1}$ are nonzero but $A^n=0$. Can you find a simple way to do so? How big must such a matrix be?
Someone voted down my this answer without a comment. What is bad with this answer? http://math.stackexchange.com/questions/2084356/the-language-accepted-by-given-turing-machine/2084375#2084375
Hey everyone, does anyone know of any references to learn Axiomatic (ZFC Set Theory), with a viewpoint towards General Topology?
I've been reading through Halmos' Naive Set Theory, as that is the only standard text I know of as a reference for elementary set theory, but I haven't heard of any standard texts for Axiomatic Set Theory
Feeling stupid today, so I'd like some linear algebra help.
Suppose I have some symmetric matrix $\mathbf A$ with the eigenvalues $\lambda_1,\lambda_2,0$. Is there an expression for the symmetric matrix with eigenvalues $a\lambda_1+b\lambda_2,b\lambda_1+a\lambda_2,0$ entirely in terms of $\mathbf A$?
(The corresponding eigenvectors are still the same.)
@J.M. I suppose rewriting $A$ as $Q\Lambda ^{t}Q$ with $\Lambda$ diagonal with the eigenvalues of $A$ on the diagonal and $Q$ orthogonal with the eigenvectors of $A$ on the columns is not an option?
Suppose your matrix is 3x3. Then given it has 3 distinct eigenvalues, $A$ can be diagonalised with its eigenvectors as its basis $B=\begin{pmatrix}v_1 & v_2 & v_3\end{pmatrix}$. That is: $A=B^TD(\lambda_1,\lambda_2,0)B$, where $D$ is your diagonal matrix. Now you want a matrix $A'$ that has the eigenvalues as shown. Notice that for a diagonal matrix, the eigenvalues are indeed the diagonal elements only. Also since $A'$ and $A$ have the same eigenvectors, $B$ is the same. Therefore observe: \begin{align}
@Secret Apparently, there's no way to escape computing the eigenvectors, then. I wanted to avoid that because the expressions for the entries of $\mathbf A$ are simple, but the eigenvalues and eigenvectors are unwieldy.
(and yes, it is $3\times3$, that's why I said it has only those three eigenvalues.)
yeah, even in the above case, you still need to worry about the details of B as you need that to transform the permutation matrix that kinda help simplify your work a bit
NB: I like abstract algebra partly because I am lazy. How one can get beautiful answer by just fiddling around symbols and not worry too much number crunching
--- Speculation on the general case: Hmm, I suspect if given a diagonalisable nxn matrix $C$ with n distinct eigenvalues, and one want $C'$ of the same eigenspace as $C$ such that the n distnict eigenvalues are polynomials of the eigenvalues of $C$, then $C'=\textrm{some polynomial involving permutation matrices and diagonal matrices of $C$}$
Ah, I forgot to mention that the nonzero eigenvalues are guaranteed distinct; thanks for the reminder.
The most general case I'm actually considering is that one has the same set of eigenvectors, but a linear combination of the eigenvalues; that is, if $\mathbf w=(\lambda_1,\lambda_2,\dots)^\top$ is the vector of eigenvalues, the desired matrix has the eigenvalues with the entries of $\mathbf P\mathbf w$, $\mathbf P$ being SPD.
problem with {x,y,0}->{ax+by,bx+ay,0} is that the operation knows which eigenvalue is 0. presumably any operation done that can be algebraically expressed purely with A would blindly do the same thing to all three eigenvalues
Let desired diagonalisable matrix be $A'$. Given $A$ with vector of eigenvalues $w$, then $A'=B^T\left(\sum_i a_iP_i\right)D(\lambda)B=B^T\left(\sum_i a_iP_i\right)BB^TD(\lambda)B=B^T\left(\sum_i a_iP_i\right)BA$. $P_i$ are permuations matrices including the identity matrix.
How to express as $Pw$ I have no idea. Perhaps it will mean $B^T\left(\sum_i a_iP_i\right)B^T$ needs to be SPD (Since $D(\lambda)$ acted by any matric on the left is basically the same as the result for Pw at the diagonal form some given SPD matrix P)
typo: form=from
I see no reason why the summation will be positive definite (if I do not mistaken what SPD means) espeically if enough $a_i < 0$. It may also not necessary symmetric in general
Consider the fabius function
https://en.m.wikipedia.org/wiki/Fabius_function
https://people.math.osu.edu/edgar.2/selfdiff/
How does one show that this function is nowhere analytic ?
Probably related , Maybe even a step in the answer : how to evaluate this function for nonreals ? Is it defined...
Hello @arctictern !! I am still looking at the problem of yesterday. We have the automorphism $\tau (x)=x^{p^n}$. To compute the order of $\tau$ do we do the following? $\tau(a)=a^{p^n}=a+1 , \tau^2(a)=a^{p^n}+1=a+2, \ldots , \tau^k(a)=\tau^{k-1}(\tau(a))=\tau^{k-1}(a^{p^n})=\tau^{k-1}(a+1)=\ldots =a+k, \ldots , \tau^{p^n}(a)=a+p^n=a$ So the order is $p^n$. Is this correct?
The question is contained in the title; I mean the standard axioms ZFC. The wiki link: Riemann hypothesis. There are finite algorithms allowing one to decide if there are non-trivial zeroes of the $\zeta$-function in the domains whose union exhausts the whole strip $0<\Re z<1$, but this does not ...
@MaryStar First off, if $\tau(x)=x^{p^n}$ then $\tau^k(x)=x^{p^{kn}}$. The trivial automorphism is $x^{p^{p^n}}$ so we must solve $kn=p^n$, i.e. $k=p^n/n$. On the other hand, the charactersitic is $p$, so solving $\tau^k(a)=a+k$ with $k$ minimal gives $k=p$.
that seems pretty restrictive on p and n...
In fact it seems to only be possible when $p=2$, $n=2$.
Caption: The idea is to construct a bunch of GUI such that the user can specify the column vectors of the matrix of interest in one of the fields. Then above them display the vector field of the resulting column matrix obtained along the n columns once it is plugged into the polynomial, so one can get some idea on given an arbitrary vector r, how it would be transformed by the polynomial matrix function
Therefore, if P(A)=0 for some A, it means the user should see all n boxes above them being zero vector fields
The n boxes idea will work because $P(A)\mathbf{r}=\sum_i,^n [P(A)]_jr_i$, so the user only need to scale up and down the n vector fields correspondingly and they will obtain the value of $P(A)\mathbf{r}$ for a given $\mathbf{r}$
say I have a function that has the same value for some x as another. But not for all. I then write equals and all the information about the other function having an x defined in other regions is no longer considered
(cont. from caption) We hope that by exploiting some version of flick animation like effect, that as the user play around with the aforementioned applet, the linked changes in all n boxes due to changes of one vector in the A box will be somwhow snapped together into some coherent whole and allow the user to perceive the matrix function in its entirety
(...which I have absolutely no idea what it would look like)
Now a nice question I already posted earlier for someone else, if you have a matrix $A$ such that $A,A^2,...,A^{n-1}$ are all nonzero and $A^n=0$ what can you say about the size of $A$ and why? @DHMO
Lets say I write $\forall x A(x)$ then I write on a new line: $\forall x A(x) = \exists x A(x)$ and continue my workings with the more limited information I now have.
I hope the latex works. Its the first time ive used it in chat
@Astyx greater or equal unless I'm off by one. The idea is that a nilpotent matrix can't have maximal rank and the best case (to construct nilpotent matrices with a big exponent) is when its kernel has dimension 1
@user400188 What you mean by domain. How can the statement $\exists$ be the same as $\forall$ ? Or do you mean we restrict the statement to a singleton for which the statement holds
@Astyx yes i mean for the case in which we restrict the statement. By domain I mean the domain the function is defined over. This question has an example of what I mean by domain if it helps: math.stackexchange.com/questions/2078380/…
Note you don't need to answer the question there just get familiar with the terminology. Its whats taught in a few high school classes.
@Astyx Originally it was how is information loss treated in mathematics or logic. I guess I'm asking: How often is it done? Is there a field of study about it? Is it frowned upon in any way?
oh so the distance is right angeled to the sides and the upper side of the square is also right angeled so they are parallel and that makes a parallelogram
@user400188 I think mathematicians try to use hypothesis that as weak as possible, meaning one of the goal of math is to limit the amount of information loss in theorems etc.
However, it's not that big of a deal to use theorems that crush some minor statements even if that implies ignoring a huge amount of what the theorem tells us since (as far as I understand it) maths is more about proving things in a certain environement (not quite sure what term to use here) and slowly but surely constructing theorems, thus the information is always kept in the hypothesis
Hope this makes some sense to you
@MartianCactus What do you mean by "which equals the sides" ?
Well you could say we study information loss when we consider when an implication is in fact an equivalence or how to stregthen the second statement to make it imply the first (or even better, make it equivalent to the first)
Even equivalence seems to have some form of information loss. For instance when we use it although the two give the same value all the time the shape or form may be different on each side.
Which means that they give the same results for different reasons. If you were to follow through the steps of evaluating equivalence you would see different structures getting restricted in different ways until they eventually give the same result.
@user400188 I don't think different form matters, formulae don't "give results", they are what they are, and equality means two formulae are in fact the same thing
@MithleshUpadhyay you should probably edit the question to contain details of what you've tried thus far or what ideas you might have about what the answer probably is.
Suppose I have the piecewise function $f(x) = \begin{cases}1, & 0<x<1 \\ 0 & x>1 \end{cases}$
and I wanted to extend it to an even function over the entire real line. How would I do that?
I know that normally for a function $\phi$ defined on the half line, the even extension would normally be ...
Consider the fabius function
https://en.m.wikipedia.org/wiki/Fabius_function
https://people.math.osu.edu/edgar.2/selfdiff/
How does one show that this function is nowhere analytic ?
Probably related , Maybe even a step in the answer : how to evaluate this function for nonreals ? Is it defined...
