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

1. Jul 3, 2017

### Math Amateur

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:

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

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2. Jul 3, 2017

### 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.

3. Jul 4, 2017

### Math Amateur

Thanks for the help, fresh_42 ... notion of $\mathbb{Q}$-Linear Independence clarifies things for me ...

Peter

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