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Suppose one doesn't know any Galois theory, but they are familiar with the basic theory of field extensions. Then how would one justify the following picture?

enter image description here

I'm pretty sure they're just saying these are the known subfields, and there could be more but they just write down those ones (which seems weird, which is why I'm asking). Or have they deduced that this is a comprehensive list of subfields just by using basic theory? This occurs on page 573 of Dummit and Foote's Abstract Algebra.

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$\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} : a,b,c,d\in \mathbb{Q}\}$

Take some arbitrary element and show that this has to be in one of those specified fields...

The point is that whether the choosen element has contribution from $\sqrt{2}$ or $\sqrt{3}$ or $\sqrt{6}$

Take $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$

  • suppose $b=c=d=0;a\neq 0$ then we just have $a$ so this element is in $\mathbb{Q}$

  • suppose $a=c=d=0;b\neq 0$ then we just have $b\sqrt{2}$ so this element is in $\mathbb{Q}(\sqrt{2})$

  • suppose $a=b=d=0;c\neq 0$ then we just have $c\sqrt{3}$ so this element is in $\mathbb{Q}(\sqrt{3})$

  • suppose $a=b=c=0;d\neq 0$ then we just have $d\sqrt{6}$ so this element is in $\mathbb{Q}(\sqrt{6})$

Doing similar calculations as taking only two/three/four of $a,b,c,d$ being non zero would conclude there is no other field.

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  • $\begingroup$ This shows that those are subfields, but it does not show that there are no more... $\endgroup$ – Sam Nead Oct 25 '18 at 14:52

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