Let $F$ be a field. If $F^*$ is a cyclic group, must $F$ be a finite field? (It's well-known that if $F$ is a finite field, $F^*$ is a cyclic group).  Thank you in advanced.
Let $F$ be a field and $F^\times$ be its group of units. If $F^\times$ is cyclic, then show that $F$ is finite.  I'm a bit stuck. I know that I can represent $F^\times = \langle u \rangle$ for some $u \in F^\times$ and that we must have that $|F^\times| = o(u)$, where $o(u)$ denotes the order of $u$ in $F^\times$.  I tried assuming $o(u) = \infty$, but I'm not sure exactly where to go from there. I was wondering if I could get a hint.
GPG can't check signature How long does the ubuntu installation take and what to do after it? Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$ Error propagation with asymmetric uncertainties I red the Wikipedia  on error propagation. I get into some trouble when I want to calculate the error propagation in a specific case. I have real numbers $A$ and $B$ with related upper ($\Delta A_u$, $\Delta B_u$) and lower ($\Delta A_l$, $\Delta B_l$) uncertainties. I have two functions calculated as follows: $$F = \sqrt{A^2+B^2}$$ and  $$G = \tan^{-1}\frac{B}{A}$$ What are the related upper and lower errors of $F$ and $G$?  I found general formulae for Gaussian error distributions ($\Delta A_u = \Delta A_l$ and $\Delta B_u = \Delta B_l$), for instance the first formula of the first answer on . Is there a way to generalize that formula for arbitrary distributions with asymmetric errors as in my example? Identifier for win64 configuration in Qmake Is there a "win64" identifier in Qmake project files?  documentation does not mention other than unix / macx / win32.  So far I've tried using:  win32:message("using win32") win64:message("using win64") amd64:message("using amd64")   The result is always "using win32".  Must I use a separate project-file for x32 and x64 projects, so they would compile against correct libraries? Is there any other way to identify between 32-bit and 64-bit environments? I installed Ubuntu 14.04.1 on one of my laptops and now I want distribute that exact installation to a few of my friends. I tried making an iso with Gparted on a live session but it makes a 500 GB file (I've only used around 8 GB) .   Thanks in Advance ! Uniform vs discrete Gaussian sampling in Ring learning with errors The Wikipedia article on  mentions two methods of sampling "small" polynomials namely uniform sampling and discrete Gaussian sampling. Uniform sampling is clearly the simplest, involving simply uniformly selecting the coefficients from the set of small coefficients, thereby guaranteeing they will be small. However, the articles mention that security proofs rely on discrete Gaussian sampling.  However, I have a bit of difficulty wrapping my head around    Why discrete Gaussian sampling would be better than uniform, security-wise?   I mean, usually uniform is preferable - it maximizes the entropy/information in each sample, making the outcome more unpredictable. I could understand it if the Gaussian sampling technique would somehow serve to avoid (at least with high probability) "weak" parts of the space of outcomes but I have found no information/justification that this is the case.  Another reason I can see why discrete Gaussian might be better is that there is, after all, a chance (albeit a small one) one would select a non-small coefficient, so the space of outcomes is actually larger. However, given that this probability of hitting this extra space is (by design) quite small but also that one would likely immediately reject such a sample (because the algorithm calls for a small polynomial), this seems to be irrelevant.  It is not because I am against discrete Gaussian sampling I just want to understand why it is more secure. Also, I want to understand if it really IS more secure or it is simply easier to prove security in that model. I would actually consider that very interesting if it is the case because it raises the question of why is it easier to prove the security under conditions that are intuitively worse (i.e. discrete Gaussian vs the intuitively better uniform case)? If it is so intuitively clear that uniform sampling is better, surely it must be possible to transform that intuition into a security proof of RWE also under uniform sampling. Otherwise, I would fear the intuition must with high likelihood be wrong or at least dubious. I recently installed Lubuntu but I have a problem with Adobe Flash Player. I always see something that looks like this when I open a video:    (source: )    The video is small, grainy, and is green and purple.  I tried to reinstall Flash and to test it in multiple browsers without success. I installed an older version of Flash but no result.  [0300]: Intel Corporation 82865G Integrated Graphics Controller [8086:2572] (rev 02)     Subsystem: Hewlett-Packard Company D530 sff(dc578av) [103c:12bc]     Kernel driver in use: i915 I'm at roughly 67% completion of the Alola Dex, with ~200 caught, ~290 seen. After scanning QR codes, I've caught a Swinub and Cyndaquil. However, neither were registered in the Pokédex, since neither are Pokémon in the Alola Dex. This brought up the following question:  When will I unlock the National Dex?   Do I need to have seen all the Pokémon in the Alola Dex? Do I need to complete some specific quest, like catching/defeating the Ultra Beasts? An element not in a field extension Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this? Steward badge can now be awarded multiple times -- backfill details As we announced in our post on  to the Review Queues, the Steward badge can now be earned more than once per queue. The purpose of this post is not to debate the merits of this decision ( already exists), but to explain the nuts and bolts of how it will be backfilled. This is how it will appear on your profile if you have earned it multiple times:  Once we push the visual design changes live across the network, this is how badge attainment will look on /review/close/stats:  For those of you who have already completed enough reviews to earn multiple Steward badges, we will be retroactively awarding them to you, starting today. The backfill script will run about twice an hour, and your achievements trophy will light up as each one is awarded. For those of you who have done thousands and thousands of reviews, things may be a bit noisy, so we thank you in advance for your patience. It may take up to two days for all of your badges to show up. When you cast invisibility on your turn, and walk away from 2 enemies, do they get an attack of opportunity? When a PC is in melee with a creature, casts the basic invisibility spell, and leaves melee, does an NPC get an attack of opportunity? Removing spaces between "words" in math mode Intuitive reason why the Fisher Information of Binomial is inversely proportional to $p(1-p)$ It confuses/blows my mind that the Binomial has variance proportional to $p(1-p)$. Equivalently, the Fisher information is proportional to $\frac{1}{p(1-p)}$. What is the reason for this? Why is the Fisher Information minimized at $p=0.5$? That is, why is inference most difficult at $p=0.5$?  Context:  I'm working on a sample size calculator, and the formula for $N$, the sample size needed, is an increasing factor of $p(1-p)$, the result of a variance estimation in the derivation. Properties of p-adic valuations Let $n$ be a integer not equal to zero, $p$ a prime and $$|n|_p = p^{-\operatorname{ord}_p(n)}; |0|_p=0$$  Proposition: $|xy|_p=|x|_p|y|_p$  We want to show: $\operatorname{ord}_p(ab)=\operatorname{ord}_{p}(a)+\operatorname{ord}_p(b)$ So if we suppose there is a largest integral $v$ such that $p^{v}|(ab)$, then we can split up v such that $p^{v_1}|a,p^{v_2}|b$. If we suppose that there exist $v_1,v_2$ with them both being the largest integers such that $p^{v_1}|a, p^{v_2}|b$,  with $v_1+v_2= : v $ then $p^v|ab$. (I don't believe this to be a formal proof but rather my best own "idea" I can come up with...)  The script also mentions that the padic valuations' sums are bounded: $|x+y|_p\le \sup\{|x|_p,|y|_p\}$, how to show this?   Edit:   According to Robert Israel and Bill Dubuques answers, the first proposition is proven, if I understood correctly:   Given $ord(a)= v$ and $ord(b) = u$ which is the same as $a:= p^{v}m, b:= p^{u}n$ with $gcd(m,p)=gcd(n,p)=1$, because $ab=p^{u+v}mn$ with $gcd(mn,p)=1$ it follows that $ord(ab)=ord(a)+ord(b)$ and so $|xy|_{p}=p^{-ord(xy)}=p^{-(ord(x)+ord(y))}=|x|_{p}|y|_{p}$  Now to show: $|x+y|_{p}\le sup\{|x|_{p},|y|_{p} \}$  and the second can be tackled like this: set $a:=p^{v}m. b:=p^{u}n$, then checking two cases $1:(u=v)$ and $2:(v&lt;u)$: $(a+b)=p^{v}m+p^{u}n= p^{v}(m+p^{u-v}n)$ so in case of (u=v) it follows that  $ord(a+b)= u = sup\{ord(a),ord(b)\}$  and in case of $(v&lt;u)$ because the biggest v is $v:= u-1$: $ord(a+b)\le p^{u-1}(m+pn) \le p^{u}(m+n)= sup\{ord(a),ord(b)\}$  It seems that I did something wrong in both cases... Proof that if $p\equiv3\,\left(\mbox{mod 4}\right)$ then $p$ can't be written as a sum of two squares I'd appreciate your help showing that if $p\equiv3\,\left(\mbox{mod 4}\right)$   then p can't be written as a sum of two squares.  Thanks! For a lot of questions here on TeX.SE and on other fora regarding MiKTeX, it turns out that the underlying problem actually lies in the not understood distinction between administrative and user modes. So let me ask these questions:  What’s the difference between administrative and user mode in MiKTeX, and in which situation should I use either one of these modes? Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$   I tried to prove it by induction  but I got a contradiction. How can I get a shapefile from a postgis query? My data is in a PostGIS database. I would like to generate a shapefile from a query. How can I do that? Unable to login after password change, having home directory encrypted