I want to prove that the polynomial $p(x) = x^{4} - 4x^{2} + 8x + 2$ is irreducible over $\mathbb{Q}(\sqrt{-2})$.  Following a series of hints given by my textbook, I start by using the fact that $\mathbb{Z}(\sqrt{-2})$ is a unique factorization domain. So then by a corollary to Gauss' Lemma, I observe that $p(x)$ is irreducible over $\mathbb{Q}(\sqrt{-2})$ iff it is irreducible over $\mathbb{Z}[\sqrt{-2}]$.  Suppose $p(x)$ has a linear factor $x - \alpha$ in $\mathbb{Z}[\sqrt{-2}]$. My hint says that I should be able to infer that $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ (this is obvious) and then conclude that $\alpha \in \{\pm 1, \pm 2, \pm \sqrt{-2}\}$. What I don't get is how it even makes sense to consider that $\alpha = \pm \sqrt{-2}$ in the integer field?
Question is :     Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]   What I have done so far is :  Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :  $p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$  i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.  so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$  I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(  Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$   i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$   $p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$ $p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$  $p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$    Once I prove that those are the only divisors then I would consider :    $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$ $p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$ $p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$ $p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$ $p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$ $p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$   So, no divisor of $2$ is a root..  Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$  suppose I have something like :  $$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$  But then, $bd=2\Rightarrow \text { b,d  are  a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$  But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  So, only possibilities are   $x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$ $x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$   But these are not possible...  This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...  So, please help me to clear this..  Thank you...   P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain.
Weighting features prior to SVM I'm building an object detector using HOG features and linear SVM. Some of the regions of the object are more "distinctive" so I would like to give more weight to the features extracted from those regions. (e.g. imagine we want to detect between mugs and glasses, then the most "distinctive" part is the mug's handle)  How could I do this? For sure there should be a more intelligent way than just replicating the features of those "distinctive" regions. I've lost my portal home, and I am invincible. How can I leave the nether? Proper way to exit iPhone application? I am programming an iPhone app, and I need to force it to exit due to certain user actions.  After cleaning up memory the app allocated, what's the appropriate method to call to terminate the application? What are the powers shown by the Rings of Power? What performance differences are there when installing with Wubi? HTML favicon won't show on google chrome Possessive of a word that is already possessive As far as I can tell downvoting is mostly useless.     For questions the only people who get downvoted are lower rep users, or incredibly stupid questions    most questions that are downvoted get closed anyway which should be sufficient downvoting doesn't have too much affect on your rep anyway a negative comment is more useful and more of a push too change or delete the question  For answers downvoting is often used incorrectly as well   people commonly competitively downvote (downvoting another answer to a question that you also answered) obviously not a good behavior downvoting (and upvoting) are contagious, meaning one downvote leads to a bunch of others even though downvotes don't effect rep that much they create a negative environment, no one wants their post downvoted, on the other hand criticizing comments, I think, are far more welcome      So basically in a nutshell a downvote serves no real purpose even if a post deserves it. Flagging, closing and commenting are much more useful to everyone including the poster than downvoting so why have it.    I now see from the answers and comments that downvotes do serve a purpose and would like to retract my request. "Fake" 3d effect in Photoshop/Illustrator with raster images I have a directory that contains the following:  x.pdf y.zip z.mp3 a.pdf   I want to delete all files apart from x.pdf and a.pdf. How do I do this from the terminal? There are no subdirectories so no need for any recursion. Formatting Keyboard Shortcuts for Comments I was thinking along the lines of using typeid() but I don't know how to ask if that type is a subclass of another class (which, by the way, is abstract) How do I execute a program X times in the terminal?  I read that I have to do a bin/bash txt, but I don't know how to execute this X times. How can I navigate my scene FPS style similar to how Unity does? 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This can be simply written as $P(Y=y) = Bin(y+m, m+n, p)$. Hence, given equal success probabilities, the sum of two independent binomially distributed random variables is binomial, but also their difference, just shifted to the left.  This question here  is actually the same as mine, but received no answer and only the comment that there would be no simple formula. But the above formula looks pretty simple to me.   Is it correct that for the case of equal success probabilities, the above equations actually describes the distribution of $Z=X-Y$? I read in a book that $Z$ could not be binomial distributed because it had negative support. Is it right to call it a shifted binomial? How do websites block selecting text and how do I unblock that? There's a website () that somehow blocks selecting text. Also it blocks Ctrl+A to select everything, there is also no “Copy” in the popup menu.  What I have tried:  Some sites use JavaScript to block selection. 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It’s pretty annoying when a website does such things and forces me to type technical terms and names into google by hand. It disturbs my workflow. Examples proving why the tensor product does not distribute over direct products. I recently read about the result that the tensor product distributes over direct sums. I was curious if it also distributes over direct products, but google tells me it doesn't.   What are some simple counterexamples to why this property isn't true? I know that there is a natural homomorphism  $$ \left(\prod M_i\right)\otimes N\to \prod (M_i\otimes N) $$ given by $(\prod m_i)\otimes n\mapsto \prod (m_i\otimes n)$ when $M$ and $N$ are modules over some commutative ring $R$. Are there standard examples where this homomorphism is not injective/surjective and hence not an isomorphism?