Basis for an extension field

  • Thread starter Prometheos
  • Start date
  • #1
13
0

Main Question or Discussion Point

I am having no luck understanding how to find the basis of a field adjoined with an element.

For example
Q(sqrt(2)+sqrt(3))
I know that if i take a=sqrt(2)+sqrt(3) that i can find a polynomial (1/4)x^4 - (5/2)x^2 + 1/4 that when evaluated at a is equal to zero.

So, from that I know the degree is 4 and the basis should have 4 dimensions.

{1, sqrt(2), sqrt(3), Y} where Y is the part I don't understand.
{1, sqrt(2), sqrt(3), sqrt(6)} is the actual basis, but how do you get sqrt(6) as Y?

I don't see how it is a linear combination of sqrt(2) and sqrt(3) as defined in the book I'm using. Since it should be a additive combination of scalars from Q times sqrt(2) and sqrt(3)... but the only way to obtain sqrt(6) is sqrt(2)*sqrt(3), in which neither are scalars as elements of Q.

Any help understanding this concept is greatly appreciated.
 

Answers and Replies

  • #2
200
0
Remember, you should not be able to get sqrt(6) as a Q-linear combination of sqrt(2) and sqrt(3) for otherwise the set {1,sqrt(2), sqrt(3), sqrt(6)} would be dependent and therefore not a basis. You know that sqrt(6) is in your field however (being the product of sqrt(2) and sqrt(3)) so if you can show it is independent from sqrt(2) and sqrt(3) over Q you will have your basis (since you already know the dimension is 4).
 
  • #3
13
0
So by putting sqrt(2) and sqrt(3) adjoined to the field, then it has to have multiplicative closure so sqrt(6) has to be there... so simple now that I think about it.

Thanks!
 

Related Threads on Basis for an extension field

Replies
12
Views
1K
Replies
1
Views
439
Replies
2
Views
573
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
654
Replies
18
Views
1K
Replies
15
Views
916
  • Last Post
2
Replies
35
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
Top