$Fr$ is the frobenius transform($x\to x \ ^p$) i showed that $<Fr> \le Gal(\overline{F_p} / F_p)$ and that it is infinite. also i showed that $\overline{F_p} \ ^{<Fr>} = F_p$. now i need to show that $Gal(\overline{F_p}/F_p)$ is abelian.
i cant show it. i tried using galois mapping but i cant because the extention is not finite
I think the infinite galois group is defined by some inverse limit over finite extensions - in which case does the claim not just follow from inverse limit of an inverse system of abelian groups is abelian?
@TobiasKildetoft, $\int _0^{\frac{\pi }{2}}\:\frac{\sin x}{x}$ is integrable because $\frac{\sin x}{x}$ has finite limit at $0$ as well as at $\frac{\pi}2$; while $\int _{0^{^{^1}}}\:\frac{x}{\ln x}$ is not integrable because limit of $\frac{x}{\ln x}$ at -1 is $-\infty$, right?
@TobiasKildetoft, $\int _0^{\frac{\pi }{2}}\:\frac{\sin x}{x}$ is integrable because $\frac{\sin x}{x}$ has finite limit at $0$ as well as at $\frac{\pi}2$; while $\int _0^1\:\frac{x}{\ln x}$ is not integrable because limit of $\frac{x}{\ln x}$ at $1$ is $-\infty$, right?
This is same question as posted here: Math.StackExchange.com but I have used a different approach to prove the question and this question is regarding that approach.
After I realise that I made a mistake in understanding definition of convex criteria, this is my second attempt:
We need to show ...
@TobiasKildetoft, $\int _0^{\frac{\pi }{2}}\:\frac{\sin x}{x}$ is integrable because $\frac{\sin x}{x}$ has finite limit at $0$ as well as at $\frac{\pi}2$; while $\int _0^1\:\frac{x}{\ln x}$ is not integrable because limit of $\frac{x}{\ln x}$ at $1$ is $-\infty$, right?
now, in the right category, that becomes a morphism $\varphi: k[X,X^{-1}] \to \mathcal O(G)$ such that $m(\varphi(X)) = \varphi(X) \otimes_k \varphi(X)$
$Fr$ is the frobenius transform($x\to x \ ^p$) i showed that $<Fr> \le Gal(\overline{F_p} / F_p)$ and that it is infinite. also i showed that $\overline{F_p} \ ^{<Fr>} = F_p$. now i need to show that $Gal(\overline{F_p}/F_p)$ is abelian.
Now, for $\sigma, \tau \in \operatorname{Gal}(\overline{\Bbb F_p}/\Bbb F_p)$ and $x \in \overline{\Bbb F_p}$, we want to show that $\sigma(\tau(x)) = \tau(\sigma(x))$
@TobiasKildetoft @MatheinBoulomenos how do we see that $\Bbb R[X,Y]/(X^2+Y^2-1)$ is another solution? I mean, I can verify it, but how would one come up with this?
the first step is adjunction because $R^\times$ is the image of the monoid $R$ under the forgetful functor, and $k[M]$ is the image of the group $M$ under the free functor
so now $R \mapsto \operatorname{Hom}_{Grp}(M, R^\times)$ is a functor represented by $\operatorname{Spec}(k[M])$
If
$$
h(k) = \begin{cases} \frac{1}{2l+1} & -l \leq k \leq l \\0 & \text{otherwise}\end{cases}
$$
Where $l \geq 0$ is some integer.
I've done some computation and the summation
$$
F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = \frac{1}{2l+1}e^{-j\omega} \frac{\sin(\omega(l+1/2))}...
@TobiasKildetoft I actually posted it on the Fb group " >implying we can discuss mathematics", it got more than 1200 reads. A few guys took the time to ask me something of it
What is true is that morphisms of $K$-schemes $T_K \rightarrow \mathbb{G}_m$ is the same as morphisms of functors $Hom(-,T_K) \rightarrow Hom(-,\mathbb{G}_m)$
But you put $\mathrm{Spec}(K)$ there, so $Hom(\mathrm{Spec}(K), T_K)$ and $Hom(\mathrm{Spec}(K),\mathbb{G}_m)$ are sets
ok what about this as an example: if you take $T_K = \mathbb{G}_m$ too, and let's say everything is over $K=\mathbb{F}_2$.
Then the final thing in your equality has size $1$, but for the first thing you are looking at morphisms from $\mathrm{Spec} \mathbb{F}_2[T,T^{-1}] \rightarrow \mathrm{Spec} \mathbb{F}_2[T,T^{-1}]$, which has more than 1 element.
$L:= Hom(T_K, \mathbb{G}_{m,K})$. So an element of $l$ gives you a morphism of schemes $T_K \rightarrow \mathbb{G}_{m,K}$, and hence a morphism of $K$-valued points $T_K(K) \rightarrow \mathbb{G}_{m,K}(K) = K^{\times}$.
Let $X$ be a set, and $\mathcal P(X)$ be the power set of $X$. Consider the operations $\Delta$ = symmetric difference (a.k.a. "XOR"), and $\bigcap$ = intersection. So, $\Delta$ behaves as addition, whose identity is $\varnothing$.
@AlessandroCodenotti, Is this true: $\mathbb Z/n\mathbb Z$ has these ideals: $d\mathbb Z/n\mathbb Z$ for every divisor $d$ of $n$. Out of these maximal ideals are those only which are $d\mathbb Z/n\mathbb Z$ where $d$ is prime.
You might have come across GameScience's seven-sided die before:
There's mixed discussion of whether it might be biased toward the 6 and 7 faces (the 6 being on the opposite side of the 7 you can see in the photo). It's a GameScience die, and they tend to market themselves on making properly f...
@AlessandroCodenotti here $I$ is ideal of $R$? There is a bijective correspondence between : set of ideals $K$ of $R$ with $I\subset K$, and ideals of $R/I$
You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game. When should one stop playing this game?
For this question, I found the answer as $n= 2$ Let me explain How I got that: Let $X$ represents the number that comes up on the die. Therefore the game continues as long as $X<6$, So, $P(X=6)=nCrp^r q^{n−r}$ where $r=1$ then we have $\dfrac{1}{6}=nC_1 \times \dfrac{1}{6} \times \biggr (\dfrac{5}{6}\biggr )^{n-1}$ Which gives $ n = 1$
@TedShifrin your book I found for $40 came in , and its high quality and definitely looks authentic! there's a label "ILLEGAL for Sale in USA" on the cover, oops
I'm not sure how it works but it's way too nice and high quality to be a bootleg , 90% sure
the website I ordered from didn't mention the "illegal for sale in usa" part for some strange reason
good thing it doesn't say "illegal for purchase" i guess
if the original is in color, then the illustrations being in black and white would partly explain the price difference. also it's paperback. but I suspect it's something something bureaucracy something pls don't send me to prison
e.g. if $M=\Bbb N$ the free monoid on one generator, then $k[M]=k[X]$ represents the functor $R \mapsto \operatorname{Hom}_{mon}(\Bbb N,(R,\cdot))=R$ where we used the free-forgetful adjunction for monoids
Where the functor $R \mapsto \operatorname{Hom}_{Grp}(\Bbb N,R^\times)=R^\times$ is represented by $k[\Bbb Z]=k[X,X^{-1}]$
there's another adjunction that explains this: $\Bbb Z$ is the Grothendieck group of $\Bbb N$
Could you give me an example of twin primes q,p such that $x^2\equiv q mod p$ and $x^2 \equiv p mod q$ have either both or non of them a solution? Or does this hold for any prime twins?
@MaryStar Suppose $q=p+2$, then $\left( \frac{p}{q} \right) \left( \frac{p}{q} \right) = (-1)^{\frac{(p-1)(q-1)}{4}}=(-1)^{\frac{p^2-1}{4}}$ Note that if $p$ is odd, then $8$ divides $p^2-1$, so this is equal to $1$
Because then the field with $2^n$ elements is a subfield of the field with $2^k$ elements, so Lagrange applied to multiplicative groups implies $2^n-1 \mid 2^k-1$
D. none of the above , Question was like this. how do I choose the domain and range? if I choose range to be complex. then function is not well defined. right? If a function is $f:\mathbb R^+ \to \mathbb R$ then A) is false. rest of the above are true. am I correct?
I think we cannot talk about convexity or concavity at $x =0$, but when $x>0$ then the second derivative of $f(x) = x^(1/3)$ is $<0$ imlying concavity! so its concave on $(0,\infty)$
It's less inspiring than you might think. Let G be a finite group and X be a space. Then we write X_{hG} for (X x EG)/G, and X^{hG} for Map^G(EG, X). Ideally everything is really done in some "stable category", so we can add maps. You can come up with a map X_{hG} -> X^{hG}, given by 'averaging'
More precisely, Map^G(EG, G) is contractible (this is the group cohomology statement that cohomology with coefficients in $\Bbb Z[G]$ is trivial)
So X_{hG} ~ (X x EG x Map(EG, G))/G. Then the map sends (x, v, f) to the map sending v to f(v)x
You then just define Tate homology to be the homology of the cofiber of this map. It's kinda the same thing as in group homology: you come up with a comparison map between homology and cohomology and Tate fits into an exact sequence
For this to work you need to work in some sort of stable category so you can add maps - of course in spaces Map(EG,G) is G, and has no fixed points
But the claim that Map^G(EG, G) = * is true in spectra, or chain complexes
That's good intuition and indeed this procedure can be carried out sort of for a compact Lie group
In that case you can identify Map^G(EG, G) with a copy of the ground ring concentrated in degree dim G; it carries an action by pi_0 G, where g acts by Det(ad(g)) in +-1
In particular you still get this description of Tate homology as the difference between group homology and cohomology, but with a shift (dim G) and a twist (the determinant of adjoint representation)
In de Rham world this should all be possible to describe as literally averaging
Hello, very quick question. In this guys answer, he says that $Q$ has smaller dimension than $P$, thus the map cannot be injective. But it seems as though his inequality right before that statement claims just the opposite: that the dimension of $Q$ is bigger than $P$. Can anyone please clarify for me?
Let $K$ be the scalar field, and $f$ be the linear map from $K^n$ to $K^n$ associated with $A$. The fact that $A$ has a $p \times q$ zero submatrix implies that $K^n$ has a $q$-dimensional subspace $Q$ and an $(n-p)$-dimensional subspace $P$ such that $f(q) \in P$ whenever $q \in Q$. Since $p+q \...
$$2\pi n^{(3/2)} \leq 2^{(n/2+10)}$$
$$\log_2 (\pi n^{(3/2)}) \leq (n/2+9)$$
$$\log_2 (\pi)+\dfrac{3\log_2(n)}{2} \leq (n/2+9)$$
$$\log_2 (\pi)-9 \leq \dfrac{n- 3\log_2(n)}{2}$$
$$n- 3\log_2(n) \geq -14.69$$
$$3\log_2(n) - n \leq 14.69$$
Let $y=3\log_2(n)-n$
To see what is the maximum value of ...
@Mathei I am largely ignorant but I think it fits into this framework. The issue is that now the "dualizing object" Map^G(EG, G) is now just some silly thing
So it is harder to state as the cofiber of a map between homology and cohomology
I learned this from an article of Klein "axioms for Tate cohomology"
This is same question as posted here: Math.StackExchange.com but I have used a different approach to prove the question and this question is regarding that approach.
After I realise that I made a mistake in understanding definition of convex criteria, this is my second attempt:
We need to show ...
If
$$
h(k) = \begin{cases} \frac{1}{2l+1} & -l \leq k \leq l \\0 & \text{otherwise}\end{cases}
$$
Where $l \geq 0$ is some integer.
I've done some computation and the summation
$$
F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = \frac{1}{2l+1}e^{-j\omega} \frac{\sin(\omega(l+1/2))}...
You might have come across GameScience's seven-sided die before:
There's mixed discussion of whether it might be biased toward the 6 and 7 faces (the 6 being on the opposite side of the 7 you can see in the photo). It's a GameScience die, and they tend to market themselves on making properly f...
$$2\pi n^{(3/2)} \leq 2^{(n/2+10)}$$
$$\log_2 (\pi n^{(3/2)}) \leq (n/2+9)$$
$$\log_2 (\pi)+\dfrac{3\log_2(n)}{2} \leq (n/2+9)$$
$$\log_2 (\pi)-9 \leq \dfrac{n- 3\log_2(n)}{2}$$
$$n- 3\log_2(n) \geq -14.69$$
$$3\log_2(n) - n \leq 14.69$$
Let $y=3\log_2(n)-n$
To see what is the maximum value of ...
