Let $L$ be the splitting field of $f=X^4-2aX^2+b\in\mathbb Q[X]$. Show that $\sqrt b \in \mathbb Q(\sqrt{a^2-b})=:K$ if $[L:\mathbb Q]=4$.  EDIT: $a,b$ are chosen so that $f$ is irreducible. Can I argue like this? If we have $\sqrt b\notin K$, then $K(\sqrt b)$ is an extension of at least $2$ of $K$. On the other hand, since $f$ is irreducible we have that $\sqrt{a^2-b}\notin \mathbb Q$ and therefor $K/\mathbb Q$ is an extension of degree $2$. So, if $[L:K]=4$ we obtain $L=K(\sqrt b)=\mathbb Q(\sqrt b,\sqrt{a^2-b})$ by degree comparison. But then $L$ is the splitting field of $g=(X^2-b)(X^2-a^2+b)\neq f$ which is in conflict with the uniqueness of splitting fields.
Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result:    For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.   My thoughts so far: The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:   For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof. For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case?
Task manager for Ubuntu Calculating MEAN and STANDARD DEVIATION confussed zeros of a polynomial Given $P(z)=z^6+6z+10$, find how many roots are in each quadrant    I have already seen that $P(z)$ has six different roots, and that none of them are real or of the form $ki$, $k\in \Bbb R$. Since the coefficients of $P(z)$ are real, then it's roots are 3 conjugate pairs, and therefore the possibilities are limited to two.  Defining $f(z)=z^6$, $g(z)=z^6+6z+10$ and applying Rouche in $|z|&lt;2$, $|f(z)-g(z)|&lt;|f(z)|$ I get that all roots are contained in the disc, but how do I go about splitting them into the four quadrants? From what I know about this polynomial, finding zeros in one quadrant is enough to determine the rest, but I'm stuck trying to apply rouche to a quarter circle (basically I'm having trouble finding a suitable f and g) I suspect that the name server provided by my ADSL modem/router is buggy. Whenever I browse to a website for the first time in ubuntu, resolving the domain name takes at least 15 seconds.  To work around that problem, I changed the nameserver configuration in /etc/resolv.conf from 192.168.1.1 (my ADSL modem) to 8.8.8.8 (google's primary DNS). This seems to fix the problem, but unfortunately my changes to /etc/resolv.conf are overwritten by "NetworkManager" at each startup.  What is the proper way to configure the name server IP in ubuntu 10.4? Does getting disconnected from an SSH session kill your programs? Single word which means "keep moving forward" I need a single word which means "keep moving forward". What are my options?  I need word as   persevere Be persistent, refuse to stop  cognizant (sometimes followed by 'of') having or showing knowledge or understanding or realization or perception  [] keep moving forward  I want to make meaningful name for firm  please help me Why can't I pass the table name to a prepared PDO statement?  $stmt = $dbh-&gt;prepare('SELECT * FROM :table WHERE 1'); if ($stmt-&gt;execute(array(':table' =&gt; 'users'))) {     var_dump($stmt-&gt;fetchAll()); }   Is there another safe way to insert a table name into a SQL query? With safe, I mean that I don't want to do  $sql = "SELECT * FROM $table WHERE 1" How can I take a site offline using Drush? Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ Using custom tables in nft I have a Fedora 31 system on which I am using iptables-nft. I need this because there is still a bunch of software that expects the legacy iptables command line tools.  This means that my nftables configuration has the corresponding set of tables to match the legacy configuration:  table ip filter table ip nat table ip6 filter table ip mangle table ip6 nat table ip6 mangle   I use a containerized VPN service that, prior to nftables, would enable masquerading on my primary ethernet interface by running something like this when the vpn comes up:  iptables -t nat -A POSTROUTING -s 172.16.254.0/24 -o eth0 -j MASQUERADE   Since upgrading to Fedora 31 and iptables-nft, this no longer works. The container (running alpine) does not have the iptables-nft compatability wrapper, but it does have the nft command itself.  I can't use the nft cli to add rules to the existing tables, because this will break iptables-nft. But I can create new tables. I was hoping I could just apply a configuration like this:  table ip vpn {     chain postrouting {         type nat hook postrouting priority filter; policy accept;         ip saddr 172.16.254.0/24 oifname "eth0" counter masquerade     }      chain forward {         type filter hook forward priority filter; policy accept;         ip saddr 172.16.254.0/24 counter accept     } }   ...but this doesn't appear to have any impact. By setting the chains in this table to priority 0 I was hoping they would match before the legacy nat table, but that doesn't appear to be the case.  Is there a way to make this work? Get index of parent div from child divs Defining my own None-like Python constant $L^p$ and $L^q$ space inclusion Let $(X, \mathcal B, m)$ be a measure space.  For $1 \leq p &lt; q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied? IS or ARE? "The only thing that I want you to hit right now IS/ARE the books" Why is it so hard to find the roots of polynomial equations? The question that follows was inspired by  question:  When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic formula.    That the general solutions to various polynomial equations are so complex and difficult to derive seems to suggest a fundamental limitation in the problem solving capabilities of the mathematical machinery. Does this intuition of mine make any sense? What should I make of it? Why is it that with each successive degree in a polynomial equation, the solution becomes so much more complex? Can I gain some intuition about what makes finding the roots so hard? According to the : "there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher." What is so special about the quintic that makes it the cut-off for finding a general algebraic solution? Prove that a function is not injective I need font size = 14 in article document, but when I change  \documentclass[a4paper,11pt]{article}   to  \documentclass[a4paper,14pt]{article}   font becomes even smaller. How to change font size? Suggested Edits Dashboard doesn't agree with Suggested Edits view My family PC has Windows Vista and lately it has became unusable, having strange errors and taking hours to do anything at all, so I'm installing Ubuntu 11.10 alongside it.  I went into GParted to resize the Vista partition but found that it had an exclamation point in a red circle next to it:   I tried to resize it but the Resize/Move button was grayed out.  I thought that this might have to do with the exclamation point so I looked at the information and found a load of errors:   What does the exclamation point mean?  How can I resize my partition?  UPDATE 1  I booted into Safe Mode with Command Prompt in Vista as suggested by eug and ranchkdsk c: /F /R (/F Fixed errors on the disk, /R Locates bad sectors and recovers readable information) and got:  Chkdsk cannot run because the volume is in use by another process.  Would you like to schedule this volume to be checked the next time the system restarts? (Y/)   When I reboot it doesn't run chkdsk, I'm selecting Safe Mode with Command Prompt again because I can't boot normally.  How can I stop it being used?  UPDATE 2  I tried doing what Luis suggested.  Here is the terminal output:  ubuntu@ubuntu:~$ sudo ntfsfix /dev/sda3 Mounting volume... OK Processing of $MFT and $MFTMirr completed successfully. NTFS volume version is 3.1. NTFS partition /dev/sda3 was processed successfully. ubuntu@ubuntu:~$ sudo apt-get install ntfsprogs Reading package lists... Done Building dependency tree        Reading state information... Done Note, selecting 'ntfs-3g' instead of 'ntfsprogs' ntfs-3g is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   I still cannot resize the partition.  Ntfsfix doesn't look like it did anything and ntfsprogs is already installed.  I can access the hard drive and do backups with the Ubuntu Live CD which I'm using now.  Any other suggestions?  UPDATE 3  I attempted to use the ntfsresize command but I got an error:  ERROR: Volume is scheduled for check. Run chkdsk /f and please try again, or see option -f.   This is because I ran chkdsk in Vista Safe Mode with Command Prompt and it couldn't do it because it was in use but offered to do it at the next reboot.  I reboot and I get to the menu to the menu to choose whether to Start Windows Normally, Safe Mode, etc instead of checking the disk.  Is there any way I can unschedule?  EDIT  I just found out that I could use the -f option to force it.  It failed and told me to run chkdsk /f in Windows and then reboot twice.  But the chkdsk program is in the partition I am trying to check so it says it is in use. Helpful flags not getting updated after getting 'close question' privilege