I have to calculate the order of the ring $\mathbb{Z}[i]/(1+i)$. This is how far I am: If $a+bi\in \mathbb{Z}[i]/(1+i)$ then there are $n,m\in \mathbb{Z}$ such that $a+bi\equiv 0+ni$ or $a+bi\equiv m+0i $. This means that $\mathbb{Z}[i]/(1+i)=\{x\in \mathbb{Z}[i]: Re(x)=0 \text{ or }Im(x)=0\}$. So the order of this ring has to be infinite. I have a feeling I have something wrong, but cant figure what. Is there something wrong in this calculation? Thanks.
Asked
Active
Viewed 3,786 times
2
-
See also https://math.stackexchange.com/questions/1639538/algebraically-why-is-mathbbzi-i-1-isomorphic-to-mathbbz-2?noredirect=1 – Anne Bauval Feb 18 '24 at 13:25
3 Answers
10
By using the "isomorphism theorems," one can deduce the structure of this ring rather easily. Follow the following sequence of isomorphisms. We know that $\mathbb{Z}[i] \cong \mathbb{Z}[x]/(x^2 + 1)$.
$\mathbb{Z}[i]/(1 + i) \cong \mathbb{Z}[x]/(x^2 + 1, 1 + x) \cong \frac{\mathbb{Z}[x]/(1 + x)}{(x^2 + 1, 1 + x)/(1 + x)} \cong \mathbb{Z}/((-1)^2 + 1) \cong \mathbb{Z}/2. $
The second to the last isomorphism comes from the isomorphism $\mathbb{Z}[x]/(1 + x) \cong \mathbb{Z}$ induced by $x \mapsto -1$.
6
Your problem is that for all $m \neq n \in \mathbb{Z}$, you haven't proved (and in fact, you can't) that the classes represented by $m + 0i$, and $n + 0i$ to be separated. So, there maybe some (and, in fact, in this case, many) cases, where $m \neq n \in \mathbb{Z}$, but $m + 0i \equiv n + 0i$.
Big hint of the day
By a simple manipulation, can you show that $\dfrac{2}{1 + i} \in \mathbb{Z}[i]$, so what this means is that $2 \in (1 + i)$, or in other words, $2 \equiv 0$.
user49685
- 4,539
-
-
So, can you finish it from here? If you still find something unclear, just shout it out. :) – user49685 Jul 29 '13 at 04:29
1
In this ring, $\hat{-1}=\hat i$ square both sides and win. ($\hat x$ is the equivalence class of $x$)
PVAL-inactive
- 8,551