1
$\begingroup$

I'm completely stuck on this question:

Find the elements and the structure of the group of units in the ring of algebraic integers of number field $\mathbb Q (\sqrt {-7})$.

Are the group of units in the ring of algebraic integers of $\mathbb Q(\sqrt 5)$ finite? Why?

For the first part, I think it might be something to do with Dirichlet and for the second part, I believe the answer is yes but I don't know how to justify it. Pretty stumped.

Any help would be great

$\endgroup$
  • 3
    $\begingroup$ For the first part you'd be looking for elements $x+\sqrt{-7}$ with $x^2 + 7y^2 =1$ For the second part units would be elements $x+y\sqrt{5}$ which are solutions to $x^2 - 5y^2 = \pm 1$ and algrebraic integers. $\endgroup$ – sharding4 Mar 16 '18 at 20:04
  • $\begingroup$ @sharding4 The ring of integers in $\;\Bbb Q(\sqrt{-7})\;$ is not $\;\Bbb Z[\sqrt{-7}]\;$ but rather $\;\Bbb Z\left[\frac{1+\sqrt{-7}}2\right]\;$ ... Something similar happens with the other number field. Observe that we have both $\;5=-7=1\pmod4\;$ . $\endgroup$ – DonAntonio Mar 16 '18 at 20:51
  • $\begingroup$ @DonAntonio Thank you for pointing that out. That whole comment is kind of sloppy and doesn't quite make sense. So what we are actually looking for is solutions in integers to $x^2+7y^2 = \pm 4$ and $x^2-5y^2 = \pm 4$. Perhaps someone should come along and give the OP additional help. $\endgroup$ – sharding4 Mar 16 '18 at 20:59
0
$\begingroup$

Since $\;K:=\Bbb Q(\sqrt5)\;$ is a totally real number, there are only two embeddings into an algebraic closure of $\;\Bbb Q\;$ , and both real of course. Thus, here we have $\;r_1=2\,,\,r_2=0\implies\;$ the (multiplicative) group of integral units is isomorphic to $\;F\times\Bbb Z\;$ , with $\;F\;$ the finite cyclic groups of roots of unit contained in $\;K\;$, which are then only $\;\pm1\;$.

In $\;L:=\Bbb Q(\sqrt{-7})\;$ we have zero real embeddings and two, conjugate, complex non-real ones, and thus $\;r_1=0,\,r_2=1\implies\;$ the group of integral units has rank equal to zero and is thus the finite, cyclic group of all the roots of unit contained in $\;L\;$.

As before (since, again, $\;-7=1\pmod4\;$ , the ring of integers in $\;L\;$ is

$$\mathcal O_L=\Bbb Z\left[\frac{1+\sqrt{-7}}2\right]$$

and since the conjugate couple of embedding is determined by $\;\sqrt{-7}\mapsto\pm \sqrt{-7}\;$ , we have that

$$\mathcal N^L_{\Bbb Q}\left(a+\frac12b+\frac{\sqrt{-7}}2b\right)=\left(a+\frac12b+\frac{\sqrt{-7}}2b\right)\left(a+\frac12b-\frac{\sqrt{-7}}2b\right)=$$

$$=a^2+ab+2b^2=\pm1\iff a^2+ab+(2b^2\mp1)=0$$

The above quadratic in $\;a\;$ has a real (in fact, we need integer) solution if

$$\Delta=b^2-8b^2\pm4\ge0\implies\begin{cases} b^2\le\frac47\\{}\\ b^2\le-\frac47\end{cases}\implies |b|\le\frac2{\sqrt7}\iff b=0$$

and thus the only units are $\;a^2=1\iff a=\pm1\;$

$\endgroup$

Your Answer

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