Let $p, q$ be distinct primes and $L := \mathbb{Q}(\sqrt[3]{p}, \sqrt{q})$.  I want to show that $L = \mathbb{Q}(\sqrt[3]{p}\cdot\sqrt{q})$. Of course one inclusion is clear, but how do I get the other one?
Let $p,q$ be primes, $p≠q$, then I have to show that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$  So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient.  This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields.  Could someone maybe give me a hint about how to proceed? Thanks a lot!!!
A projection operator $P$ is defined as $P^2$=$P$. Use this definition to find the eigenvalues of this operator.   In this question is it necessary to define what the projection operator is? And won't the eigenvalue just be zero? I need assistance in integrating $ \frac{x \sin x}{1+(\cos x)^2}$ Use of inequality $1 - \cos (x) \leq x^2 /2$ In  answer the value of $1 - \cos(x)$ has to be evaluated in order to find its upper limit, if it exists.  In particular, $x = 2 \pi / n$. The answer is related to the length of a side of a regular $n$-gon inscribed into a unit-radius circumference; because the perimeter of the $n$-gon is always less than $2 \pi$, the single side must always be less than $2 \pi / n$.  The inequality  $$1 - \cos (x) \leq \displaystyle \frac{x^2}{2}$$ (1)  is used and the proof is completed with   $$2(1 - \cos(x)) \leq (2 \pi / n)^2$$  $$\sqrt{2(1 - \cos(x))} \leq 2 \pi / n$$  But it is well known that the cosine is a function $f(x) \in [-1;1]$, so $1 - \cos (x) \in [0,2]$. By using this information, we would obtain  $$1 - \cos (x) \leq 2$$ (2)  The proof would provide  $$2(1 - \cos(x)) \leq 4$$  $$\sqrt{2(1 - \cos(x))} \leq 2$$  which is a completely different result.   Why in that case it is preferable to use (1) instead of (2)? How to choose when it is convenient to use (1) and when to use (2) in a proof? I was thinking about writing a partial derivative macro that has a similar syntax to the . ()  The syntax could be like this:   \D{f}{x} expands to \frac{\partial f}{\partial x} \D{f}{x,y} expands to \frac{\partial^2 f}{\partial x\,\partial y} \D{f}{{x,2},y} expands to  \frac{\partial^3 f}{\partial x^2\,\partial y} et cetera for any number of derivatives.   Bonus points for an optional argument so that for instance   \D[subscript]{f}{{x,2},y} expands to f_{xxy}   I looked at the  package but the syntax is a little tedious.  For instance, to get \frac{\partial^3 f}{\partial x^2\,\partial y} with that package I would need to do \md{f}{3}{x}{2}{y}{}.  And it doesn't generalize to three variables.  Before I dive in I thought I'd ask the hive and see what's been done. How to compare the output of \ref to a number? I want to compare the output of \ref to a number in LaTeX. I thought it would be easy but after spending lots of hours surfing the net I still don't know the answer. I found similar questions on this site but I would prefer a solution that does not require another package to be included, so that I can see clearly how the \ref command (and some others) works. I tried a lot of things, some of which I share below, therefore my question is rather long, I'm sorry about that. First, I tried the following codes (the resulting error messages are shown after a % sign):  \begin{equation} \label{eq:myeq}   a=b   \end{equation}  \ifnum\ref{eq:myeq}=1 true\else false\fi % error: "Missing number, treated as zero."  \expandafter\ifnum\ref{eq:myeq}=1 true\else false\fi % error: "Missing number, treated as zero."  \edef\blabla{\ref{eq:myeq}} % error: Incomplete \iffalse; all text was ignored after line 42. \ifnum\blabla=1 true\else false\fi \expandafter\ifnum\blabla=1 true\else false\fi   Of course, \ref{eq:myeq} gave the expected result, namely 1.  Then, I found the definition of \ref using the latexdef ref command:       \ref:   macro:#1->\expandafter \@setref \csname r@#1\endcsname \@firstoftwo {#1}   Further examining \@setref and \firstoftwo, I concluded that the equation number (together with the page number) is stored in a control sequence called \r@eq:myeq. More precisely, the content of \r@eq:myeq is {1}{1} where the first number is the equation number and the second number is the page number. (The output of \expandafter{\csname r@eq:myeq\endcsname} is 11.) If I understand well, in order to obtain the equation number we have to expand \csname r@eq:myeq\endcsname by only two levels to get {1}{1} (the first level of expansion yields \r@eq:myeq), and then we choose the first argument by \@firstoftwo. Since there is no simple way to expand a control sequence by only one level, it is done in a rather circuitous (say tricky) way by \ref. This I could also reproduce by the following code:  \def\sth#1#2{\expandafter #2#1} \makeatletter \expandafter\sth\csname r@eq:myeq\endcsname\@firstoftwo \quad \ifnum1=\expandafter\sth\csname r@eq:myeq\endcsname\@firstoftwo true\else false\fi \makeatother   which resulted in 1   true. So far so good but there are still a few questions. First of all, if the above code works then why does the code with \ifnum1=\ref{eq:myeq}... fail (see above)? Seemingly, the two "algorithms" are doing the same. Second, the above code works only if the auxiliary file (with aux extension) contains the label definition. If it doesn't (the reference is undefined) then LaTeX gives the following error message: Missing number, treated as zero. How can I write a code that works even if the reference is undefined? Thank everybody who read all my above struggle and I would appreciate if someone shed light on the points that were not clear to me and who corrected me where I was wrong and confirmed where I was correct. Unfortunately the whole command syntax of (La)TeX is still very strange to me. Problem with MGI CRS on the Balkan How to import sextante(processing) for QGIS 1.9 (2.0) High CPU usage by "System" and "System interrupts" (caused by ACPI.sys) I have a laptop, which was running Windows 8.1 x64 without any problems. Now with Windows 10 x64 installed, Task Manager constantly shows unusual CPU usage by "System" and "System interrupts". To solve this, I already tried the following, without success:   Disabling and uninstalling all non-essential drivers. Installing newer drivers than the ones that were automagically installed (if available). Disabling/enabling fast boot option. Disabling all of the non-essential services. Sysprep. Resetting BIOS to defaults and various combinations of settings. Flashing BIOS to the latest available version. Clean install from the same media that I use for other PCs. Installing all of the updates offered in Windows Update to this day. Windows Performance Recorder / Analyzer.   I'm not very familiar with Windows Performance Analyzer, so I'm hoping someone here can point me in the right direction - what exactly should I look for, to figure out which device/driver is the culprit. Or, if there's any other approach to figuring out this problem?  For the brave souls, here's my  and a screenshot of the problem: Firefox doesn't launch in Ubuntu 13.04 Firefox starts when I use sudo in the terminal, but not from the launcher. All the other apps work like a charm, just not Firefox. How can I fix this? Proof for maximal ideals in $\mathbb{Z}[x]$ I have been trying to prove the following theorem:      Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where $p$ is prime integer and $f$ is primitive integer polynomial that is irreducible modulo $p$.    Idea: I tried to establish a homomorphism $\phi: \mathbb{Z}[x] \rightarrow \mathbb{F}$. Since $\mathbb{F}$ is a field it has characteristic $p$ and so integer prime p are mapped to 0 in $\mathbb{F}$. Hence $p\in \ker \phi$. Next we consider $\phi': \mathbb{Z}[x] \rightarrow \mathbb{Z_p}[x]$ and pick an arbitrary maximal ideal $M\in \mathbb{Z}[x]$. So, $\phi'(M)$ is maximal as long as $p \in M$ by correspondence. 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