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A Computational Introduction to Number Theory and Algebra (shoup.net)
25 points by echair on May 30, 2008 | hide | past | favorite | 13 comments


I'm soon starting my trek through every problem in the algebra text that Harvard's PhD prelim recommends for study:

Abstract Algebra by Dummit and Foote http://www.amazon.com/Abstract-Algebra-David-S-Dummit/dp/047...

I've started the first section of the first chapter, but that was only in a few hours of spare time. I'll be posting solutions by chapter soon and post my stories/insights on Hacker News. Here's section 1.1 (except the last problem, 36):

http://therobert.org/alg/1.1.pdf

Comments are appreciated. Better now than when I start the real journey. :)


D&F was my sophomore-year college algebra textbook so all I can say is, good luck!

It gets very hard very quick and some of the questions are intentionally above virtually everyone's head.


Cool, you're from the University of Chicago. What (math) field was your thesis in?


Chicago math students don't have to write theses. I guess they figure it's enough just to get through the curriculum.


Do you want a hiking partner? How fast are you planning on going?


That'd be great! I'm planning on spending a few weeks full-time on this (8 hours a day, or more if I can handle it). That should be a minimum of one section a day, but hopefully more, even a whole chapter on some days. It doesn't matter how fast, though, as if either falls behind we can just compare those solutions. Email me at technoguyrob[at]gmail[dot]com.


At that pace I think I would definitely be holding you back. Maybe I'll try to tackle something a bit less steep, like Mount Fraleigh.

Have you already studied some abstract algebra, or would you mainly be learning it through this project?


You can still read through my solutions and point out all my errors if you like, though. ;)

I've already had a couple classes, but I've never felt intimately comfortable with it. With several programming languages, I don't even need the occasional googling, nearly all the libraries (and of course basic syntax) are already in my head. When doing commutative algebra and Noetherian rings, I found myself looking back at ring theory for various properties about rings. Same goes for homological algebra: I had to keep going back to properties of modules. I understand it in more in terms of theorems and facts rather than mathematical intuition and maturity, and my goal is to change that around.


Note on the problem that a group G of even order contains an element of order 2. Partition G into classes {{x,x^{-1}|x\in G}. This is a partition since inverses are unique. For x\in G, call {x, x^{-1}} the class of x. At least one other element x besides the identity has a class of size one. Otherwise, the order of G would be 1 + 2*(# of classes of size 2) which is odd. Hence there is an x with x != 1 and x = x^{-1}; i.e., an element of order 2.


Thanks, that's exactly the kind of comments I was looking for. That was exactly what I thought in my head, but it came out very distorted and ugly on paper. I'll update the solution later. Do you want credit?


Credit for this is optional. Another way to see obtain a partition of G is to observe that the relation x ~ y if and only if x = y or x = y^{-1} is an equivalence relation on G. The equivalence classes induce the partition above. The point is that any two equivalence classes {x, x^{-1}},{y,y^{-1}} are either disjoint or equal, and their union is G. But this is obvious (the word 'obvious' means "I thought of it.")


hmmm why are you posting the solutions online? Apparently, this book happens to be a popular textbook in many schools http://www.google.com/search?hl=en&q=Abstract+Algebra+by...


I assume you meant to reply to me. Thanks, I'll set the robots file appropriately.

And I'm posting it for the same reason that guy posted his poker bot earlier. If you're determined enough, you can always cheat and that won't change whether there is a program or solutions manual online, but hackers and autodidacts can enjoy the intellectual sharing.




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