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Show that $h(Q(\sqrt{10})) = 2$

use this theorem

Theorem: Let $K = Q(\sqrt{d})$ where $d$ not in $(0; 1)$ is a square free integer. If $W:=\{\sqrt{d} \text{ whenever } d \equiv 2\text{ or }3 \bmod{4};\}$ or $W:=\{1+ \sqrt{d}/2\text{ whenever }d \equiv 1 \bmod{4} \}$ then $OK = Z[W]$.

I am trying to work it out and this is my solution could you please help with using the theorem:

Note that this has discriminant $4 * 10 = 40$. The degree of this over $Q$ is $n = 2$, and there are $s = 2$ real embeddings and $t = 0$ imaginary pairs of embeddings.

So, every class of ideals contains one with norm at most

$(4/\pi)^0 * (2!/2^2) * \sqrt{40} = \sqrt{10}$

(which is just above 3).


Hence, we need to factorize the primes at most 3.

Since $10 \not\equiv 1 \bmod{4}$, we know that the number ring is $Z[\sqrt{10}]$, and $t^2 - 10$ is the minimal polynomial for $\sqrt{10}$ over $Q$.

(i) Working mod $3$, $t^2 - 10 = t^2 - 1 = (t + 1)(t - 1)$ (mod 3).

$\Rightarrow \rm<3\rm> = \rm< 3, \sqrt{10} + 1 \rm> * \rm< 3, \sqrt{10} - 1 \rm >$

(ii) Working mod 2, we have $t^2 - 10 \equiv t^2$ (mod 2). $\Rightarrow \rm<2 \rm> = \rm <2, \sqrt{10} \rm >^2 $

Suppose that $\rm<2, \sqrt{10}\rm>$ were principal; that is, $\rm< 2, \sqrt{10} \rm> = \rm <a + b\sqrt{10}\rm >$ for some $a, b$ in $\mathbb{Z}$. So, $N(\rm <a + b\sqrt{10} \rm>^2) = N( \rm <2 \rm>) = 4$. $==> N(\rm <a + b\sqrt{10} \rm>) = 2$ $==> a^2 - 10b^2 = 2 or -2$, neither of which have solutions (try reducing mod 10). So, we have a contradiction, and $p := \rm <2, √10 \rm>$ is not principal.


Next, check that $\rm <2, √10 \rm> * \rm <3, \sqrt{10} + 1 \rm> = \rm <-2 + \sqrt{10} \rm>$, and so $[\rm <-2 + \sqrt{10} \rm >] = [p]^{-1}$. (We have a similar result for the other prime ideal above 3.)

That means that every class of fractional ideals either contains a principal ideal or $p$; thus the class group equals $\{[O], [p]\}$. So, $h = 2$, as required.

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You really need to clean up the formatting of this question. In particular, it is very hard to tell what you are asking. It seems you just copied your solution to a homework problem into the question. As it stands, I'm voting to close. –  Potato Mar 11 '13 at 1:08
2  
@Potato, seems to me OP is asking whether the solution presented is correct. –  Gerry Myerson Mar 11 '13 at 1:45
    
@GerryMyerson Sorry, that's more clear with the new formatting. I didn't see the relevant sentence before. –  Potato Mar 11 '13 at 2:08
    
@Potato, I didn't see the original version --- I'm sure your comment was pertinent when you made it. –  Gerry Myerson Mar 11 '13 at 2:35
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