Let $R$ be a ring of prime characteristic, $p &gt; 0$ and $x,y \in R$. Show that:  $$(x+y)^p = x^p + y^p$$    To recall definition; a characteristic of $R$ is $n\in\mathbb N \cup \{0\}$, such that $\forall a \in R, na = 0.$ If such a number does not exist then $n=0$.  Any help is much appreciated.
If $p$ is prime, then $(x+y)^p=x^p+y^p$ holds in any field of characteristic $p$. However all the proofs I have seen use induction and some relatively nasty algebra despite how fundamental this fact seems.  What is the nicest, "highest level proof" you know?
How do I copy vertex groups from one object to another? Does one work in or on an aeroplane? In an exam paper, there was a picture of an air stewardess in the aeroplane serving passengers. One of my pupils wrote the following:      The air stewardess works on an aeroplane.    Shouldn't it be the air stewardess works in an aeroplane? How to export data files using specific number format Can I answer my own questions, even if I knew the answer before asking? Fundamental group of the torus I'm trying to get my head round the following calculation of the fundamental group of the torus, using Seifert Van-Kampen (I know it's easier to do this by considering covering spaces, but I'm trying to learn the Seifert Van-Kampen method).   Consider the torus $T$ as the unit square in $\mathbb R^2$ with opposite edges identified. Let $A$ be an open disc (say of radius $1/4$ about the origin), and $\overline{A}$ be it's closure. Let $B$ be complement of $A$, so that $A \cap B = S^1$. Then we have $T = A \cup B$ where $A$ and $B$ are a neighbourhood deformation retract pair, so we can use Van Kampen. It is clear that $\pi_1(A) = \{e \}$ and $\pi_1(A \cap B) = \mathbb Z$. So we need to find $\pi_1(B)$. Observe that $B$ deformation retracts onto the boundary of the square which, after identification, is homeomorphic to $S^1 \vee S^1$; in this way we see that $B$ and $S^1 \vee S^1$ are homotopy equivalent (inclusion in one direction, the retraction in another), and so $\pi(S^1 \vee S^1) = F_2$ (which can be calculated by Van Kampen, or by using the theory of covering spaces). So, by Van Kampen, $\pi_1(T) \cong F_2 *_{\mathbb Z} \{e\}$. This is the quotient of $F_2$ by the normal closure of the image of $l_* : \pi_1(A \cap B) \to \pi_1(B)$ (where $l_*$ is the homomorphism induced from the inclusion map $l : A \cap B \hookrightarrow B$). The following is causing me confusion: The generator of $\pi_1(A \cap B)$ is homotopic in $B$ to a loop of the form $u.v.\overline{u}.\overline{v}$, where $u$ and $v$ represent the two generators of $F_2$. So $\pi_1(T) \cong \langle a,b ; aba^{-1}b^{-1} \rangle$, the free abelian group on two generators.  Any explanation or geometric intuition would be really appreciated. Thanks What is causing disabled options when adding a mesh/curve object to the scene? Sometimes when adding an object to the scene I see that all options (on toolbar or in window after pressing F6) are disabled. Can anyone tell me when can this happen? What is causing it and how to revert it - make it enabled again. It worked for me when I restarted Blender and reloaded the scene. However I would really like to know what can cause it? Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Is it possible to use an Ubuntu Live CD and run GParted to partition a my HDD (including the partition that is usually mounted on a given system)? I'm encountering a problem with my HDD's partitions and can't resize it. I've described my problem and my system here: .  Now, from a more general perspective, I was thinking on whether it is possible to use an Ubuntu Live CD and use GParted in a live session to possibly go around that problem. Is that possible? I made a separate question as I'm interested in the possibility independently from my current problem.  You see, I only have one partition and that's mounted. When I use GParted through Ubuntu I don't get any error reports (unlike the case reported in the question linked above).  Thanks beforehand. Gaussian Integers and Quotient Rings Show that $\mathbb{Z}[i]/(2+3i)$ is a finite field.    By applying some isomorphisms and realizing $\mathbb{Z}[i]$ as the quotient ring of $\mathbb{Z}[x]/(x^{2}+1)$, I found that:  $\mathbb{Z}[i]/(2+3i) \cong \mathbb{Z}[x]/(x^2+1,2+3x)$  However, I run into trouble as I can't divide $x^{2}+1$ by $2+3x$ in $\mathbb{Z}[x]$. What would be another way to approach the problem?  Thanks for the help. I have to determine how many six-integer subsets of $\{1, 2, 3, \ldots, 20\}$ are possible if no two consecutive integers are in a set. What does the 'cosmological constant' represent? Newtons theory of gravity involves a gravitational constant $G$, however one does not refer to it directly, we speak instead of gravity or the force of gravity.   Now Einsteins introduced a  in his equation for GR, and ever since it's been referred to as such; however, this usage seems strange to me, given the above parallel; it seems to me we ought to be referring to something physical, which can then be quantified and whose quantification involves the cosmological constant.   Does the cosmological constant then refer to a cosmological force? A force that is, contra gravity, repulsive? And which is even weaker than gravity, kicking in on the scale of galaxies rather than solar systems? I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:  $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$  I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other? Is there any Stack Exchange site where the main topic is literature?  I have tried the English Language site, but it seems that it is not the appropriate site.  Perhaps the writers site would work? But, then again, the main point is different than the one I am searching for. System of congruences with variable I am looking over , and I came across a solution containing this:  Problem Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$. For example, if $N = 749$, Bernardo writes the numbers $10,\!444$ and $3,\!245$, and LeRoy obtains the sum $S = 13,\!689$. For how many choices of $N$ are the two rightmost digits of $S$, in order, the same as those of $2N$? $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ Solution 1 First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. Say that $N \equiv a \pmod{6}$ also that $N \equiv b \pmod{5}$ Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b &lt; 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$. $N \equiv a \pmod{6}$, $N \equiv  a \pmod{5}$, $\implies N=a \pmod{30}$, $0 \le a \le 4$ Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$ Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$, ; Also, we have already found which digits of $y$ will add up into the units digits of $2N$. Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work) Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation.[N \equiv 30x \pmod{36}][N \equiv 30x \equiv 5x \pmod{25}]Both of those must add up to[2N\equiv60x \pmod{100}] $N\equiv{30x\mod{36}}\\N\equiv{30x}\equiv{5x\mod{25}}$ Both of these must add up to $2N\equiv{60x\mod{100}}$ [...]  I'm not really sure which modular arithmetic rule to apply in the final three congruences, or where to start. How would I go about understand this? What Easter eggs do the chat sites have? According to , the SE chat sites may have individual Easter eggs, depending on the site. If you find them, please post. Because it's so easy to make this stuff up, a screenshot as proof would be nice. (Although Photoshop isn't that hard either...) What does *~ mean? At the end of a makefile I saw   rm -f *~ *.class   I understand the *.class, but what's *~? how to copy or backup files ignoring dot files Can the editor for editing tag descriptions get updated to match the normal questions/answers editor? Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why? The annual student intake of Hogwarts seems about 100. Since in book 7 Hogwarts is declared mandatory for all children, this means that the total magical population of Britain cannot be much more than 10,000.  How do these people make a living?  All jobs mentioned in the books are non-productive occupations (government, stores, sports &amp;c) - not farmers or engineers. There are exceptions - makers of wands/robes/broomsticks - but where do they get the raw materials from?  Who and how "creates goods" in the wizarding world?  Note that if you say "wizards exchange their gold for Muggle money and buy stuff from Muggles (through intermediaries, if necessary)" then you would have to explain where they get their gold: they would either have to mine it, or produce something of value to the Muggles (in which case the separate financial system of galleons/sickles/knuts does not make sense).  PS. A simple issue of maintaining a steam locomotive for Hogwarts express is pretty much insurmountable for 10k people - without Magic, that is. So, I can imagine that the skilled tradesmen can transfigure dirt into raw materials for their trade - except for food, which "is the first of the five Principal Exceptions to Gamp's Law of Elemental Transfiguration".