For a finite commutative ring $R$, $a \in R$ is a root of $p(x)$ iff $p(x)$ can be written as $p(x) = (x-a)g(x)$
Prove that $p(x)=(x-a)q(x)$ for some $q(x)\in R[x]$ of degree= (degree of $p)-1$.
Collection of short stories; in one a spy attempts to escape detection on Mars's moon by moving faster than the spaceship can scan Why does the collada exporter not export texture references? I've been having trouble exporting collada objects from blender, so I stripped everything back and textured a single cube, and exported it.  This is a completely trivial model, one cube, with one material assigned to it, the material has a single 'g.png' texture assigned to it.   The exported collada data looks like this:    &lt;library_materials&gt;     &lt;material id="Material-material" name="Material"&gt;       &lt;instance_effect url="#Material-effect"/&gt;     &lt;/material&gt;   &lt;/library_materials&gt;   Where:    &lt;library_effects&gt;     &lt;effect id="Material-effect"&gt;       &lt;profile_COMMON&gt;         &lt;technique sid="common"&gt;           &lt;phong&gt;             &lt;emission&gt;               &lt;color sid="emission"&gt;0.2879998 0.2879998 0.2879998 1&lt;/color&gt;             &lt;/emission&gt;             &lt;ambient&gt;               &lt;color sid="ambient"&gt;0 0 0 1&lt;/color&gt;             &lt;/ambient&gt;             &lt;diffuse&gt;               &lt;color sid="diffuse"&gt;0.2163267 0.1845119 0.2121569 1&lt;/color&gt;             &lt;/diffuse&gt;             &lt;specular&gt;               &lt;color sid="specular"&gt;0.5 0.5 0.5 1&lt;/color&gt;             &lt;/specular&gt;             &lt;shininess&gt;               &lt;float sid="shininess"&gt;50&lt;/float&gt;             &lt;/shininess&gt;             &lt;index_of_refraction&gt;               &lt;float sid="index_of_refraction"&gt;1&lt;/float&gt;             &lt;/index_of_refraction&gt;           &lt;/phong&gt;         &lt;/technique&gt;         &lt;extra&gt;           &lt;technique profile="GOOGLEEARTH"&gt;             &lt;double_sided&gt;1&lt;/double_sided&gt;           &lt;/technique&gt;         &lt;/extra&gt;       &lt;/profile_COMMON&gt;       &lt;extra&gt;&lt;technique profile="MAX3D"&gt;&lt;double_sided&gt;1&lt;/double_sided&gt;&lt;/technique&gt;&lt;/extra&gt;     &lt;/effect&gt;   &lt;/library_effects&gt;   This seems to be missing the texture data associated with the object.   By comparsion, exporting a wavefront object generates this .mtl file:  # Blender MTL File: 'basic.blend' # Material Count: 1  newmtl Material Ns 96.078431 Ka 0.000000 0.000000 0.000000 Kd 0.790000 0.790000 0.790000 Ks 0.500000 0.500000 0.500000 Ni 1.000000 d 1.000000 illum 0 map_Kd g.png   &lt;----- ie. Blender is aware of this texture   So... is the collada exporter for blender just broken, or does it require some kind of special config in order to function correctly?  If the latter, how/what is required to make this work?  (NB for what it's worth, the UV coordinates are exported in the .dae, abit they obviously are completely useless without a texture assigned to them; I've also tried adding two different textures to see if the exporter needs that because it naively assumes 'one texture = don't bother, it's one texture for the whole model'. That didn't work either) This is Spivak's Calculus on Manifolds ex. 2-25, he says     Define $f:\mathbb{R}\to \mathbb{R}$ by $f(x) =  \left\lbrace   \begin{array}{l}      e^{-x^{-2}} &amp;\text{ if } x \neq 0\\      0 &amp;\text{ otherwise } \\   \end{array}   \right.$. Show that $f$ is $C^{\infty}$ anf $f^{(i)}(0)=0\; \forall i$.   I haven't found a way to get a general solution. I only managed to prove:  $(1)$ $f$ is continuous at $0$.  This follows from $f(0)=0$ and $\lim_{x\to0} f(x)=\lim_{x\to 0}e^{-x^{-2}}=\lim_{x\to 0}\displaystyle\frac{1}{e^{1/x^2}}=0$ since $\displaystyle\frac{1}{x^2}\to+\infty$ if $x\to 0$ then $e^{1/x^2}\to+\infty$ if $x\to 0$, which means that $\displaystyle\frac{1}{e^{1/x^2}}\to 0$ if $x\to 0$ .  $(2)$$f'(0)=0$.  I got this by calculating the limit $f'(0)=\lim_{h\to 0}\displaystyle\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\displaystyle\frac{f(h)}{h}=\lim_{h\to 0}\displaystyle\frac{e^{-h^{-2}}}{h}=\lim_{h\to 0}\displaystyle\frac{1/h}{e^{h^{-2}}}$.  The last one follows from L'Hôpital: $\lim_{h\to 0}\displaystyle\frac{\frac{d}{dh}(1/h)}{\frac{d}{dh}(e^{h^{-2}})}=\lim_{h\to 0}\displaystyle\frac{(-1/h^2)}{(e^{h^{-2}})(-2h^{-3})}=\lim_{h\to 0}\displaystyle\frac{h}{2e^{h^{-2}}}=\displaystyle\frac{1}{2}\left(\lim_{h\to 0}\displaystyle\frac{1}{e^{h^{-2}}}\right)\left(\lim_{h\to 0} h\right) = 0$.  $(3)$ $f''(0)=0$  I found that $\displaystyle\frac{df}{dx} =  \left\lbrace   \begin{array}{l}      2x^{-3}e^{-x^{-2}} &amp;\text{ if } x \neq 0\\      0 &amp;\text{ otherwise } \\   \end{array}   \right.$, then  I can get $f''(0)$ computing the limit $\lim_{h\to 0}\displaystyle\frac{f'(0+h)-f'(0)}{h} = \lim_{h\to 0}\displaystyle\frac{f'(h)}{h} = \lim_{h\to 0}\displaystyle\frac{2e^{-h^{-2}}}{h^4}= 2\lim_{h\to 0}\displaystyle\frac{1/h^4}{e^{h^{-2}}}=-2\lim_{h\to 0}\displaystyle\frac{-4h^{-5}}{-2h^{-3}e^{h^{-2}}}\\=4\lim_{h\to 0}\displaystyle\frac{1}{h^2e^{h^{-2}}}=4\lim_{h\to 0}\displaystyle\frac{h^{-2}}{e^{h^{-2}}})=-4\lim_{h\to 0}\displaystyle\frac{-2h^{-3}}{e^{h^{-2}}(-2h^{-3})}=-4\lim_{h\to 0}\displaystyle\frac{1}{e^{h^{-2}}}=0$, after using L'Hôpital twice.  But still I don't have a clue to prove that $f^{(i)}(0)=0$ for every $i$. One of the ideas I managed was to find a general form for $f^{(i)}$ but after I got $f'$ doesn't seem possible to do so considering that I would have to apply the product rule each time. $f_x$ is Borel measurable and $f^y$ is continuous then $f$ is Borel measurable Ok, I just bought a brand new original iPad. Sounds impossible, right? Well, we just opened the box, and it is not even charging. There is no indication that this iPad even works even though it is technically "new." Is this something that Apple will need to fix or do I just need to let it sit on the charger overnight and hope it wakes up from a 2 year sleep in its box? How do I load the contents of a text file into a javascript variable? I have a text file in the root of my web app  and I'd like to load it into a variable in javascript.. in groovy I would do this:  def fileContents = 'http://localhost/foo.txt'.toURL().text; println fileContents;   How can I get a similar result in javascript? I have to find an easy ring which this ring is isomorphic with.  I can prove that $5 = (2+i)(2-i) \in (2+i)$ but this doesn't help me further.  My book says the answer is $\mathbb {Z_5} $ A generalisation of the fundamental theorem of algebra We already know the following theorem by d'Alembert and Gauss, often called fundamental theorem of algebra.  Theorem Let $P$ be in $\mathbb C[X]$ of degree $1$ or greater. There exists $\alpha\in \mathbb C$ such that $P(\alpha)=0$.  Can we give the following generalisation for polynomials with several variables?  Generalisation Let $n$ be in $\mathbb N^*$ and $P$ be in $\mathbb C[X_1,\ldots,X_n]$ of degree $1$ or greater. There exists $\alpha\in \mathbb C^n$ such that $P(\alpha)=0$.  Any references, hints or solutions would be greatly appreciated. What do the different colors mean in ls? What do the different colours in Ubuntu's ls command mean? For example, when I type the ls command in one of my folders, I get one of the files in light green, the other (which is a folder) in blue with green highlighting.   What do those colours mean, and there is any manual about all the colours? For usability reasons, for example in , the 'viewed' counter should show as  ...instead of... . It's much harder to decipher that there are a million views on the question just by looking at it.  Some people in the comments pointed out internationalization concerns with it. I do think that since there might be a problem with it, that spaces would probably be a better idea. Set of left cosets is a group Let $N$ be a normal subgroup of $G$. Let $G/N = \{gN : g \in G \}$. Show that the $(G/N, \circ, 1_{G/N})$ is a group, where:  $\circ : G/N \times G/N \rightarrow G/N$ is defined by $(gN \circ hN) = ghN$ and $1_{G/N} = 1*N$.  Thanks for your help. How to loop the particle simulation indefinitely? Should a "thank you" comment, for instance Thanks for your answer, be flagged as non-constructive?  If a "thank you" comment is flagged as non-constructive what would the moderator reaction to this flag be? Would it be marked as helpful, or not? "Dummy Output" No sound in Ubuntu 16.04 None of the HDMI output are working, all saying "dummy output", it was working yesterday, but when I use the headphones it became soundless, with or without the headphones.  Thank you already! Normal subgroup of prime index Generalizing the case $p=2$ we would like to know if the statement below is true.  Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal. How can I find unanswered questions with zero answers in some tag? I'm looking at buying a second-hand Mac, probably one that does not include an OS installation disk.  How can I scan it to make sure that it has not had security threats installed on it?  Is there a CD or DVD image I can boot it up with to perform this scan offline?  Edit: prospective machine would be a 2011- or 2012-vintage MacBook Pro.  Also note that I don't have reliable access to high-speed internet, so solutions that involve downloading 5GB over the Internet are not ideal.  For others, though,  may be helpful. Print LIST of unicode chars without escape characters Cannot connect to internet after using airmon-ng check kill In new version of Kali 2.0 with updated "aircrack-ng" tool, there is a problem between Internet connection and using cracking tools.   I have my wireless card as wlan0. And the Internet connection is fine too. Then, what I do is:  airmon-ng start wlan0 airodump-ng wlan0mon   Now the airodump-ng doesn't work, saying "Device or resource busy".  In order to run this I have to kill some processes, using:  airmon-ng stop wlan0 airmon-ng check kill airmon-ng start wlan0 airodump-ng wlan0mon   Now the cracking process is up.   I can start cracking, but now I cannot connect to the Internet after the kill process.  On typing iwconfig in terminal, I get this:  root@kali:~#iwconfig wlan0mon IEEE 802.11bgn  ESSID:"myessid"         Mode:Managed  Frequency:2.412 GHz  Access Point: 00:00:00:00:00:00          Bit Rate=108 Mb/s   Tx-Power=20 dBm          Retry short limit:7   RTS thr:off   Fragment thr:off       Encryption key:off       Power Management:off       Link Quality=42/70  Signal level=-68 dBm         Rx invalid nwid:0  Rx invalid crypt:0  Rx invalid frag:0       Tx excessive retries:2  Invalid misc:9   Missed beacon:0  eth0      no wireless extensions. lo        no wireless extensions.   And, on running ifconfig on terminal, I get this:  root@kali:~#ifconfig lo        Link encap:Local Loopback         inet addr:127.0.0.1  Mask:255.0.0.0       inet6 addr: ::1/128 Scope:Host       UP LOOPBACK RUNNING  MTU:65536  Metric:1       RX packets:5604 errors:0 dropped:0 overruns:0 frame:0       TX packets:5604 errors:0 dropped:0 overruns:0 carrier:0       collisions:0 txqueuelen:0        RX bytes:452904 (442.2 KiB)  TX bytes:452904 (442.2 KiB)   There is no wlan0 or wlan0mon. I also restarted the network manager by:      service network-manager restart   But nothing changed. Though it displays wlan0 when checked using ifconfig this time. Nicely discouraging serial upvoting Probably everybody who is very active on SO has experienced cases of random, serial upvoting as the expression of thanks from a (usually newbie) user. The pattern is usually 5 or more upvotes on your latest answers. The upvotes will be detected within 24 hours, and removed. (Or so I assume.)  No harm done except maybe a minor irritation for people after one of the &quot;rep cap&quot; badges because they are shown a higher reputation than they actually have. Still, I was thinking whether, in an attempt to channel the positive attitude into more productive areas, it would be worth showing a tooltip on the third upvote for the same person within time frame x:  Please use one upvote, a comment, or some quality content of your own, to say thank you Large numbers of upvotes for the same user may become invalid as part of our spam detection mechanisms.  Does this make sense? Maybe it is too localized and rare, but it has happened to me three or four times in the last two weeks alone. You decide.  Clarification: I am not advocating putting up any kind of barrier on upvotes. . I'm suggesting informing people of the barrier and the futility of their ten or so upvotes for a single person because those votes are likely to be revoked.  Update: @devinb makes the good point that &quot;saying thanks&quot; is not the only possible reason for serial upvoting, and the message wouldn't make sense then. The point still stands: Your votes aren't going to count, and when it comes to up voting (which is always some sort of positive behaviour, however misguided if done indiscriminately) I think it's fair to inform people of this.