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I Finite Extensions - A&F Example 44.2 ... ...

  1. Jul 3, 2017 #1
    I am reading Anderson and Feil - A First Course in Abstract Algebra.

    I am currently focused on Ch. 44: Finite Extensions and Constructibility Revisited ... ...

    I need some help in fully understanding Example 44.2 ... ...


    Example 44.2 reads as follows:



    ?temp_hash=1ccd86d5e2cab800447e497946771ed4.png





    I am trying to fully understand EXACTLY why

    ##\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6} \}##

    is the basis chosen for ##\mathbb{Q} ( \sqrt{2}, \sqrt{3} )## ... ...


    I can see why ##1, \sqrt{2}, \sqrt{3}## are in the basis ... and I understand that we need ##4## elements in the basis ....

    ... BUT ... why EXACTLY do we add ##\sqrt{6} = \sqrt{2} \cdot \sqrt{3}## ... an element that is already in the set generated by ##1, \sqrt{2}, \sqrt{3}## ...

    ... indeed, what is the rigorous justification for adding ##\sqrt{6}## ... why not add some other element ... ... for example, why not add ##\sqrt{12}## ...


    Hope someone can help ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Jul 3, 2017 #2

    fresh_42

    Staff: Mentor

    ##\sqrt{12} = 2\sqrt{3}## which means, they are ##\mathbb{Q}-##linear dependent. We need an element, that is ##\mathbb{Q}-##linear independent of ##\{1,\sqrt{2},\sqrt{3}\}##. It doesn't have to be ##\sqrt{6}##, but in any case we have to incorporate ##\sqrt{6}## somehow, as it is contained in ##\mathbb{Q}[\sqrt{2},\sqrt{3}]##. ##\sqrt{24}## would work.
     
  4. Jul 4, 2017 #3
    Thanks for the help, fresh_42 ... notion of ##\mathbb{Q}##-Linear Independence clarifies things for me ...

    Peter
     
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