The following exercise appeared in An Invitation to Arithmetic Geometry by Dino Lorenzini (AMS Graduate Studies in Mathematics, Volume 9, page 32, Exercise 1) Prove that ${\bf Z}(\sqrt{d})$ is not a finitely generated abelian group for $d\in{{\bf Q}-{\bf Z}}$. I am trying to understand why this group must contain rationals whose denominators come from an infinite set of integers which are relatively prime in pairs. This would seem to prove the assertion. But my attempts to find such a set feel contrived!
When $d$ is in $\mathbb{Z}$, $\mathbb{Z}[\sqrt{d}] = \{a_0 + a_1 \sqrt{d} \mid a_0, a_1 \in \mathbb{Z}\}$  However, when $d$ is in $\mathbb{Q} \setminus \mathbb{Z}$, I cannot prove it.  Of course, when I see an example, I can understand that this is not a finitely generated abelian group, but I want to know the proof in general.
What color codes can I use in my PS1 prompt? I used several colors in my PS1 prompt such as   \033]01;31\] # pink \033]00m\]   # white \033]01;36\] # bold green \033]02;36\] # green \033]01;34\] # blue \033]01;33\] # bold yellow   Where can I find a list of the color codes I can use?  I looked at  but it didn't answer my question about a list of the actual codes.  It would be nice if there was a more readable form also.  See also Disable the "You've gone full screen" notification in Chrome HTML 5 Video stretch How to properly write multiple authors in bibtex file? How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$? Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$  Apart from induction, I tried with  to check the validity, but I can't think of an easy (manual) alternative.  Please suggest an intuitive/easy method. I was wondering how one can construct a Borel set that doesn't have full measure on any interval of the real line but does have positive measure everywhere.   To be precise, if $\mu$ denotes Lebesgue measure, how would one construct a Borel set $A \subset \mathbb{R}$ such that $$0 &lt; \mu(A \cap I) &lt; \mu(I)$$ for every interval $I$ in $\mathbb{R}$?   Moreover, would such a set necessarily have to contain infinite measure? Getting polygon boundaries of City in JSON from Google Maps API? When searching Google Maps for a city, the city's administrative boundary is displayed.    Is there a way to get the polygon boundaries of City in a JSON format using google maps api? How is canonicity of derivative works determined for Star Wars? I love R. Courant and F. John's Introduction to Calculus and Analysis because of its wide coverage, precise description and friendly written style.  Are there any classic textbooks like it on ODE/PDE?  thanks. How to run long time process on Udev event? I want to run a ppp connection when my USB modem is connected, so I use this udev rule:  ACTION=="add", SUBSYSTEM=="tty", ATTRS{idVendor}=="16d8",\     RUN+="/usr/local/bin/newPPP.sh $env{DEVNAME}"  (My modem appears in /dev as ttyACM0)  newPPP.sh:  #!/bin/bash /usr/bin/pon prov $1 &gt;/dev/null 2&gt;&amp;1 &amp;   Problem:  The udev event fires, and newPPP.sh is running, but newPPP.sh process is killed after ~4-5s. ppp does not have time to connect (its timeout is 10s for dial up).  How can I run a long time process, that will not be killed?  I tried using nohup, but it didn't work either.  System: Arch Linux  Update  I found a solution , thanks to .  I use at now to run my job detached from udev process.  But the one question remains unanswered: Why do nohup and &amp; not work? Parameter Sniffing (or Spoofing) in SQL Server A while ago I had a query that I ran quite a lot for one of my users. It was still being evolved and tweaked but eventually it stablised and ran quite quickly, so we created a stored procedure from it.   So far, so normal.   The stored procedure, though, was dog slow. No material difference between the query and the proc, but the speed change was massive.   [Background, we're running SQL Server 2005.]  A friendly local DBA (who no longer works here) took one look at the stored procedure and said "parameter spoofing!" (Edit: although it seems that it is possibly also known as 'parameter sniffing', which might explain the paucity of Google hits when I tried to search it out.)   We abstracted some of the stored procedure to a second one, wrapped the call to this new inner proc into the pre-existing outer one, called the outer one and, hey presto, it was as quick as the original query.   So, what gives? Can someone explain parameter spoofing?   Bonus credit for    highlighting how to avoid it  suggesting how to recognise possible cause discuss alternative strategies, e.g. stats, indices, keys, for mitigating the situation Does the sum of reciprocals of primes converge? Is this series known to converge, and if so, what does it converge to (if known)?  Where $p_n$ is prime number $n$, and $p_1 = 2$,  $$\sum\limits_{n=1}^\infty \frac{1}{p_n}$$ Show "Insufficient Privileges" directly when clicking 'Edit' (override), not on 'Save' If a user clicks 'Edit' on an opportunity which he isn't allowed to edit, the "Insufficient Privileges" page directly appears. This is good. If a user clicks 'Edit' on a custom object (which has an edit override that points to a visualforce page), he first sees the form without problems, and the "Insufficient Privileges" page only appears when he tries to save it. This behaviour is misleading for the users, because they think they can manipulate the data and put effort into it, and only afterwards they realize that this wasn't possible after all.  An idea was to check at page load if the user has edit permissions and otherwise redirect him to the "Insufficient Privileges" page.  Might there be another way? Thanks! Notation of the second derivative - Where does the d go? In school I was taught that we use $\frac{du}{dx}$ as a notation for the first derivative of a function $u(x)$. I was also told that we could use the $d$ just like any variable. After some time we were given the notation for the second derivative and it was explained as follows: $$ \frac{d(\frac{du}{dx})}{dx} = \frac{d^2 u}{dx^2} $$ What I do not get here is, if we can use the $d$ as any variable, I would get the following result: $$ \frac{d(\frac{du}{dx})}{dx} =\frac{ddu}{dx\,dx} = \frac{d^2 u}{d^2 x^2} $$ Apparently it is not the same as the notation we were given. A $d$ is missing. I have done some research on this and found some vague comments about &quot;There are reasons for that, but you do not need to know...&quot; or &quot;That is mainly a  notation issue, but you do not need to know further.&quot; So what I am asking for is: Is this really just a notation thing? If so, does this mean we can actually NOT use d like a variable? If not, where does the $d$ go? I found  related question, but it does not really answer my specific question. So I would not see it as a duplicate, but correct me if my search has not been sufficient and there indeed is a similar question out there already. I'm Australian and have obtained my Canadian working visa. However rather than flying direct to Canada I am meeting an American friend in LA and driving across the border. I have a return flight booked back to Australia from LA in March of next year too. Will the esta visa will be enough to enter the USA because I have my Canadian working visa and plan to travel there? Or will there be a problem because my return flight is booked later then the 90 days allowed on the visa. What are the rations Rey eats? In The Force Awakens, we see Rey pouring some flour-like polystarch rations into water so that the flour amazingly transforms into something that looks like bread. What are these rations? Is there any documentation of the fictional chemistry and biology that makes the starch rations behave as they do? "They are quite full up this year." When and where can I find the Tower's exotic item vendor (Xur Agent of the Nine)? $|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here. showing that a group of order 45 is abelian