Let $F=\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3})$. Show that over $\mathbb{Q}$ this is an extension of degree 9.  I see that this is equivalent of showing that $\sqrt[3]{2}\notin \mathbb{Q}(\sqrt[3]{3})$. I don't know any other way of showing this except going through $\sqrt[3]{2}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$ which doesn't lead me anywhere.
We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this.  To be specific, it gives us 3 diophantine equations  $$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$  and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.   So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.   Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$.
How to setup Windows from an external hard drive? Is it possible to install Windows from an external hard drive?  I have Windows XP installed on my laptop, and I want to remove it and install Windows 7.  My Windows 7 installation is in an .iso format on an external USB hard drive.  Is it possible to install it or will I need to put in a DVD before I start installing? Book set underground everyone lives in pods woman wants to see her son Can anyone help me identify this book please? I read a book in the 80s about a woman living underground in a pod. Everyone had a big tv which gave them news etc (but it was starting to go wrong) and their food was delivered direct to their pod.  They weren't allowed to go to the earth surface as it was supposedly contaminated (post-war?). She gets a message from her son on the other side of the world and she takes the tube train from outside her pod and then some sort of aircraft to see him and eventually gets onto the surface of the earth and breathes the air and meets free people. Incorrect icons on hot network questions What are the grammatical rules determining whether to use which or what?  I usually go by intuition. It's "What car are you looking at?" and "What bicycle do you like most?" but "Which song did you sing?" Quite often, I don't know whether to use what or which.  How can I tell? Help finding this set Lets define the following:  Let A be a set. A is innumerable if and only if there exists a bijective function from A to $\mathbb{N}$  Proof that there exists an innumerable set $B \subseteq \mathcal P \left({\mathbb{N}}\right)$ such that it has the following properties:  1) $X\cap Y =\emptyset$ for all $X,Y \in B$ such that $X \neq Y$  2) $\mathbb{N}= \bigcup B$  3) All elements of B are innumerable.  Any help is appreciated. I don´t know how to find this set ... Add a dependency in Maven I've recently set up the compiz cube again, and I was very happy to find that that flickering window glitch is gone! However, the cube reflection and deformation plugins are also gone.  I am running a fresh 12.10 install, and I have every compiz plugin package installed (including compiz-plugins-extra, which supposedly contains everything that compiz-fusion-plugins-extra package used to have), but those two plugins are still missing. I haven't been able to find anyone with the same problem.   Any ideas? (thanks!) Why do clocks measure arc-length? Apologies in advance for the long question.  My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$   t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.    Why is this so?     Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$.  Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$.  If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$.  If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.  I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$   T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$.  Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$   I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align}   I_{ba}  &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &amp;= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.  How do others feel about this argument?  I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.  I imagine that in GR a similar argument could be made by invoking local flatness of the metric. Drawing a constant randomly distorted circle I am trying to draw a special and contradictive plot : A circle with random distortions which have to remain constant regardless of how many times I build the document.  An example of what I am trying to achieve can be seen below    This figure was produced with the following code  \documentclass{standalone} \usepackage{tikz}    \usetikzlibrary{decorations.pathmorphing}   \begin{document}     \begin{tikzpicture}         % axes         \draw[ultra thin] (-2.5, 0) -- (2.5, 0);         \draw[ultra thin] (0, -2.5) -- (0, 2.5);         % undistorted sphere         \draw[ultra thin, dashed] (0, 0) circle (2.25cm);         \draw[ultra thin, dashed, -&gt;] (0:0) -- (125:2.25);         \node at (-1, 1) {$R$};         % distrorted sphere         \draw[very thick] plot[domain=0:350,smooth cycle ] (\x:2+rnd*0.5);         \draw[very thick, -&gt;] (0:0) -- (45:2.1);         \draw[ultra thin, -&gt;] (1, 0) arc (0:45:1);         \node at (1.2, 0.4) {$\theta$};         \node at (0.7, 1.25) {$R(\theta)$};     \end{tikzpicture} \end{document}   The problem with the code is that whenever I make a change to it and rebuild the document, the "distorted" circle will have a different shape, since I am using the rnd function. The reason I would like to have a constant-random shape is because I would like to denote its radius at an angle θ, therefore it has to touch the outline of the circle.  Any idea on how to achieve such a thing? Images for Area51 proposal community promotion on imgur.com? For each proposal on Area51, a promotional image is automatically generated and it shows the current state, e.g. number of followers/supporters or comitted. Example:    They’re obviously useful for  on existing sites. Alas, the current requirements for those say that the image needs to be hosted on imgur.com. I can easily provide the URL https://area51.stackexchange.com/ads/proposal/{proposal-number}.png in the image upload dialog, to get a cryptic imgur address back, but I’m afraid that it won’t update dynamically.  Please note that I was able to use the area51 URL on some sites until right now, when the poster generator seems to be broken or off-line, but on other sites the software stopped me from posting any image not hosted on imgur. Can we please have the dynamic posters working everywhere (and in “HiDPI”) or be assured that the external service will update them regularly? If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.  The question can be simplified to:  Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.  As $1+n^2+n^4$ forms a GP, $$t_n=\frac{n(n^2-1)}{n^6-1}\,.$$ But I can't figure out how to solve further. It would be great if someone could help. Has the concept of non-integer $(n+m)$-dimensional spacetime ever been investigated by theoretical physicists? The following image serves to aid the reader in understanding the "privileged character" of $3+1$-spacetime.    The wikipedia article on , and the  "The priveleged character of $3+1$-dimensional spacetime" in particular, made me think a bit about the possibility that we might live in a non-integer amount of spatial and/or time dimensions.   The notion of attaching a non-negative real number to a metric space has at least mathematically already been described by such concepts as "" and "".   This may sound silly/ignorant/absurd to professional practicing theoretical physicists. To me (a layman), however, it doesn't sound much stranger than the idea of wrapping up six extra dimensions (which is, from what I understand, considered to be a serious possibility by those who study $10$-dimensional String Theory) into intricate shapes called "Calabi-Yau Manifolds".   Has any research on $(a +b)$-dimensional spacetime (where $a,b \in \mathbb{R}_{\geq 0} $) ever been done? If so, what where the findings? If not, why not? I was reading the following article by Mathew Might.  I saw the following the paragraph  What doesn't matter GPA? I don't care if it's 2.0 or 4.0. I won't even look at it. The school you went to? I'll judge you the same whether you went to Nowhere State U or a top-ten school. Transcripts? Never seen one. GREs? Irrelevant. Where you work/worked? Unless it's a research lab, it's not important. I don't think these items have much predictive capacity as to whether or not someone can complete a Ph.D.  Is Mathew Might telling the truth? For me it's very hard believe that GPA, GRE scores and undergraduate school don't matter. I used to think that they play an extremely important role in grad school admission. How to represent it with UNIQUE indexes and foreign keys? Is "laic" an alternative word to the word "layperson"? Ubuntu 20.04 Font and UI elements too big on login I'm running Ubuntu 20.04 on an ASUS X99-E-10G WS w/ an NVIDIA 1080Ti and an LG 4k 32&quot; monitor which persistently gets recognized as 27&quot;. When I log in, the text of certain UI elements is ~2x the normal size. After I go to settings/universal access and turn large text on and then off again, everything is the normal size I expect. Screenshot after logging in  After turning &quot;large text&quot; on  After turning &quot;large text&quot; off  How do I get the conditions I have after cycling the &quot;large text&quot; feature to be present immediately after booting? I have the 440 version of NVIDIA's drivers installed. How do I get past the "do not enter" traffic sign that stops me from installing "Sierra" instead of "High Sierra"?  I booted into rescue mode and deleted the original "High Sierra" installation, but booting from the USB stick with the "Sierra" installer on it still gives me the "no". Can I remove a digital signature from a DLL? Why does a minimal prime ideal consist of zerodivisors? Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.  This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)  Can this be proved without using localization? Infinite algebraic extension of a finite field I have recently started studying algebraic field extensions and I got to know that algebraic closures $\overline{F}$ of finite fields $F$ are infinite. Therefore, I've asked myself the following question and couldn't come up with an answer.  Suppose $F$ was a finite field, i.e. $|F| &lt; \infty$. Is it possible to construct an algebraic extension $F'$ of $F$ sucht that $|F'| = \infty$ and $F' \subsetneq \overline{F}$? Meaning, that $F'$ is not algebraically closed.