12
$\begingroup$

The text of the exercise is the following:

Show that $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain, and that the identities $21 = (4+\sqrt{−5})\cdot(4 − \sqrt{−5})$ and $21 = 3 · 7$ represent two factorizations of $21$ into pairwise non-associate irreducible elements.
How does the ideal $(21)$ factor into prime ideals in $\mathbb{Z}[\sqrt{-5}]$?
Determine the order of the subgroup of $\textrm{Cl}(\mathbb Z[√−5])$ generated by the classes of the primes dividing $(21)$.
Can you find an ideal in $\mathbb{Z}[\sqrt{-5}]$ whose class is not in this subgroup?

Got lost while trying to factor $(21)$. Any clues?

$\endgroup$
  • $\begingroup$ Maybe try ideals generated by two of the four irreducibles? $\endgroup$ – Nishant Oct 14 '14 at 21:30
10
+100
$\begingroup$

Note that $(21)=(3)(7)$, so that it suffices to factorize $(3)$ and $(7)$. This is easier because $3$ and $7$ are primes in $\mathbb Z$.

Is $(3)$ a prime ideal ? Inspection reveals it is not : if $z_1=1-\sqrt{-5}$ and $z_2=1+\sqrt{-5}$ are both not in $(3)$ but $z_1z_2=6$ is. Straightforward computations show that $(3)=(3,z_1)(3,z_2)$. Note that $(3,z_1)$ and $(3,z_2)$ are the same thing as $\lbrace x+y\sqrt{-5} \ | \ x,y\in{\mathbb Z}, y\equiv -x\ ({\sf mod} \ 3) \rbrace$ and $\lbrace x+y\sqrt{-5} \ | \ x,y\in{\mathbb Z}, y\equiv x\ ({\sf mod} \ 3) \rbrace$ respectively, and those two ideals are easily seen to be prime.

Similarly, one obtains the factorization $(7)=(7,3-\sqrt{-5})(7,3+\sqrt{-5})$. In the end, the complete Dedekind factorization of $(21)$ is

$$ (21)=(3,1-\sqrt{-5})(3,1+\sqrt{-5})(7,3-\sqrt{-5})(7,3+\sqrt{-5}) \tag{1} $$

Call those factors $J_1,J_2,J_3,J_4$ in that order. For an ideal $J$, denote its ideal class by $c(J)$ and let $c_i=c(J_i)$. Straightforward computations show that $J_1^2=(2+\sqrt{-5})$, $J_3^2=(-2+3\sqrt{-5})$, $J_1J_3=(1+2\sqrt{-5})$, so $c_1=c_2=c_3=c_4$ and the subgroup generated by the $c_i$ is a two-element group.

One can also show that the whole class group consists only of two elements, but that’s a little harder.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.