## 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.

———————————–

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.

### 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?
———————————————————————

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.