Monthly Archives: February 2010

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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Cubic curves

Dear Professor Kim,

Thank you for your reply. I have another question, this time regarding the group law on a cubic, C. I am happy with showing that the points of C form a group. However I am a little unsure of the ‘simplified’ group law as described in the textbook. I believe that if we take O=(0,1,0) as the identity element we can treat the cubic as an affine curve. However as this corresponds to the ‘line at infinity’ in projective 2-space, what point (x,y) does it correspond to in affine 2-space?
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Reply:

IF C is the curve Y^2Z=X^3+axZ^2+bZ^3, in P^2, then it meets the line Z=0 in the point O=(0:1,0) as explained in the book. So we then express C as a union

C=\{O\}\cup C_0

where C_0=C\cap \{(X:Y:Z) | Z\neq 0\}. We then note that
\{(X:Y:Z) | Z\neq 0\}\simeq A^2 by the map

(X:Y:Z)\mapsto (X/Z,Y/Z).

Then C_0 goes to the curve in A^2 with equation

y^2=x^3+ax+b.

Of course the origin does not lie on this curve, because this affine curve corresponds exactly to the complement of the origin in C.

Does this answer your question? Let me know again if you’re still curious.

Question on canonical forms for quadrics

Here is a question from Dominque Miranda.

And here is my reply.