In Hartshorne, ex I.2.12 we see that the image of the d-uple embedding, $\varphi_d$ is exactly $Z(Ker \theta)$ where $\theta: K[y_0,\dots,y_N]\rightarrow K[x_0,\dots,x_n]$ is defined by sendind $y_i$ to $M_i$ where $M_0,\dots, M_N$ are all the monomials of degree d in $x_0,\dots x_n$). I was wondering: what is exactly $Ker\theta$? Reading online I found this: Let $w_I$ where $I=(i_0,\dots,i_n)$ such that $i_0+\dots+i_n=d$, be the homogeneous coordinates of $\mathbb{P}^N$ the image of $\varphi_d$ is $Z(w_Iw_J-w_Kw_L | I+J=K+L)$ here $I+J=(i_0+j_0,\dots,i_n+j_n)$. So if I rewrite $\theta: K[\{w_I\}]\rightarrow k[x_0,\dots,x_n]$ sending $w_I$ to $x_0^{i_o}\dots x_n^{i_n}$ Is $Ker\theta=(w_Iw_J-w_Kw_L | I+J=K+L)$? Of course $(w_Iw_J-w_Kw_L | I+J=K+L)\subseteq Ker\theta$. I'm having some trouble with the other inclusion: let $f\in Ker\theta$ and let $M=w_{I_1}\dots w_{I_n}$ a monomial of degree $n$ appearing in $f$ with some non zero constant coefficient. Since $\theta(f)=0$, there must be another monomial $M'$ of degree $n$ appearing in $f$ with the same coefficient multiplied by $-1$ such that $M'=w_{J_1}\dots w_{J_n}$ and $I_1+\dots+ I_n=J_1+\dots+J_n$. I would like to show that $M-M'\in (w_Iw_J-w_Kw_L | I+J=K+L) $. By induction on $n$: If n=2 then $M-M'$ is exactly a generator of $(w_Iw_J-w_Kw_L | I+J=K+L)$. Assume $n&gt;2$ I would like to write $M-M'$ as a combination of $(N-N')$, monomials of degree $n-1$, and generators of $(w_Iw_J-w_Kw_L | I+J=K+L)$ but I don't know how to do this. I don't know if it is the right way...
I am trying to mimic the interesting proof from  to give a generating set of the ideal correspond to the $d$-uple embedding: We fix the notation $\Delta = \big\{ (\nu_0, \dots, \nu_n) \in {\mathbb{N}}^{n+1}_0 ~:~ \nu_0 + \dots + \nu_n = d \big\}$. The polynomial map is $\theta : k[y_\nu : \nu \in \Delta] \rightarrow k[x_0, \dotsc, x_n]$ given by $$ \theta ( y_\nu ) := x^{\nu_0}_0 \cdots x^{\nu_n}_n $$ and we know that ${\mathrm{ker}}~\theta$ is the ideal of the $d$-uple embedding, i.e., $\rho_d({\mathbb{P}}^n) = Z({\mathrm{ker}}~\theta)$ (cf. Exercise I.2.12 Hartshorne). Now let $W \leq k[y_\nu : \nu \in \Delta]$ denote the ideal $$ W := \Big\langle y_{\tau_1} y_{\tau_2} - y_{\tau^{\prime}_1} y_{\tau^{\prime}_2} ~:~ \tau_1, \tau_2, \tau^{\prime}_1, \tau^{\prime}_2 \in \Delta, \tau_1 + \tau_2 = \tau^{\prime}_1 + \tau^{\prime}_2 \Big\rangle $$ Clearly, $W \subseteq {\mathrm{ker}}~\theta$. Using the lemma in the above link, we need to produce a $k$-subspace $T \subseteq k[y_\nu : \nu \in \Delta]$ with the conditions $$ T + W = k[y_\nu : \nu \in \Delta], \hspace{.2in} \theta \lvert_{T} = {\mathrm{injective}} $$ After a few thoughts it seems reasonable to think of the following subspace as a candidate for $T$: First define the equivalence relation on $\Delta \times \Delta$ given by $(\tau_1, \tau_2) \sim (\tau^{\prime}_1, \tau^{\prime}_2)$ if $\tau_1 + \tau_2 = \tau^{\prime}_1 + \tau^{\prime}_2$. Let $\Omega \subseteq \Delta \times \Delta$ be a subset with a unique representative from each class of $\sim$. Now let $T$ be the $k$-span of the monomials $y^{\alpha_1}_{\tau_1} \dotsc y^{\alpha_r}_{\tau_r}$ (with the distinct elements $\tau_1, \dotsc, \tau_r \in \Delta$ and $\alpha_1, \dotsc, \alpha_r \geq 1$) such that $(\tau_i, \tau_j) \in \Omega$ for each $i \neq j$. Clearly, $\theta$ is injective if restricted to $T$. I got lost in the step : Each monomial $M$ of $k[y_\nu : \nu \in \Delta]$ satisfy $M \equiv b$ mod $W$ for some $b \in T$. Did I do anything wrong?
́╗┐' is not recognized as an internal or external command When executing certain files (mainly batch) using PsExec I get these weird ́╗┐' symbols before my command. My initial thought was, that I were using wrong encoding, but after checking, I realized that all my files were using UTF-8. Can every game of Klondike-Solitaire be solved? I play a lot of Solitaire games on my Android phone and love to keep an eye out for the statistics.  Given that the Solitaire version lets you restart the game endlessly, I usually play until I solve it. But I never managed to solve more than 80% of the games played (1000+).  So now I wonder, is every Solitaire game solvable? fitch style proof problem regarding rule of inferences Okay I have the following fitch proof almost figured out, from here I know I have to move forth in conlcuding but I don't know what I am missing  (2 premises)  S → (R ∨ P)  P → (¬R → Q)   my work below  new scope assume S   Rule: > elim  R ∨ P  New scope assume R  Q∨R     Rule: V intro  End scope – (new scope assume P)  ¬R → Q         Rule: > Elim  my confusion comes where I go from here  Q∨R            rule Taut Con ??  Goal:├ S → (Q ∨ R) What's the name for the relativistic paradox with the train/car travelling over a gap? There is a relativistic paradox I was introduced by a friend in Ithaca (Courtenay Footman). It involves a vehicle travelling over a ravine relativistically, so it's in the general family of ladder paradoxes. There are many ways to phrase it, but here's a simple one.   A three-car train with each car of length $L$ is travelling over a bridge across a ravine. (Assume that the couplings between cars are stiff: this train does not bend.) Unfortunately, some evil genius has precisely removed a length $6L$ of the track. The train however is relativistic, travelling at a speed such that $\gamma \approx 12$.  In the reference frame of the ground, the train should appear to be 1/24th the size of the gap, which suggests that it will fall a little and should crash into the bridge opposite. But in the reference frame of the train, each car of the train is at least twice as long as the gap and aside from the very front wheel and the very back wheel, which need to be supported by rigidity, everything is constantly well-supported. This suggests that the train makes it. Who is right?  I vaguely remember that as an undergraduate knowing only a little relativity, I trusted the train's perspective because it's at rest in that perspective, and Courtenay said that this was ultimately correct after a lot of details got resolved. But I do not remember all of these details such that I could work it out for myself. So I'm looking for some discussion that's a little more well-sourced or authoritative.  What is this paradox called, and what literature has addressed the issue? How Can I View All The Reputation Gained/Lost For One Specific Question If I would like to view all of the reputation I have gained or lost due to one specific question of my own, how can this be done?  Is it possible for someone with, say, 421 rep on the site in question? Six simple tips to get Stack Overflow reputation fast Is "remove" a reserved keyword in Google Chrome? What are your thoughts on epigraphs in theses? I am wondering if there is a general consensus as to the suitability of epigraphs in a thesis, either a single one at the start of the document, or an appropriate quote to begin every chapter.  For example, at the start of a technical chapter describing some code, I could write     Beware of bugs in the above code; I have only proved it correct, not tried it. - Donald E. Knuth   I am lucky to have complete freedom over how I choose to typeset my thesis (Master's thesis in astronomy) and I like the idea of a thematic quote to begin a chapter. However, I am wary of seeming unprofessional or tacky so I am curious to get some more thoughts on the matter. How to remove zsh alias? I have an alias which is predefined by an oh-my-zsh plugin.  -&gt; % alias gcm gcm='git checkout master'  I'd like to remove it, i.e. I'd like alias | pcregrep "\bgcm\b" return 1. I've tried with alias gcm='' but after that the alias is still existent. Take the following function:  DataTable go() {     return someTableAdapter.getSomeData(); }   When I set a breakpoint in this function, is there a possibility to inspect the returned value? go() is directly coupled to a datagrid in an .aspx page.  The only way to inspect the returned datatable is to use a temporary variable. However, that's a bit inconvenient. Isn't there another way? Android backup/restore: how to backup an internal database? From what items can a wizard learn his spells? I'm now wondering after googling for a while, what are the different sources a Wizard can learn his spells from? Is it only limited to scrolls and tomes, or can a Wizard learn spells from other magic items?  The following quotes are from the PHB, page 114 Your spellbook      You might find other spells during your adventures. You could discover a spell recorded on a scroll in an evil wizard's chest, for example, or in a dusty tome in an ancient library.      Copying a Spell into the Book   When you find a wizard spell of 1st level or higher, you can add it to your spellbook if it is of a level for which you have spell slots and if you can spare the time to decipher and copy it.      For each level of the spell, the process takes 2 hours and costs 50gp. The costs represents material components you expend as you experiment with the spell to master it, as well as the fine inks you need to record it. Once you have spent this time and money, you can prepare the spell just like your other spells.   The way this is worded and considering the few examples we're given, shouldn't it be possible to learn a spell from more than just written sources?  An example would be Wand of Magic Missile: would it be possible for a Wizard to learn the spell Magic Missile by studying the spell when used by the item and write down his research and thus learning how to use the spell on his own and eventually master it? If I wave around a bar magnet, the magnetic field in the space around it changes. Is this enough to go through the whole speed of light derivation implying that the motion creates an electromagnetic wave? If so how do I determine the wavelength and direction of the resulting wave? Hello,  I'm getting this error every time when I boot my computer. My Ubuntu installation is fresh and I was got a few error on last step while install Ubuntu (because i have problem with my CDROM driver). Anyway, installation is continued .   For example when i tried to open ADDITIONAL DRIVER and LANGUAGE SUPPORT (on System Settings) nothing happened, it's not opening.  When i try this in terminal gnome-language-selector i'm getting this error  Traceback (most recent call last):   File "/usr/bin/gnome-language-selector", line 8, in &lt;module&gt;     from LanguageSelector.gtk.GtkLanguageSelector import GtkLanguageSelector   File "/usr/lib/python2.7/dist-packages/LanguageSelector/gtk/GtkLanguageSelector.py", line 34, in &lt;module&gt;     from aptdaemon.gtk3widgets import AptProgressDialog   File "/usr/lib/python2.7/dist-packages/aptdaemon/gtk3widgets.py", line 61, in &lt;module&gt;     class AptStatusIcon(Gtk.Image):   File "/usr/lib/python2.7/dist-packages/aptdaemon/gtk3widgets.py", line 66, in AptStatusIcon     def __init__(self, transaction=None, size=Gtk.IconSize.DIALOG): AttributeError: type object 'GtkIconSize' has no attribute 'DIALOG'   Also getting this error for jockey-gtk  Traceback (most recent call last):   File "/usr/bin/jockey-gtk", line 418, in &lt;module&gt;     sys.exit(u.run())   File "/usr/lib/python2.7/dist-packages/jockey/ui.py", line 461, in run     self.ui_show_main()   File "/usr/bin/jockey-gtk", line 81, in ui_show_main     col_icon.set_sizing(Gtk.TreeViewColumnSizing.AUTOSIZE) AttributeError: type object 'GtkTreeViewColumnSizing' has no attribute 'AUTOSIZE'   I'm worrying about are there any lack . I don't know what's wrong with my installation.  Are there any way, check my installation (packages, files etc..) . How do I increase the open files limit for a non-root user? This is happening on Ubuntu Release 12.04 (precise) 64-bit Kernel Linux 3.2.0-25-virtual  I'm trying to increase the number of open files allowed for a user. This is for my eclipse java application where the current limit of 1024 is not enough.  According to the posts I've found so far, I should be able to put lines into   /etc/security/limits.conf like this:  soft nofile 4096 hard nofile 4096   to increase the number of open files allowed for all users.  But that's not working for me and I think the problem is not related to that file.  For all users, the default limit is 1024, regardless of what is in /etc/security/limits.conf (I rebooted after changing that file)  $ ulimit -n 1024   Now, despite the entries in /etc/security/limits.conf I can't increase that:  $ ulimit -n 2048   -bash: ulimit: open files: cannot modify limit: Operation not permitted The weird part is that I can change the limit downwards, but can't change it upwards - even to go back to a number which is below the original limit:  $ ulimit -n 800 $ ulimit -n 800  $ ulimit -n 900  -bash: ulimit: open files: cannot modify limit: Operation not permitted   As root, I can change that limit to whatever I want, up or down. It doesn't even seem to care about the supposedly system-wide limit in /proc/sys/fs/file-max  # cat /proc/sys/fs/file-max 188897  # ulimit -n 188898 # ulimit -n  188898   But even I get eclipse to run as root, my application still crashes because of "Too Many Open File" exception!   So far, I haven't found any way to increase the open files limit for a non-root user.  How should I properly do this? I have looked at several other posts but no luck! When does a Möbius transformation map $\Im(z)>0$ to itself? Show Möbius transformation which maps $\Im(z)&gt;0$ to itself iff  $$ f(z)= \frac{az+b}{cz+d}\,,\,\,ad-bc&gt;0$$ and $a,b,c,d$ are real. How can I remove all unused materials from a .blend file? Show that $u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$ is in $W^{1,n}(U)$, where $U=B(0,1)\subset\mathbb{R}^n$. The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback! My regex is matching too much. How do I make it stop? I have this gigantic ugly string: J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM J0000010: Project name: E:\foo.pf J0000011: Job name: MBiek Direct Mail Test J0000020: Document 1 - Completed successfully  I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time). Here's the regex I've been playing with: Project name:\s+(.*)\s+J[0-9]{7}:  The problem is that it doesn't stop until it hits the J0000020: at the end. How do I make the regex stop at the first occurrence of J[0-9]{7}? $ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $, prove $ab + bc + ca \le 3$ If $a,b,c$ real numbers such that $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$,  prove that $ab + bc + ca \le 3$.     Attempt:  Assume that $a,b,c &gt; 0$.   $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1  $$ $$ \implies  (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1) + (a^{2} + b^{2} + 1)(a^{2}+c^{2}+1) +  (a^{2} + c^{2} + 1)(b^{2}+c^{2}+1) \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  $$ \implies  ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1) + ((ab)^{2} + (ac)^{2} + (bc)^{2} + b^{2} + 2 a^{2} + c^{2} + a^{4} + 1) + ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 c^{2} + b^{2} + c^{4} + 1) \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  $$ \implies  3((ab)^{2} + (ac)^{2} + (bc)^{2}) + 4(a^{2} + b^{2} + c^{2}) + (a^{4}+b^{4}+c^{4}) + 3  \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  Next, notice  $$(a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1) = ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1)(c^{2} + a^{2} + 1) $$ $$ = (abc)^{2} + a^{2}c^{4} + b^{2}c^{4} + (ac)^{2} + 2(bc)^{2} + c^{4} + b^{4}c^{2} + c^{2} + a^{4} b^{2} + a^{4}c^{2} + (abc)^{2} + a^{4}  + 2(ab)^{2} + (ac)^{2} + b^{4}a^{2} + a^{2} +  (ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1 $$ $$ = 2(abc)^{2} + 3(ac)^{2} + 3(bc)^{2} + 3(ab)^{2} + c^{4}a^{2} + c^{4} b^{2} + b^{4}a^{2} + b^{4} c^{2} + a^{4}b^{2} + a^{4} c^{2} + a^{4} + b^{4} + c^{4} + 2(a^{2} + b^{2} + c^{2}) + 1  $$  So we have  $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$ $$ \implies  2(a^{2} + b^{2} + c^{2}) +  3  \ge 2(abc)^{2}  + c^{4}a^{2} + c^{4} b^{2} + b^{4}a^{2} + b^{4} c^{2} + a^{4}b^{2} + a^{4} c^{2} + 1 $$