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Let $R=\mathcal{O}(K)$ be the ring of the integers of $K=\mathbb{Q}[\zeta_8]$, where $\zeta_8=e^{2\pi i/8}=\sqrt{2}/2(1+i)$ is a primitive eighth root of unity in $\mathbb{C}$. It can be shown that $R$ is a P.I.D.

Let $\mathscr{P}$ be the ideal $\langle \zeta_8-1\rangle$ and let $$ \mathscr{P}^{-2}=\{x\in K\mid (\zeta_8-1)^2x\in R\}. $$

Claim: $\mathscr{P}^{-2}/R\cong R/\mathscr{P}^2\cong \mathbb{Z}/4\mathbb{Z}$.

I have absolutely no idea how to prove this!

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Which isomorphism do you have trouble with? Can you put words to what is giving you difficulty? –  Hurkyl Oct 10 '12 at 2:54
    
I don't understand either isomorphism. What are the four cosets of $\mathscr{P}^{2}$ in $R$? The four cosets of $R$ in $\mathscr{P}^{-2}$? –  Clinton Boys Oct 10 '12 at 3:16
    
Would also like to know how to show $R/\mathcal{P}\cong \mathbb{F}_2$. –  Clinton Boys Oct 10 '12 at 3:36
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It might help you to get started to show that 2 is in $\mathscr{P}$. –  David Loeffler Oct 10 '12 at 10:02

1 Answer 1

up vote 1 down vote accepted

Well, ${\zeta_8}^2$ is a fourth root of unity, and namely it is $i$ (at $\pi/2$), prefer to denote it $\zeta_8=\sqrt i$. This field $\Bbb Q(\sqrt i)$ is then a 4d vectorspace over $\Bbb Q$ with a standard basis $1,\sqrt i,\,i,\,i\sqrt i$. (The next power, ${\zeta_8}^4= -1$ is already dependent on them. Note also that $\Bbb Q(\sqrt i)=\Bbb Q(i,\sqrt 2)$ as field extension.) So that $\Bbb Z[\sqrt i] =\{a+b\sqrt i+ci+di\sqrt i \mid a,b,c,d\in\Bbb Z\}$.

  1. What are the integers in $\Bbb Q(\sqrt i)$? It is going to be $\Bbb Z[\sqrt i]$, but it needs to be thought over.
  2. As David mentioned in a comment, $2\in\mathscr P=\left((1-\sqrt i)\right)$, because (writing '$a\equiv b \pmod{\mathscr P}$' for $b-a\in\mathscr P$), we have $$1\equiv\sqrt i \overset{()^2}\implies 1\equiv i \overset{()^2}\implies 1\equiv -1 \overset{+1}\implies 2\equiv 0 \pmod{\mathscr P} $$
  3. Then, for $\mathscr P^2$, it is the ideal generated by $(\sqrt i -1)^2=i-2\sqrt i+1$. So, basically the relation $$ 1+i \equiv 2\sqrt i \pmod{\mathscr P^2}$$ generates it. From this, we also have $2i\equiv 4i $, implying $0\equiv 2i$ then (multplying by $(-i)$:) $\ 0\equiv 2 \pmod{\mathscr P^2}$, so again $-1\equiv 1$. Also, $1+i\equiv 2\sqrt i\equiv 0\sqrt i=0$, that is, $1\equiv -i=(-1)i\equiv i$. And, because $\sqrt i-1\notin\mathscr P^2$, we also have $\sqrt i\not\equiv 1 \pmod{\mathscr P^2}$. So, finally, we can conclude that $R/\mathscr P^2$ is represented by the following set: $$\{ 0,1,\sqrt i,1-\sqrt i \}$$ And, it is not $\Bbb Z/4\Bbb Z$, but still a ring with 4 elements..

I'm sure there is a more sophisticated and simpler solution, but until we find it, you can play around with $\sqrt i$..

About the other statement, $\mathscr P^{-2}/R$, well.. the problem is that $R$ cannot be an ideal in there, because $1\in R$.

Edit: But, as Hurkyl noted, they are both $R$-modules, so it can make sense anyway.

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On your last sentence, while they aren't a ring and ideal, they are two abelian groups -- they're even two $R$-modules. So the quotient should be interpreted as a quotient of those structures. –  Hurkyl Oct 10 '12 at 17:11
    
Ah, yes, you're right, $R$-modules, thanks.. –  Berci Oct 10 '12 at 20:08
    
Thanks for this. One final question: why is $\mathscr{P}^4=\langle 2\rangle$? –  Clinton Boys Oct 12 '12 at 3:30

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