How to prove normality of a field extension with Q(\sqrt{2+\sqrt{2}}):Q?

  • Thread starter Thread starter gonzo
  • Start date Start date
  • Tags Tags
    Field Normal
Click For Summary
To prove the normality of the field extension Q(√(2 + √2)):Q, it is essential to show that the conjugate √(2 - √2) is also included in the field. The discussion highlights various approaches, including solving for √(2 - √2) using its minimum polynomial and manipulating radicals to express it in terms of known elements. The Galois group perspective is also mentioned, but participants express confusion about its application without confirming that the necessary elements are in the field. Ultimately, understanding the relationships between the roots and their expressions is crucial for establishing normality. The conversation reflects the complexity of the problem and the need for a solid grasp of Galois theory concepts.
  • #31
gonzo said:
To put it another way:

\beta=\sqrt{2}/\alpha
\alpha\mapsto\beta
\beta\mapsto\sqrt{2}/\beta

This last one doesn't follow, you're assuming that this map fixes \sqrt{2}, but it doesn't. To find what this map does to \beta, use \beta=\alpha^3-3\alpha, or work out where sqrt(2) goes using \sqrt{2}=\alpha^2-2.
 
Physics news on Phys.org
  • #32
im puzzled as to what you all are doing. doesn't the remark that b = sqrt(2)/a prove that the field Q(a) also contains b? hence is normal?

or am i way behind? certainly my earlier comments were incorrect, since i confused a biquadratic extension, namely one where both square roots adjoined were of elements in Q, with this situation where one adjoins the square root of 2, i.e. of an element of Q, and then a square root of an element of the larger field Q(sqrt(2)).

Indeed this is a standard way to get a non normal extension, generated by the real 4th roots of 2, for instance.


But if you want the galois group elements here, it seems they are:

1) identity

2) sqrt(2) fixed, hence a goes to the other root of its minimal polynomial over Q(sqrt(2)), namely -a. thus b goes to -b.

3) sqrt(2) goes to the other root of its minimal polynomial over Q, namely -sqrt(2). a goes to some root of its transformed minimal polynomial under the previous map on sqrt(2), say a goes to b. then since a field map takes quotients to quotients, b = sqrt(2)/a goes to -sqrt(2)/b = -a.

4) This one also takes sqrt(2) goes to the other root of its minimal polynomial over Q, namely -sqrt(2). Then a must go to the other root of its transformed minimal polynomial under the previous map, namely a goes to -b. then b = sqrt(2)/a goes to -sqrt(2)/-b = a.

the orders of these elements seem to be 1,2,4,4. In particular the group is cyclic of order 4.


the key observation being that of gonzo, that b = sqrt(2)/a.


I have not been able to follow all your discussion but to compute galois groups of an extension like Q(c,a) by hand, one simply sends c to some other root of its minimal polynomial over Q, then follows what happens to the minimal polynomial of a over Q(c) and sends b to some root of that transformed polynomial.


am i in the ball park? i admit i always was weak at algebra. which is why i wrote the book on my webpage. my book is full of thorough details, and shares everything i myself understand, but to understand things well i recommend reading a book by someone who gets it, like artin, or jacobson.

:smile:
 
Last edited:
  • #33
Thanks, that was what I was confusing ... if a goes to b, then b goes to -a, I missed that. That makes it easy.

Thank you.
 
  • #34
By the way, Mathwonk, can you give me a link to your web page, I would be interested in looking through it if you have some of this stuff explained. It always helps to get a different perspective.
 
  • #35
goto http://www.math.uga.edu/~roy/

download the algebra course notes 843-2, and read pages 43-46 for a 2 calculations of the galois group of X^4-2.

the whole section 19, pages 34-46 is relevant.:smile:
 
  • #36
Thanks, I'll take a look at it.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
48
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K