Galois Theory for Dummies - Part I
September 27th, 2007
Galois Theory is an algebraic theory providing a powerful connection between fields and groups. Many complicated problems involving fields can be converted to (possibly) simpler problems involving groups.
Galois Theory, albeit being extremely beautiful (it answers some very elementary questions, which were all open problems until its arrival), is far from being wildly known (i.e. by non-mathematicians). One of the reasons for this, in my opinion, is that is it a vast subject and most books on the topic are filled with too many details and so are inadequate for a quick read.
I wrote this short article as an introduction to Galois Theory. It is not very easy to read, but as it is very short (only 4 pages), I believe mathematically oriented readers, without excessive mathematical background will be able to read it. I chose to write it as a Latex document, instead of simple HTML residing directly in my site, as it contains a lot of complex mathematical notations.
The article consists of only the first part of the introduction to the topic. I believe that perhaps two more parts (of about equal length) will be necessary to complete the introduction.
If you like this article, let me know (this will enhance the chances of me actually writing down the remaining parts). If anything is not clear - please ask.
Enjoy!
October 9th, 2007 at 12:19 am
I liked the article. Quite informative, and offers a nice introduction. On a related note, do you perchance know which of the ‘Algebra B’s touches on Galois theory?
A little criticism: The proof that you provided in part 3, as well as the related lemmas, are quite elegant. However, I feel that it is much easier for the reader to ‘believe’ in the existence of such a polynomial without proof than in the theorem about it. Therefore, unless it was necessary for developing the rest of the theory, I think that part 3 might have been better fit in a later installment. Nonetheless, it made for an enjoyable read by itself.
October 9th, 2007 at 12:50 pm
Dan,
The topic is covered in the (excellent) Algebra B 2 course.
For the second part of your comment: most of the algebra books I read did postpone the proof of the existance of the said polynomial until the end. Actually, as is mentioned in the introduction of the article I chose to present many thing in the reverse order. Now, I completely agree with you that the existance of such polynomials is not as important to the subject as the fundamental theorem, for example, but I really liked the proof!
November 19th, 2007 at 1:47 am
Thank you very much for the great pleasure I had reading the article. I try to imagine the level of excitement I could have gotten should I understood it.
August 11th, 2008 at 10:56 am
There is also a really nice proof for Eisenstein’s Criterion (with coefficients in Z) in Wikipedia.
http://en.wikipedia.org/wiki/Eisenstein%27s_criterion#Basic_proof
June 2nd, 2009 at 11:16 pm
Hi
Some observations:
1.-
Page 2, line 2 reads:
“The problem is that g(x) is not irreducible.”
I think the text must be
“The problem is that g(x) is irreducible over Q.”
or perhaps
“The problem is that g(x) is not reducible over Q.”
2.- I think that the symbols “” are swapped on the
first line of Gauss´s Lemma proof
“()”
That is:
(=>) If f(x) splits on Z, obviously it splits on Q, since Z is contained on Q.
3.- Whitout loss of generality I must assume that
“WLOG”
stands for
“Whitout Loss Of Generality”,
right?
4.- I can cope with the arguments up to page 3 where is claimed that
“Let v = \sigma^{-1}(s). It is clear that v \times u = 1″
justifying the claim with
“(if this is not clear to you at this point try to prove it)”
well I recognize that I tried to prove it and was unable to do it, so:
can you sketch the proof giving working hints to a standard proof?
5.- About the “Fields Automorphisms” sections: elsewhere I found a
figure relating what seemed to me as extension fields (perhaps splitting fields)
and Galois Groups, but there was not any explication and I can not
reformulate the underline relation myself. So, it would be great if you
can explain that relation.
Regards
César
June 4th, 2009 at 12:51 pm
Cesar,
1. The original phrasing is correct - g(x) is not irreducible. We are using it in order to create the polynomial f(x) which will be irreducible.
2. You are right, thanks!
3. Right.
4. v*u = sig-1(sig(u)*sig(v)) = sig-1((p+f(x))*(s+f(x))) = sig-1(p*s+f(x)) = sig-1(1+f(x)) = 1.
5. As the article says - the Galois group is the automorphism group of E/F. Part 2 of the article will explain the importance of the Galois group.
June 4th, 2009 at 5:24 pm
Thanks for your answer…
At+
César