I'm trying to solve the following exercise. Let $a$ and $b$ be relatively prime integers, and $i^{2}=-1 .$ Prove that the natural map $\mathbb{Z} \rightarrow \mathbb{Z}[i] /\langle a+b i\rangle$ induces an isomorphism of $\mathbb{Z} /\left(a^{2}+b^{2}\right) \mathbb{Z}$ with $\mathbb{Z}[i] /\langle a+b i\rangle$. And I'm totally lost. How should I proceed? I have a feeling that I should use the universal property, and take the quotient by the kernel of the natural map. This would induce an isomorphism with the image of the natural map. But I don't know what the kernel would be, and I honestly don't know how to work with rings like $\mathbb{Z}[i] /\langle a+b i\rangle$. Would anyone have a rigorous (and readable!) explanation?
I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following:     For any $\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1$, we have a ring isomorphism   $$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.$$   Could someone show me an isomorphism between these rings to prove this?
Transit visa requirements at Narita Airport Problem to launch java at Debian: "error while loading shared libraries: libjli.so" Is there a word for when an action has an effect opposite to the one intended? Is there a word or phrase to express the concept of an action having the opposite effect of the expected outcome?  For example, a drug taken to cure headaches that actually causes headaches, or an advertising campaign designed to deter smoking that leads to an increase in smoking. I'm trying to install Windows 7 with a 4GB flash drive, but my error that comes up is that my hard drive I'm trying to install on is in ext4. I need to format it to read NTFS. I can't seem to find any topics on how to format an active hard drive. I found a topic that explains how to move Ubuntu to a new drive, but it's a bit confusing to me. Choosing coordinate system to use in South Africa? What are the principal datums and projections used in South Africa including the names of geodetic datums and their projections? There are a few "automatically protect questions" requests out there, but I'd like to add another twist.  I'd like to request that questions that have three or more deleted answers by users with &lt;50 rep be automatically protected.  The reasoning behind this is that we have lots and lots of "not an answer" flags in the mod queue.  But we cannot protect a question from the mod queue, so we have to open each answer, evaluate if the question needs protecting, then protect it.    This process takes a long time, and can definitely be automated, as the only thing you really need to check is how many new users are posting non-answers or are spamming their awful products. Easy is to beginner as medium is to I want to display three levels in my game...but I don't know what the text should say.  easy -&gt; Beginner  medium -&gt; ?  hard -&gt; Expert   Basically, I want the form that describes the person playing the game, instead of the game itself. What adjective is best?  I have considered adept.  I doubt my users will know what adept means. Also, I need something in between expert and beginner. Adept falls into the expert category. Is there anything more common that I can use, that has the same impact as the other two words I mentioned?  Proficient would work, theoretically, but in day-to-day term, the levels "beginner - proficient - expert" won't sound great. Remove HTML tags from a String Editing Gsettings unsuccesful when initiated from cron Let $G$ be an abelian group of order n.   Let $\mathbb{C}^*$ be the multiplicative group of non-zero complex numbers, and suppose $f:G \rightarrow \mathbb{C}^*$ is a homomorphism. Prove that $\sum_{g \in G} f(g) = 0$ or $\sum_{g \in G} f(g) = n$.  I can prove this for cyclic groups but am unable to generalise to all abelian groups. Including the interaction but not the main effects in a model What are Harry, Ginny, Ron and Hermione's professions? After the end of the Second Wizarding War Harry Potter later married Ginny Weasley, and the couple had three children of their own: two sons James Sirius, Albus Severus, and one daughter Lily Luna. What are Harry and Ginny's professions? I heard that Harry became an Auror, but what about Ginny, Ron, and Hermione? Additional Boxes - Criteria, Longevity, Manipulability This feature has existed for a while (I'm not sure since when exactly), but I'm mainly interested in the Gen VI implementation (hence the tag).  Usually, at the start of a game, you have some number of Boxes available. In B/W, those were 8, in X/Y I think it was 15. Under certain conditions though, you receive additional Boxes that just appear at the end of your current Boxes list. Which brings me to Part 1 of my question:     1) What are the exact conditions for additional Boxes to appear and do you always get 8 additional Boxes? (Using Pokemon Bank reveals that in X/Y, the maximum number of Boxes is 31, which I have experienced to be unlocked in 2 steps of 8 Boxes each.)   Since I'm a person who really likes to order and organize stuff, my Boxes have names and are used to categorize my Pokemon. This results in some Boxes ending up empty for a while most of the time. So here's to     2) Do additional Boxes you receive stay forever or do they disappear if the conditions for them are no longer met?      3) If they disappear, is there a way (such as renaming or putting any Pokemon in the last Box) to make them stay no matter what, even if you released all your other Pokemon? Is there anyway to sync Outlook's calendar with Google's calendar? I've tried  before but it doesn't work properly. I'm using Windows 7 and Outlook 2010.  The thing is, it's really easy to check my Google Calendar on my phone since I have an Android terminal, but on my computer I prefer just opening Outlook instead of opening a web browser and logging into Gmail. I have an equation contained inside \[...\], which automatically makes a \sum with sub- and superscripts turn big--so that the summation sign looks awkward inside parenthesis.  Any idea how to make the parenthesis completely enclose the whole summation?  \documentclass{article} \begin{document} We have: \[ \sum_{i=1}^n i = (\sum_{i=1}^{n-1} i) + n = \frac{(n-1)(n)}{2} + n = \frac{n(n+1)}{2} \] \end{document} Is it possible to find a smooth function $p$ such that $$p(x)=0 \text{ if } x\le -1$$ $$p(x)=0 \text{ if } x&gt; 0$$ $$0\le p(x) \le 1 \text{ if } -1&lt;x&lt;1$$ $$p(x) = 1 \text{ if } -\frac{1}{2}\le x \le \frac{1}{2}$$ Hint: The following function has continuous derivatives of all orders $$f(x) = 0 \text{ if } x\le 0$$ $$f(x) = e^{-\frac{1}{x^2}} \text{ if } x&gt;0$$  My thought: Considering the left-half of the function $p$, I tried something like $e^{-\frac{1}{\tan^2(x\pi)}}$. The shape looks good but I am not sure about the smoothness at $-1$ and $-\frac{1}{2}$. For $x=-1$, since $\tan$ is infinitely differentiable so we have smoothness there. But for $x=-\frac{1}{2}$, I am not sure about the smoothness after &quot;gluing&quot; with $y=1$. Some webpages forces mobile version even I change user-agent. I changed user-agent in built-in browser and "Dolphin Browser HD",  I am checking it on page www.user-agent-string.info  Main problem is engadget.com, always in mobile version. I tried to clean cookies, restart browsers etc.  This is my results:   in built-in browser, user-agent - "desktop"   () doesn't work, redirects to mobile version   in "Dolphin Browser HD", user-agent - "desktop"    (NSFW: redtube.com) doesn't work, redirects to mobile version  (allegro.pl) works fine  in "Dolphin Browser HD", user-agent - "IPad"   () doesn't work, redirects to mobile version  (allegro.pl) works fine (NSFW: redtube.com) works fine  in "Dolphin Browser HD", user-agent - "custom": Mozilla/5.0 (Windows NT 6.0; rv:7.0.1) Gecko/20100101 Firefox/7.0.1   (engadget.com]) doesn't work, redirects to mobile version  (NSFW: redtube.com) doesn't work, redirects to mobile version  (allegro.pl) works fine    Maybe this pages using "adobe flash" user-agent detection ?  (I'am using Asus Transformer TF101, Android 3.2.1, Adobe Flash Player 11) Activity Monitor in Mavericks "Show Memory Usage" dock icon gone? I just purchased a new MacBook Pro with Mavericks installed and noticed that Activity Monitor doesn't have the "Show Memory Usage" dock icon option anymore.  I really liked this option since I used it to prevent disk swapping by closing applications when I had too many open.  What gives? Why is the MacTeX distribution so large? Is there anything smaller for OS X? I would like to know how to install LaTeX on Mac 10.6 or higher.  But the MacTeX is some 1.28 GB file! Is there a smaller distribution I could install?  What exactly is TeX Live? Show that if $|f(z)| \leq M |z|^n$ then $f$ is a polynomial max degree n I can't prove this statement, can anybody show me how to prove it?    $$f:\mathbb{C}\rightarrow \mathbb{C} \in \mathcal{O}(\mathbb{C}), \exists n\in \mathbb{N}, R &gt;0 , M&gt;0 : |f(z)| \le M|z|^{n} \ \ \forall |z|&gt;R \Rightarrow \deg(f)\le n $$  To show is that if there exists such an $M$, that then $f$ is a polynomial of max degree $n$. I started like this:   $$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ So if I put this into the inequality: $$|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .$$