I do not make sense of the following argument. We want to determine the ring of integers of $\mathbb{Q}(\sqrt{d})$. An element $\alpha = a + b \sqrt{d}$ is an algebraic integer iff its minimal polynomial over $\mathbb{Q}$  has coefficients in $\mathbb  Z$. Expanding the minimal polynomial, this translates into the two conditions $$2a \in \mathbb Z \quad \text{and} \quad a^2 - b^2 d \in \mathbb Z.$$ How do we get the usual conditions depending on $q \equiv 1$ mod $4$ or not?
I'm currently reading Marcus' "Number Fields," and I'm having difficulty proving the following result:  Corollary 2.2: Let $m$ be a squarefree integer. The set of algebraic integers in the quadratic field $\mathbb{Q}(\sqrt{m})$ is $$ \left\{ a + b \sqrt{m} : a,b \in \mathbb{Z} \right\} $$ if $m \equiv 2,3 \pmod{4}$; and$$ \left\{ \frac{a+b\sqrt{m}}{2} : a,b \in \mathbb{Z} ~,~ a \equiv b \pmod{2} \right\}$$ when $m \equiv 1 \pmod{4}$. $\square$  It is easy enough to see that the claimed sets do, indeed, consist of algebraic integers. The other half of the proof proceeds something like this: suppose that $\alpha = r + s \sqrt{m}$ is an element of $\mathbb{Q}(\sqrt{m})$. Then $\alpha$ is a root of the polynomial $$ x^2 - 2rx + r^2 - s^2m $$ In particular, this means that $\alpha$ is an algebraic integer if and only if both $2r$ and $r^2 - s^2 m$ are integers. At this point, the proof stops, and says that this implies the result. And therein lies the problem: I don't see how to make the connection. I do know, for example, that if one of $r,s$ is an integer, then the other one must be, too, i.e., either both are integers, or both are not. However, I cannot get any further than that. Can anyone give me a gentle nudge in the right direction? Thank you very much!
Are spells learned from feats considered to be associated with your class? The multiclassing spellcasting rules (PHB p. 164, or ) state:  Each spell you know and prepare is associated with one of your classes, and you use the spellcasting ability of that class when you cast the spell.  According to the above, every spell you know or prepare should be associated with a class.  However, for spells learned from feats like Drow High Magic or Wood Elf Magic are these actually associated with a class? If so, what class? If not, is the PHB statement above wrong/outdated? Are the spells learned in these feats (and others like it) associated with something else (like your race, in these cases)?  For reference, the  racial feat (XGtE p. 74) states:  You learn more of the magic typical of dark elves. You learn the detect magic spell and can cast it at will, without expending a spell slot. You also learn levitate and dispel magic, each of which you can cast once without expending a spell slot. You regain the ability to cast those two spells in this way when you finish a long rest. Charisma is your spellcasting ability for all three spells.  The  racial feat (XGtE p. 75) states:  You learn the magic of the primeval woods, which are revered and protected by your people. You learn one druid cantrip of your choice. You also learn the longstrider and pass without trace spells, each of which you can cast once without expending a spell slot. You regain the ability to cast these two spells in this way when you finish a long rest. Wisdom is your spellcasting ability for all three spells.  In regards to the third and fourth questions, the fact that the PHB quote says &quot;use the spellcasting ability of that class when you cast the spell&quot; and the fact that the feats specify what ability the spell uses in these examples (independent of class) suggests to me that they are not associated with class. How to modify existing, unpushed commit messages? I wrote the wrong thing in a commit message.  How can I change the message? The commit has not been pushed yet. What changes when an Alienist becomes an outsider? Yu Yuphoria bricked after installing custom recovery and boot file My YU YUPHORIA got bricked while I was working with it. I want to install custom ROM into it for that I am installing custom recovery and boot file but the file that is found on Internet was not working and that's why I am stuck with company logo. I can't go into recovery mod, neither it boot, and USB debugging mode is turned off. I tried to install fresh ROM from fastboot but after installing every driver properly command prompt is showing "waiting for the device". Please sugest some way to get rid of this problem. List of GUI Based Torrent clients (I'm aware of  but it has no answers…)  Let be $u_0 \in \mathbb{R}$ and the sequence $(u_n)_n$ defined by: $u_{n + 1} = \frac12 \arctan(u_n)$.  I define also: $v_n = 2^n u_n$, so I can show that: $\lim (u_n)_n = 0$ (by studying $x \mapsto \frac12 \arctan(x)$), thus, I can show that $(v_n)_n$ is monotone and converges because it is bound.  Now, I conclude: $u_n \sim \dfrac{l}{2^n}$, I'd like to determine $l$ more precisely.  Here is what I tried, I suspect $l$ to be something like $f(\pi)$ for some $f$ :   push the asymptotic expansion of $\arctan$ to the 2nd order and reinject it ; use $\arctan(u_n) + \arctan(1/u_n) = \dfrac{\pi}{2}$ ; use series techniques to look for $\sum v_{n + 1} - v_n$, maybe conclude using Cesaro summation Arbitrary products of quasi-coherent sheaves? Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer) Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer)  I saw no example of this problem anywhere.. I saw an example how to solve it without calculus but I'm afraid that'll be deductions on my test.. I have the volume of a cylinder and the method of area between two curves but this is out of my reach.... I don't know what I can try... I am thinking about a wizard who has a rat familiar (from the  spell), and has the  spell prepared. Can he use his familiar's senses to teleport into another room (where the familiar is hiding) by looking through the eyes of the rat and casting misty step?  The wizard could use his action to switch to the familiar's senses and then use his bonus action to cast the spell. This would greatly reduce the "blind and deaf" time he experiences from doing this. How to draw arrows between circled coefficients of polynomials using tikz After reading Gonzalo's answer to , I thought  I would be able to modify it so that I could have the corresponding coefficients circled and an arrow going from one circle to the next. For example, I would like the coefficient of x^2 on the left, a, circled and the coefficient of x^2 on the right, 5 circled, then a curved arrow drawn between the two circles. How might I achieve this? Also, how would I add some text along the path of the arrow? I reproduce his code here.  \documentclass{article} \usepackage{amsmath} \usepackage{tikz} \usetikzlibrary{calc}  \newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}  \begin{document}  \begin{equation}   a\tikzmark{a}x^2 + bx + c = 5\tikzmark{b}x^2 + bx + c.   \begin{tikzpicture}[overlay,remember picture,out=315,in=225,distance=0.4cm]     \draw[-&gt;,red,shorten &gt;=3pt,shorten &lt;=3pt] (a.center) to (b.center);   \end{tikzpicture} \end{equation}  \end{document} Making ffmpeg process first audio stream differently than all others I'm trying to write a command such that ffmpeg executes a given instruction for the first audio stream in a file and a different instruction for any and all others.  In the test case below, for example, I want the first stream copied to the output file and all others converted to mono.  If I write it this way  ffmpeg -i INPUT -map 0 -c copy -map 0:a:0 -c:a:0 copy -map -0:a:0? -c:a libfdk_aac -ac 1 OUTPUT   the first audio stream is ignored when the mapping is overridden later in the command line.  But if I move the instruction for that stream to after the others'  ffmpeg -i INPUT -map 0 -c copy -map -0:a:0? -c:a libfdk_aac -ac 1 -map 0:a:0 -c:a:0 copy OUTPUT   it's no longer ignored, but becomes the last audio stream in the file.  I know I'm close here; I just can't see what I'm doing wrong. A maximal ideal among those avoiding a multiplicative set is prime Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime. If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with a ring homomorphism $R\to F$. Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f\colon R\to F$, where $F$ is a field, such that $f(x)=0$ for all $x\in I$ and $f(y)\neq 0$ for all $y\in S$.   [You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0\notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]  Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.   For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that? Trying to identify a short sci-fi story about a human that is "resurrected" by malignant aliens For example,      I like chocolate, vanilla, and lemon and orange ice cream.    Indicating "lemon and orange" is a combined flavor, as an item in the list needing an initial and. How do I create a villager with custom trade offers? Beamer -- not indenting itemize in a single environment The output in the MWE posted at the end of this post is shown below. Notice how the default indentation from itemize gives an asymmetric slide w.r.t the spacing to the left and right of the itemize environment. How can I push the bullet points to the left so that the spaces to the left and right are the same? Such a solution would ideally be:   implementable on a one-off basis, so that it is applied when necessary (since some itemize environments should be given the default indentation) independent of the margin size, so that the solution doesn't need to be changed when the margin size is changed     \documentclass{beamer}   \begin{document}  \begin{frame}{A title} \begin{itemize} \item Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod \item tempor incididunt ut labore et dolore magna aliqua. \item Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi \item ut aliquip ex ea commodo consequat. \begin{itemize} \item Duis aute irure dolor in reprehenderit in voluptate velit esse \item cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat \end{itemize} \end{itemize}    \end{frame} \end{document} Problem in notation: $(\vec{v}\cdot\vec{\nabla})\vec{v}$ versus $L_v(\vec{v})$ Old-style/Antique typesetting in LaTeX/TeX Light "diode" and 2nd law of thermodynamics If I had a light "diode" - an object that only allowed light (at least for a range of frequencies) to travel through it in one direction, would this necessarily allow violations of the 2nd Law of Thermodynamics?  Or would it be possible to restrict other properties of the device such that it could be okay.  Why I think it should violate 2nd law:   I can place such devices at temperature T in a bath of thermal radiation at temperature T, such that it forms a box which only lets the light in or out.  This would appear to always allow one to create a thermal gradient.  Once a thermal gradient is formed, useful energy can be extracted.  This seems to allow useful energy to always be extracted from a thermal bath and thus give perpetual motion. Release one such device at temperature T in a bath of thermal radiation at temperature T.  Since the photons are absorbed or reflected in one direction, and undisturbed in the other, the device will start moving.  It can use a thermal bath to directly generate useful motion.   Why I think I'm missing something and such a device could potentially co-exist with the 2nd Law:   The above arguments seem valid if the diode was only 'partially' rectifying as well.  However there exist sunglasses which largely reflect on one side, but not from the other side. Researchers have succeeded in using meta-materials to create a "blackhole" for microwaves.   Can anyone help sort this out? Editing/Creating files in C:\Program Files\WindowsApps