My original problem is the following: given the ring $$ R:=\left\{\begin{bmatrix}q &amp; 0\\ r &amp; s\end{bmatrix}\;:\;q\in\Bbb Q,\;r,s,\in\Bbb R\right\} =:\begin{bmatrix}\Bbb Q &amp; \Bbb 0\\ \Bbb R &amp; \Bbb R\end{bmatrix} $$ I have to show that it is a left artinian ring, i.e. seen as a left module on itself, $_RR$, is artinian, i.e. the lattice of its submodules $\mathcal{L}(_RR)$, partially ordered by inclusion $\subseteq$ is an artinian poset.  So I tried to figure out how the submodules of $_RR$ are, and after some elementary computation I deduced that $$ \mathcal{L}(_RR)=\left\{\begin{bmatrix}A &amp; 0\\ B &amp; C\end{bmatrix}\;:\;A\;\mbox{is a $_{\Bbb Q}\Bbb Q$-submodule},A\subseteq B,\\ \;B,C\;\mbox{are $_{\Bbb R}\Bbb R$-submodules}\right\} $$ thus a descending chain of submodules of $_RR$ is of the type $$ \begin{bmatrix}A_0 &amp; 0\\ B_0 &amp; C_0\end{bmatrix}\ \supseteq \begin{bmatrix}A_1 &amp; 0\\ B_1 &amp; C_1\end{bmatrix}\ \supseteq\cdots\;\;\;\;\;\;\;\;\;(*) $$ where $\{A_i\}_{i\ge0}$ is a descending chain of $_{\Bbb Q}\Bbb Q$-submodules and $\{B_i\}_{i\ge0},\;\{C_i\}_{i\ge0}$ are descending chains of $_{\Bbb R}\Bbb R$-submodules. Thus if $_{\Bbb Q}\Bbb Q$ and $_{\Bbb R}\Bbb R$ were artinian, then $(*)$ would be stationary, thus my ring $R$ would be left artinian. Is my argument correct? How can I prove that $_{\Bbb Q}\Bbb Q$ and $_{\Bbb R}\Bbb R$ are artinian modules?
Let $R$ be the subring of $M_2(\mathbb{R})$ defined by $$R=\left\{\begin{pmatrix}a&amp;b\\0&amp;d\end{pmatrix}\mid a\in\mathbb{Q},b,d\in\mathbb{R}\right\}$$  Show that $R$ is right-Artinian but not left-Artinian.  The "obvious" approach to show it is not left-Artinian is to find an $A\in R$ such that $$RA\supset R(A^2)\supset R(A^3)\supset\dots$$  is an infinite decreasing chain of proper left submodules of $R$ (as a module over itself). However, after spending a lot of time in this endeavor I've come to the conclusion that no such element $A$ exists (I can post a proof if anybody is interested but it'd be long and not worth the time for me otherwise). Does anybody have an idea for another approach I can take to this? I have no idea what else submodules of $R$ would look like.
Concatenating a string and primary key Id while inserting Unable to Change Apple ID and password I have tried repeatedly to change my Apple ID and password without success. I have followed all of the directions trying both my emails and asked for new passwords----no luck whatsoever. Ok, so at my school, everybody was supplied with a Macbook Pro. I set up the Messages app and it was all great until the county decided to block it. Now, instead of receiving text messages on my phone they go to my Mac. I have no idea how to fix this because I can't open the Messages app. Please help?! I played around with it for a while, but I simply can't figure it out.  I made a tank that fires missiles, and when the missiles hit the walls, I want them to bounce off, but I want them to bounce off to the right angle.  Right now I haven't got any obstacles, the missiles just bounce off when they get outside the viewportRectangle I made.  Is the solution I'm looking for quite advanced?  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Regardless which one I start first, I can only open that one, and get the above error if I attempt to sidestep.  What makes this an especially annoying problem is that I often have to respond to emails that I need to access the database to deal with. So either I close and open many windows. How to show loading spinner in jQuery? In Prototype I can show a "loading..." image with this code:  var myAjax = new Ajax.Request( url, {method: 'get', parameters: pars,  onLoading: showLoad, onComplete: showResponse} );  function showLoad () {     ... }   In jQuery, I can load a server page into an element with this:  $('#message').load('index.php?pg=ajaxFlashcard');   but how do I attach a loading spinner to this command as I did in Prototype? Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an module over $A$. Is it true that $\operatorname{Ann}(M/IM) = \operatorname{Ann}(M) + I$?    One inclusion is certainly true, but I don't know about the other one. 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In fact, in recent updates it's required for the player who just can't start their mornings without bacon and eggs!  I've tried pulling two pigs together in a single pen, but all they do is snort at me. And the chickens? Well, there's plenty of eggs, but no hatching.  How do I get them to make more bacon and eggs? I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The  contains a delta function in the formula, but  gives only the formula without the delta function,  $$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$  Is it intentionally left out?  My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge. How to remove the WordPress version from some .css/.js files I know that I can use the following function to remove the version from all .css and .js files:  add_filter( 'style_loader_src', 'sdt_remove_ver_css_js', 9999 ); add_filter( 'script_loader_src', 'sdt_remove_ver_css_js', 9999 ); function sdt_remove_ver_css_js( $src ) {     if ( strpos( $src, 'ver=' ) )         $src = remove_query_arg( 'ver', $src );     return $src; }   But I have some files, for instance style.css, in the case of which I want to add a version in the following way:  function css_versioning() {     wp_enqueue_style( 'style',         get_stylesheet_directory_uri() . '/style.css' ,         false,         filemtime( get_stylesheet_directory() . '/style.css' ),         'all' ); }   But the previous function removes also this version. So the question is how to make the two work together? I've set up the physics of a swinging wrecking ball and set the animation to run for 240 frames.  This was not long enough for the entire "swing" of the ball-and-chain.   I'm unable to reset the timeline, but I can set the end frame to (say) 1000 frames, but when I try to play the animation, the animation progress halts at frame 240 and I cannot get it to go any further.   There are no markers or keyframes in the timeline.  What should I do to "force reset" the animation-timeline, to allow changes to the length? Why is rendered viewport faster than rendering individual frames? When I'm using rendered viewport it initially takes a while before rendering starts, but once it does, I can freely move around and rendering a new view happens almost instantaneously. When on the other hand I'm rendering a movie, for each frame Blender takes a significant amount of time before rendering even starts, usually doing the following:  Synchronizing object, Updating images, Updating meshes, Updating Scene BVH, Packing BVH,  Why does it do this? And is it possible to disable this to increase movie rendering speed? Is  structured data (in Microdata, JSON-LD, or RDFa) relevant for SEO?   I know that structured data can provide benefits in other areas, but here I’m only asking about its significance for general web search engines. Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$ Given six numbers $x,y,z,a,b,c$ which satisfy the following relations $y^2+yz+z^2=a^2$ $z^2+zx+x^2=b^2$ $x^2+xy+y^2=c^2$ Express $x+y+z$ in terms of $a,b,c$    My attempt: $\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-x^3)}{(z-x)}=b^2,\dfrac{(x^3-y^3)}{(x-y)}=c^2$ But this step didn't seem to help. So I tried adding and expressing $k(a+b+c)^2$ but some other unmanagable elements came along. I could not think of any other way to approach the problem. Please help. Force a Blender operator to always start with default values I'm writing an operator for which it is crucial that every time it gets called all the properties are set to their default values.   The ops.object.delete has 1 param. use_global. It is always false, no matter what the user put in last time it got called. I would like to replicate exactly that.   (For example: I want to modify delta scale. The default val for my_factor is 1, so nothing happens until the user alters that number. If the user calls the operator again, it will have stored whatever number the user has used last, however I want it to always start with 1.) Wrong place for hyperref link in scaled tikz picture I have a TikZ picture with a node containing a minipage with an itemize environment; each item is a hyperref to other sections of the document.   When scaling the image (for example with \begin{tikzpicture}[scale=.7, transform shape] I've a strange behavior: links in itemize are not in the right place: they present a somewhat vertical offset, as if they don't scale as the rest of image.   Here the MWE:  \documentclass[a4paper]{article} \usepackage{tikz} \usepackage[pdftex]{hyperref} \begin{document}  Here something interesting\label{text:interesting-one} \newpage  Here something interesting, too\label{text:interesting-two} \newpage  and here there are very very interesting stuff\label{text:interesting-three} \newpage  \newpage Now the problem: links rectangles have some offset, making links unusable in case of long itemize  \begin{figure}[htb] \begin{tikzpicture}[scale=.7, transform shape] \node at (0,0) (mynode) { \begin{minipage}{4cm} \begin{itemize} \item \hyperref[text:interesting-one]{First link} \item \hyperref[text:interesting-two]{Second link} \item \hyperref[text:interesting-three]{Third link} \item \hyperref[text:interesting-one]{First link again} \item \hyperref[text:interesting-two]{Second link again} \item \hyperref[text:interesting-three]{Third link again} \end{itemize} \end{minipage} }; \end{tikzpicture} \end{figure} \end{document}   And here a snapshot of the problem    Note: This is evident using the default settings of hyperref package, which draws the red rectangle around links, but it took me some time to found it when only the text is coloured, i.e. when using \usepackage[linkcolor=blue,colorlinks=true]{hyperref} what is a thyristor As part of my 2330 lv3 am required to know about thyristors. I just want to confirm my understanding is correct. A thyristor is basically similar to a transistor, only it stays switched on when it's trigged for as long as there isn't a break in the circuit it's switching. Am I correct?