1

I have been asked to prove 3 things about the Lipschitz Quaternion $A = \mathbb{Z}\ +\ i\mathbb{Z}\ +\ j\mathbb{Z}\ +\ k\mathbb{Z}$.

$(a)$ I have to prove that for every non-zero left ideal $I \subset A$. $$|A/I| = \begin{cases} n^2 &\text{if and only if $I$ is principal}\\ 2n^2 &\text{otherwise} \end{cases}$$

$(b)$ For every prime $p \neq 2$, we have $A/pA \cong \text{Mat}_2(\mathbb{F_p})$

$(c)$ For every prime $p$, there exists an ideal $I \subset A$ with $|A/I|=p^2$. How many such ideals are there?

I believe that $(c)$ will follow from $(a)$ and $(b)$

I know posting multiple questions is frowned upon. But since they arise from a common structure, I decided to post it.

Looking for any sort of hint to help me solve $(a), (b)\ \& \ (c)$. Very, very stuck.

I'm also aware that $|\text{Cl}(A)|=2$ (the class number). As mentioned here: Ideal class “group” of Lipschitz (fully-integer) quaternions

Naweed G. Seldon
  • 2,635
  • While flooding the site with questions is frowned upon, especially when are homework questions, asking a couple in the same day is fine. On the other hand, here it isn't clear what your question is... – A.P. Dec 25 '17 at 21:46
  • 1
    Fortunately, they're not assignment questions, not exactly. This post-finals let's-solve-questions-that-I-didn't-before phenomenon Questions are $a,\ b,\ c$ – Naweed G. Seldon Dec 25 '17 at 21:50
  • When you say $I$ is PID in item (a), what do you mean exactly? That $I$ is a principal ideal? To say $I$ is PID seems to mean $I$ is a domain, which to me means $1 \in I$; but then $I = A$. – Robert Lewis Dec 25 '17 at 23:41
  • Yes, I do mean it's Principal Idea. But I don't get your point about PID means that $I$ is a domain, which means $1 \in I$ – Naweed G. Seldon Dec 26 '17 at 00:23
  • @junkquill: that is my point. If $I$ were a domain, $1 \in I$ by definition of a domain as I know it, so then $I = A$. "PID" is standard terminology for "principle ideal domain", which is a commutative ring with unit, no zero divisors, and every ideal of the form $(a)$. So I guess you mean $I$ is a principle ideal, but not a PID. – Robert Lewis Dec 26 '17 at 01:28
  • 1
    Oh, crap. You're absolutely right. Grave typographical error on my part – Naweed G. Seldon Dec 26 '17 at 01:34

0 Answers0