Let $K \subset L$ be a field extension and $A:=\{l \in L\,:\,a\text{ algebraic over }K\}$. I already have shown that $A$ again is a field and $K \subset A$ is algebraic. Also: If $l \in L$ is algebraic over $A$ it's also algebraic over $K$.  I now have to prove the following statement:  Suppose $L$ is algebraically closed. Show that $A$ is algebraically closed and that $K \subset A$ is algebraic.  As mentioned the latter I already have proven. Any hints how to proceed? I thought about considering a polynomial $a(x) \in A[x] \setminus A$ which decomposes into linear factors of $L[x]$ since $L$ is closed. From there on it would've been quite nice to show that these linear factors indeed are contained in $A[x]$ but I unfortunately do not know how to proceed.  Thanks for checking in! :)
Let $\mathbb F$ be a field and $\mathbb  K $ be an extension field of $\mathbb F$  such that $\mathbb K$ is algebraically closed.   Let $\mathbb L$ be the field of all elements of $\mathbb K$ which are algebraic over $\mathbb F$. Then $\mathbb L_{|\mathbb F}$ is an algebraic extension.   My question is : Is $\mathbb L$ algebraically closed ?  I am trying to prove the existence of algebraic closure, so please don't assume that every field has an algebraic closure.
Are there any COGO tools in QGIS 1.9+? The "Azimuth and Distance" plugin does not appear to be maintained any longer, and has not been transitioned to the new QGIS 2.0 plugin repository. What are the current options for entering COGO in QGIS? Does the roadmap include these tools, or is this something that will still depend on a plugin? I have CSV data loaded into a multidimensional array. In this way each "row" is a record and each "column" contains the same type of data.  I am using the function below to load my CSV file.  function f_parse_csv($file, $longest, $delimiter) {   $mdarray = array();   $file    = fopen($file, "r");   while ($line = fgetcsv($file, $longest, $delimiter))   {     array_push($mdarray, $line);   }   fclose($file);   return $mdarray; }   I need to be able to specify a column to sort so that it rearranges the rows.  One of the columns contains date information in the format of Y-m-d H:i:s and I would like to be able to sort with the most recent date being the first row. How to delete (invalid) files with colon in their name under Windows? I've got a rather large number of files on my Windows drive that have a colon in their name. (These came from unpacking some Unix archives.)   When I try to delete them, Windows (XP) complains that the file does not exist, and refuses to delete it. This happens when trying to delete or rename it from explorer or the command line. Still, chkdsk does not complain about those files nor fix the problem.   Any ideas on how to get rid of them? When can a (finite) group be written as the quotient of some other group by its center? So if I'm given $H$, when I can conclude there is a group $G$ such that $H\cong G/Z(G)$? It's easy to show that non-trivial cyclic groups are not of this form. More generally, any group with the property that all finitely generated subgroups are cyclic (e.g. $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$) cannot be of this form. Are there are any others? Proving with induction $(1-x)^n<\frac 1 {1+nx}$ Prove using induction that $\forall n\in\mathbb N, \forall x\in \mathbb R: 0&lt;x&lt;1: (1-x)^n&lt;\frac 1 {1+nx}$   My attempt:   Base: for $n=1: 1-x&lt;\frac 1 {1+x}\iff 1-x^2&lt;1$, true since $0&lt;x&lt;1$.  Suppose the statement is true for $n$, prove for $n+1$:  $(1-x)^{n+1}=(1-x)(1-x)^{n}\overset{i.h}&lt;\frac{(1-x)}{1+nx}$  Now I got stuck, maybe another induction to show that $1+nx+x&lt;1+nx$? Is there another way?  Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $\frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove? Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ I can't crack this one.      Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ (the only constraint is that  $a,b,c,z \in \mathbb{Z}$ and $c\neq 0)$ Can you enter UK later than intended date of travel on the visa application? How do I check which shell I am using? I read that terminal is nothing but shell, and Unix provides different flavors of shells:   Bourne shell (sh) C shell (csh) TC shell (tcsh) Korn shell (ksh) Bourne Again shell (bash)   Questions:   When I open a terminal window, which shell is opened by default? How do I check how many shells are installed? How do I change the shell used from my account? Polynomial $P(a)=b,P(b)=c,P(c)=a$ Let $a,b,c$ be $3$ distinct integers, and let $P$ be a polynomial with integer coefficients.Show that in this case the conditions $$P(a)=b,P(b)=c,P(c)=a$$ cannot be satisfied simultaneously.  Any hint would be appreciated. I've got the following question:  $A = \{\sqrt{n} - \left\lfloor\sqrt{n}\right\rfloor : n \in \mathbb{N}\}$  Find inf A, sup A    inf A is no big deal, but I have a hard time proving sup A is 1  I've tried to prove that: $\sqrt{n^2 - 1} -\left\lfloor\sqrt{n^2-1}\right\rfloor &gt; \sqrt{n} -\left\lfloor\sqrt{n}\right\rfloor $  but couldn't find a way to do that Where did they go / do they went? Where did they go?   Where do they went?   Can you please advise which of above forms is correct?  I'm pretty sure that first one is fine but wondering about second one.   Can I use a second form as well or it's incorrect? Function with same name but different signature in derived class Using a foreign driver's license in California when staying with a J-1 visa Limit of sequence in which each term is defined by the average of preceding two terms We have a sequence of numbers $x_n$ determined by the equality   $$x_n = \frac{x_{n-1} + x_{n-2}}{2}$$  The first and zeroth term are $x_1$ and $x_0$.The following limit must be expressed in terms of $x_0$ and $x_1$ $$\lim_{n\rightarrow\infty} x_n $$  The options are:   A)$\frac{x_0 + 2x_1}{3}$ B)$\frac{2x_0 + 2x_1}{3}$  C)$\frac{2x_0 + 3x_1}{3}$ D)$\frac{2x_0 - 3x_1}{3}$  Since it was a multiple choice exam I plugged $x_0=1$ and $x_1=1$. Which means that all terms of this sequence is $1$,i.e, $$x_n=1, n\in \mathbb{N} $$  From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence's limit? Individual community preferences vs. SE network policy: who wins? In the recent question , the OP asks where the following question would be appropriate:     What would be the best way to approach someone that works for me about bad odor?   CHAOS team member Abby Miller  by suggesting the proposal  &#8212; which is in commitment as I'm writing this &#8212; because     One of the up voted example questions there is about dealing with an irritating coworker. This question will be perfectly on topic for that site.   In my opinion,  qualifies for the not constructive close reason*, which applies to all sites in the network, and therefore, the question shouldn't be allowed, no matter how much it has to do with workplaces.  I know Abby isn't the only one who disagrees with me about this. I once had a talk with a mod of some beta about the presence of poll/GTKY questions on his site. His position, in short, was that his community had different rules from SO and its members enjoyed such questions. I don't remember which site it was, it might even be graduated now, but I never did end up participating there.  Anecdotal examples are nice and all, but here's the real question: to what extent can individual communities/sites override network policies? To what extend should they be able to do so?  *: not constructive comes with the description "This question is not a good fit to our Q&amp;A format. We expect answers to generally involve facts, references, or specific expertise; this question will likely solicit opinion, debate, arguments, polling, or extended discussion." A gold reviewer badge? Suppose $u$ is an element of a ring with a right inverse. I'm trying to understand why the following are equivalent.   $u$ has at least two right inverses $u$ is a left zero divisor $u$ is not a unit   If $v$ and $w$ are distinct right inverse of $u$, then $u(v-w)=0$, but $v-w\neq 0$, so $u$ is a left zero divisor. It's also clear that if $u$ is a left zero divisor, it cannot be a unit (else I could cancel $u$ from $ub=0$ to see $b=0$).  I'm having a heck of a time seeing why $u$ is not a unit implies $u$ has at least two right inverses. I tried the contrapositive, but saw no good approach. What am I missing? passthrough graphics card to virtualbox So I'm trying to get my laptops NVIDIA fx 880m to pass to a virtual box running window seven on a linux mint 17 install.  So far everything seems to be (maybe) heading in the right direction:  the device passed through happily, when I booted the virtual box it installed a bunch of new drivers, but when I try to install the NVIDIA driver on the guest it can't find the card. I looked under the device manager and there's no listing for the nvidia card under the pci bus, my guess would be that I need to disable it in the host so that it can be passed through to the guest (the virtual box manual said they can't be shared) but I'm not sure how to do so. Can anybody help with this? This question does not seem to me to be a duplicate of , because here the given solution was that it was not possible with a windows host, however in this case the host is linux Do Legendary Pokemon Respawn After The Elite Four? Do I have to clear customs when transit via the US on my way to / coming from somewehere else? I shall arrive at Washington Dulles Airport (IAD) by Emirates flight to catch a united airlines flight to Detroit after 3.5 hours. What steps shall I have to go through? Is the connection time enough? I have US F1 visa.