Tower theorem

Dear Prof Kim

I can show that [Q(\sqrt{2},\sqrt{3},\sqrt{5}):Q]=8. But [Q(\sqrt{6},\sqrt{10},\sqrt{15}):Q] = 4 not 8. Why is that?

Can you explain why and give me instruction to prove it please?

Thanks very much.

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Reply: It’s important at this point to understand precisely what goes into the tower theorem.

For the first case, we have

Q\subset Q(\sqrt{2})\subset Q(\sqrt{2},\sqrt{3})\subset Q(\sqrt{2},\sqrt{3},\sqrt{5}).

But the important point is that the square root we’re throwing in at each stage is not in the previous field. That is, \sqrt{3}\notin Q(\sqrt{2}) and \sqrt{5}\notin Q(\sqrt{2},\sqrt{3}). Of course, these statements must be proved.(Try it!)

Now let’s examine the tower

Q\subset Q(\sqrt{6})\subset Q(\sqrt{6},\sqrt{10})\subset Q(\sqrt{6},\sqrt{10},\sqrt{15}).

In this case, it’s true that \sqrt{10}\notin Q(\sqrt{6}), so that the degree of
Q(\sqrt{6},\sqrt{10}) is 4. But note that

\sqrt{10}/\sqrt{6}=\sqrt{5/3} and 3\sqrt{5/3}=\sqrt{15}. So in fact,

Q(\sqrt{6},\sqrt{10})= Q(\sqrt{6},\sqrt{10},\sqrt{15}).

This explains the degree 4.

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