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

[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:
https://www.physicsforums.com/attachments/6862

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

 

[Euge]

Hi Peter,

It looks like the authors are illustrating the tower law of fields in this example. Check to see if they mentioned the tower law before it. The law states that if $K\subset F \subset L$ is a tower of fields, then $[L : F] = [L : K] [K : F]$. The standard proof of this theorem (in the case $[L : F] < \infty$) shows that if $\{u_1,\ldots, u_m\}$ is a $K$-basis for $F$ and $\{v_1,\ldots v_n\}$ is an $F$-basis for $L$, then the set $\beta = \{u_iv_j: 1\le i \le m, 1\le j \le n\}$ is an $F$-basis for $L$.

Going back to the example, the tower $\Bbb Q \subset \Bbb Q(\sqrt{2})\subset \Bbb Q(\sqrt{2},\sqrt{3})$ is used in the first paragraph. Noting that $\{1, \sqrt{2}\}$ is a $\Bbb Q$-basis for $\Bbb Q(\sqrt{2})$ and $\{1, \sqrt{3}\}$ is a $\Bbb Q(\sqrt{2})$-basis for $\Bbb Q(\sqrt{2},\sqrt{3})$, the set $$\beta = \{1\cdot 1, \sqrt{2}\cdot 1, 1\cdot \sqrt{3}, \sqrt{2}\cdot \sqrt{3}\} = \{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}$$ is a $\Bbb Q$-basis for $\Bbb Q(\sqrt{2}, \sqrt{3})$.