from my tree diagram I can get 6 from $\Bbb Q(t,p,\omega)/\Bbb Q(t,\omega)$ or $\Bbb Q(t,p,\omega)/\Bbb Q(t,p^3)$ or $\Bbb Q(t,p^3,\omega)/\Bbb Q(t)$ or $\Bbb Q(t,p)/\Bbb Q(t)$
another one: if $R$ is a noncommutative ring, is it necessarily true that $\forall x \in R, x^n$ has the same value independently of the order of operations?
Are you sure you mean <(1 2 3 4 5 6), (1 4)> and not <(1 2 3 4 5 6), (1 4)(2 5)(3 6)>? If you swap $p$ and $q$, you have to swap $\omega p$ and $\omega q$ as well
Hi. Does anyone have some references or experience with dealing with samples of random size? Mostly I'm interested in CLT's, LLN's... Having problems with a random denominator (sample size)
Well, there are multiple definitions of normal extensions. One possible definition is that $L/K$ is a normal extension, if for any polynomial with coefficients in $K$ that has a root in $L$, all roots must lie in $L$
We can write the roots as $\zeta^k$ where $0 \le k < 12$.
Then, let $r$ be an automorphism of $\Bbb Q(\zeta)/\Bbb Q$ that takes $\zeta$ to $\zeta^s$. We know that $r(\zeta^k) = r(\zeta)^k = \zeta^{sk}$.
Therefore, automorphisms of $\Bbb Q(\zeta)/\Bbb Q$ are automorphisms of $\Bbb Z_{12}$, which...
$x^{n-1} = x \times x \times \dots \times x \implies x \times x^{n-1} = x \times x \times x \times \dots \times x = (x \times x \times \dots \times x) \times x = x^{n-1} \times x$
You cannot in general deduce the irreducibility of $f(X^n)$ from the irreducibility of $f(X)$.
(Consider $f$ as polynomial over $\mathbb Q[t]$ and reduce modulo the prime element $2t+1$, then we get $\bar f = X^6+X^3+1$ over $\mathbb Q[t] / (2t+1) \cong \mathbb Q$. But the irreducibility of thi...
We can write the roots as $\zeta^k$ where $0 \le k < 12$.
Then, let $r$ be an automorphism of $\Bbb Q(\zeta)/\Bbb Q$ that takes $\zeta$ to $\zeta^a$. We know that $r(\zeta^k) = r(\zeta)^k = \zeta^{ak}$.
Therefore, automorphisms of $\Bbb Q(\zeta)/\Bbb Q$ are automorphisms of $\Bbb Z_{12}$, which...
I said it's $V_4$ here but somehow I can't believe it
Once you know that the extension has degree 4, this means that the Galois group can't be $\mathbb{Z}/4\mathbb{Z}$, because $\mathbb{Z}/4\mathbb{Z}$ has only one subgroup of index $2$
DogAteMy: Suppose I take a codimension-2 real subspace of $\Bbb C^3$ and project it down to $\Bbb CP^2$. Of course, when it's a complex $2$-dimensional subspace, we get a projective line downstairs. What if it's not closed under complex scalar multiplication?
Say I want to maximize $f(x,y)$ and you want to minimize it, and I have control of $x$ and you have control of $y$. A reasonable idea would be that at each time step, I increment $x$ by an infinitesimal multiple of $\partial_xf$ and you increment $y$ by an infinitesimal multiple of $-\partial_yf$.
Mr anon, I'm being a dope. Can you answer this? Suppose I take a codimension-2 real subspace of $\Bbb C^3$ and project it down to $\Bbb CP^2$. Of course, when it's a complex $2$-dimensional subspace, we get a projective line downstairs. What if it's not closed under complex scalar multiplication?
The precise statement is: If $L_1/K$ and $L_2/K$ are galois extensions and we have that $L_1 \cap L_2 = K$, then the Galois group of the compositum of these extensions is the product of the individual Galois groups
@TedShifrin in the hopf fibration $S^3\to S^2$, if you take the preimages of latitudes then stereographically project, you get nested tori in $\Bbb R^3$. if you intersect these with any open half-plane along the axis in $\Bbb R^3$, you get nested circles. is there a good reason off the top of your head these are precisely the circles with prescribed center in the upper half-plane model of hyperbolic geometry?
This is something I should know off the top of my head, but I never did much with hyperbolic geometry in higher dimensions, @anon (cuz I did projective complex algebraic geometry mostly).
Have you looked at Cecil's book on Lie sphere geometry?
in general, given any collection of circles nested around a given point, there's no reason to believe all of their centers are that one single point (interpreted as circles in the upper half plane model of hyperbolic geometry), so the situation with the hopf fibration is special in that we should expect some explanation to exist
Maybe this is related to looking at the $S^3$ at infinity in $\Bbb H^3$ and thinking about it there. One of my former colleagues at UGA would know this immediately.
Another cool book you should check out for various classic geometry things, @anon, is Marcel Berger's 2-volume book Geometry I,II (meant for high school teachers in France, but there's a ton of sophisticated stuff in there).
Oh, I was wrong about that area question. My affine geometry approach was not valid.
Is that what you're recalling? I said you can obtain an ellipse from a circle by a linear mapping on the plane. Define $T(x,y) = (ax,by)$ and it sends the unit circle to the ellipse $x^2/a^2+y^2/b^2=1$.
In my text, I have the following change of variables formulas:
A function $f: \Omega \mapsto \mathbb{R}$ is integrable with respect to the induced measure $\mu (T^{-1})$ iff $f(T)$ is integrable with respect to $\mu$.
In this case, it holds that $$ \int_{\Omega}f(T(\omega))\mu(d \omega) ...
@TedShifrin if that's what it is, I'm unaware that it's called that. The Radon-Nikodym theorem appears way, way, way at the end of our text, and therefore, probably can't be used here.
@ALannister. For starters, it would help if you defined $T$ as a mapping from $\Omega_1$ to $\Omega$. And I have no idea what you're writing when you write $\mu(T^{-1}) d\mu_1. So I'm not going to try to spend hours thinking about this.
If there's no (easy) way to get the inverse of this ugly function, I'll stick with the brute force method, as I only need to calculate it to an integer accuracy.
We can write the roots as $\zeta^k$ where $0 \le k < 12$.
Then, let $r$ be an automorphism of $\Bbb Q(\zeta)/\Bbb Q$ that takes $\zeta$ to $\zeta^s$. We know that $r(\zeta^k) = r(\zeta)^k = \zeta^{sk}$.
Therefore, automorphisms of $\Bbb Q(\zeta)/\Bbb Q$ are automorphisms of $\Bbb Z_{12}$, which...
You cannot in general deduce the irreducibility of $f(X^n)$ from the irreducibility of $f(X)$.
(Consider $f$ as polynomial over $\mathbb Q[t]$ and reduce modulo the prime element $2t+1$, then we get $\bar f = X^6+X^3+1$ over $\mathbb Q[t] / (2t+1) \cong \mathbb Q$. But the irreducibility of thi...
In my text, I have the following change of variables formulas:
A function $f: \Omega \mapsto \mathbb{R}$ is integrable with respect to the induced measure $\mu (T^{-1})$ iff $f(T)$ is integrable with respect to $\mu$.
In this case, it holds that $$ \int_{\Omega}f(T(\omega))\mu(d \omega) ...
