## 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.