How can I show that $\Bbb Q(i)$ and $\Bbb Q(\sqrt2)$ as fields are not isomorphic. Is there an element of order $4$ in $\Bbb Q(\sqrt2)$?
Page 105 of D. Burton's A First Course in Rings and Ideals reads     It is not difficult to show that if (I'll call them $n$ and $m$ instead of $n_1$ and $n_2$) $n$, $m$ are square free integers, then $\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ if and only if $n=m$.   Well, it is getting difficult for me, any help or would be appreciated.    My progress so far: Suppose $\phi :\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ is an isomorphism. Then $\phi(u)=u$ for every $u \in \mathbb{Q}$, so that $\phi(\sqrt{n})^2=\phi(n)=n$. If $\phi(\sqrt{n})=a+b\sqrt{m}$ for some $a,b\in\mathbb{Q}$, then $(a+b\sqrt{m})^2=a^2+b^2m + 2ab\sqrt{m}=n$ implies $$a^2+b^2m = n$$ and $$2ab=0.$$  Then $b=0$ cannot happen since that would imply that $\sqrt{n}\in\mathbb{Q}$, where $n$ is a square-free integer. Also $a=b=0$ cannot happen. Hence $a=0$ and we have $$b^2m=n.$$ If $\psi:\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ were an isomorphism then there would exist some $s\in\mathbb{Q}$ such that $$s^2m=n,$$ then $s=b$ or $s=-b$. Therefore the only isomorphisms from $\mathbb{Q}(\sqrt{n})$ to $\mathbb{Q}(\sqrt{n})$ are $\phi(u+v\sqrt{n})=u+vs\sqrt{m}$ and $\psi(u+v\sqrt{n})=u-vs\sqrt{m}$.    I don't know if there's an easier way and I don't know if I'm in the correct way here. Thanks beforehand :)
Sharepoint 2013. Add JS to NewForm.aspx of a list Let $F,G,H$ be groups such that $F\trianglelefteq G \trianglelefteq H$.  I am asked whether we necessarily have $F\trianglelefteq H$. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true? For which $f,g \in k[t]$, $k[f,g]$ is integrally closed? Let $f,g \in k[t]$, $k$ is a field of characteristic zero, $\deg(f)=n \geq 2$, $\deg(g)=m \geq 2$.     Is it possible to characterize all such $f$ and $g$ for which $k[f,g]$ is integrally closed in its field of fractions $k(f,g)$?   I do not mind to further assume that $k(f,g)=k(t)$ (but I do not want to assume that $k[f,g]=k[t]$, which makes my question trivial).  Thank you very much!  Edit: (i)  is relevant.  (ii) If also $k(f,g)=k(t)$, then being integrally closed is equivalent to $k[f,g]=k[t]$. Indeed: (a) If $k[f,g]=k[t]$ then $k[f,g]$ is integrally closed (since $k[t]$ is). (b) If $k[f,g]$ is integrally closed then, in particular, since $t \in k(t)=k(f,g)$ (= the field of fractions of $k[f,g]$) and obviously $t$ is integral over $k[f,g]$ (for example, $t$ is a root of $f(T)-f \in k[f,g][T]$) hence $t \in k[f,g]$, so $k[f,g]=k[t]$.  For example: $f(t)=t^2+1$, $g(t)=t^3-4t$. We have, $t= \frac{t(t^2-4)}{t^2+1-5}=\frac{g}{f-5}$, so $k(f,g)=k(t)$. By Abhyankar-Moh-Suzuki theorem, $k[f,g] \neq k[t]$, so $k[f,g]$ is not integrally closed. Similarly, for every $f,g$ such that $k(f,g)=k(t)$ and one of the degrees does not divides the other, then $k[f,g]$ is not integrally closed.  (iii) If also $k(f,g)=k(t)$ and $k$ is algebraically closed, then being integrally closed is equivalent to $H(t)=(f(t),g(t))$ is injective and $H'(t)=(f'(t),g'(t)) \neq (0,0)$ for every $t \in k$. Indeed, $k[f,g]=k[t]$ is equivalent to the new conditions (it is important that $k$ is algebraically closed!!).  Edit: What if we replace $k$ by an integral domain $D$ (which is a $k$-algebra)? See also  question. Chi-squared test when two vectors have different lengths I want to compare two groups (vectors) having different lengths to see if their distributions differ considerably. However, when I use chisq.test, I get an error due to inequalities of lengths:  &gt; chisq.test(factor(Group1), factor(Group2))   results in:  Error in chisq.test(factor(Group1[, col]), factor(Group2[, col])) :  'x' and 'y' must have the same length   Could someone help me with this? Thanks. How can I add reflection to a C++ application? I'd like to be able to introspect a C++ class for its name, contents (i.e. members and their types) etc. I'm talking native C++ here, not managed C++, which has reflection. I realise C++ supplies some limited information using RTTI. Which additional libraries (or other techniques) could supply this information? Taylor's polynomial uniqueness proof - why are these limits inferable? So I'm viewing a short proof on the uniqueness of Taylor polynomials.     Uniqueness of Taylor polynomial:   Let $f:]a,b,[ \rightarrow \mathbb{R}$ $n$ times continuously   differentiable and $x_0 \in ]a,b,[$. If $p$ is $n$th degree polynomial   function for which       $$f(x)-p(x)=o(|x-x_0|^n), \space \text{when} \space x \rightarrow x_0$$  then      $$p(x)=T_{n, x_0} f(x)$$   Proof starts like:  $$\frac{p(x)-T_{n, x_0}f(x)}{(x-x_0)^n} = \frac{p(x)-f(x)}{(x-x_0)^n}+\frac{f(x)-T_{n, x_0}f(x)}{(x-x_0)^n}$$  then   $$\frac{p(x)-f(x)}{(x-x_0)^n} \rightarrow 0 \space, when \space x-x_0 \rightarrow 0$$  $$\frac{f(x)-T_{n, x_0}f(x)}{(x-x_0)^n} \rightarrow 0 \space, when \space x-x_0 \rightarrow 0$$  Now my problem is, how can one actually say that the above approach $0$ if the denominator approaches $0$? I got 2 ghast tears after I thought I deflected a fireball back at the ghast (with my looting 3 sword).  But I was pretty close, so I might have just hit it. I'm not sure.  If I deflect a fireball with a looting sword and it kills the ghast, do I still get the looting bonus? Quantum Entanglement - What's the big deal? Bearing in mind I am a layman - with no background in physics - please could someone explain what the "big deal" is with quantum entanglement?  I used to think I understood it - that 2 particles, say a light-year apart spatially, could affect each other physically, instantly.  Here I would understand the "big deal".  On further reading I've come to understand (maybe incorrectly) that the spatially separated particles may not affect each other, but in knowing one's properties you can infer the other's.  If that it the case, I don't see what the big deal is... 2 things have some properties set in correlation to each other at the point of entanglement, they are separated, measured, and found to have these properties...?  What am I missing? Is it that the particles properties are in an "un-set" state, and only when measured do they get set?  (i.e. the wave-function collapses).  If this is true - why do we think this instead of the more intuitive thought that the properties were set at an earlier time? Series $\frac{x^{3n}}{(3n)!} $ find sum using differentiation Find sum of the series $$\sum_{n=1}^{\infty}\frac{x^{3n}}{\left(3n\right)!}$$ using differentiation. So far I found that $$S(x)+1=S'''(x)$$ but it does not help. Then I tried different interesting ideas  like $$S(x)+S'(x)+S''(x)=e^x-1\,.$$Maybe if I get the third equation it will allow me to construct a kind of differential equation. Then, by solving it, obtain $S(x)$. arbitrary large finite sums of an uncountable set. Let $T$ be a linear transformation $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that $T^{2}=\lambda T$ for some $\lambda\in\mathbb{R}.$ Then which of the following is/are true?  $1.\|T(x)\|=|\lambda|\|x\|$ for all $x\in\mathbb{R}^n.$  $2.$ If $\|T(x)\|=\|x\|$ for some non-zero $x\in\mathbb{R}^{n}$, then $\lambda=\pm 1.$  $3.T=\lambda I$ where $I$ is the identity transformation on $\mathbb{R}^{n}.$  $4.$ If $\|T(x)\|&gt;\|x\|$ for a nonzero vector $x\in\mathbb{R}^{n},$ then $T$ is singular.  According to me since $T^{2}=\lambda T$ we have $T$ satisfies the polynomial $x^{2}-\lambda x.$ So choices for its minimal polynomial are $x,x-\lambda,x(x-\lambda).$ So $3$rd option is clearly incorrect and $4$th option is not satisfied if $T=I$ and hence correct one. I am confused in first two options . Please help me. Thanks in advance. This is almost surely an Arzela-Ascoli question, since it comes from an old exam of which such problems are quite common. Unfortunately, I can't seem to get it though.  $\{ f_n \}$ is a sequence of functions $[0,1] \to \mathbb{R}$ satisfying $|f_n'(x)| \leq \frac{1 + |\ln (x)|}{\sqrt{x}}$ and $-10 \leq \int_0^1 f_n(x) dx \leq 10$ for every $n$. The question is to show the existence of a uniformly convergent subsequence.  In these kinds of situations, the bounds given usually turn into something nice  to force uniform boundedness. Maybe I could hope to establish equicontinuity using the integral somehow, but I'm not told that the $f_n$ are always positive, so the absolute values would probably screw things up anyway.  Is it an easy problem I'm missing or is this totally the wrong track? If so, what's the right track? I have some troble understanding the Kernel density estimation.  If a consider the next example:  s&lt;-0 z&lt;-density(rnorm(25)) f&lt;-function(i)approx(z$x,z$y,xout=i)$y for(i in seq(min(z$x),max(z$x),length.out = 10000)) s&lt;-s+f(i) s   I expected getting that s is approximately equals to one since I am considering an approximation for the integral of the Kernel density function (that I suppose is also a density). However, I don't get the expected result. Why it doesn's sum one?  The idea behind this is that I wanted to use the Kernel as the density of my data points. I was interpolating to get my explicit function and use with other x values. Or do you know another way to get a density function estimated by fitting? What do Family Tree Maker users do, now that FTM has been discontinued? I was considering purchasing Family Tree Maker only to find it has been discontinued.    What do they plan for a replacement ? Is there a way of creating a filesystem object akin to this:  mknod files p cat file1 file2 ... fileN &gt; files   but such that it can be seeked in, as if it were a regular file? How do I fix Outlook only showing the first line of every message? I'm experiencing a strange problem with Outlook 2019. When viewing a message it only shows the first line. On an HTML message, it stops at the first close paragraph tag. On a plaintext message, it stops at the first newline character. This is true if viewing a message in the reading pane as well as if double-clicking the message to open in its own window. I have the preview pane set to show 3 lines and it does show more of the message so it seems like Outlook can see more of the message than it's rendering in the reading pane. If I use View Source, the underlying data seems fine and the whole message is there. I've tried the following:  Deleting and re-creating the user's mail profile Uninstalling &amp; re-installing Office Creating a new Windows account &amp; connecting Outlook to a different Exchange account than the one that originally exhibited the problem Checked for Windows updates Checked for Office updates  Any ideas what might be causing this? I feel like I've tried everything obvious &amp; I'm striking out on Google. I´m posting this because I have some active questions in SO and I´m leaving for the weekend and wont have internet until monday or tuesday.  SO is a very active site and I hate no to answer/thank/comment people who are kind enough to answer my questions.  I think an "away" indicator like an out of office reply would be a really nice feature to add for "politeness" sake. Count the number of fish caught in total if we are given the number of fishermen who caught at least $n$ fish. Some fishermen caught some fish. No one caught more than 20 fish. $a_i$ is the number of fishermen who caught at least $i$ fish. How many fish were caught?   So my guess is that the number of fish caught has to be $$ a_1+a_2+\ldots + a_{20}$$  I reason that if a fisher man caught $n$ fish then he will appear in the tally $n$ times.   Is this logic correct? How can my module add xml sitemap entries? Wrong view count for questions? I have a question with 1508 views, but in the personal page list it gets reported as 2k.  Bug or aggressive approximation?