Clarification about field cosets Let $R=Z[i]/\langle 2-i\rangle$ be a ring. So each element of $R$ looks like $a+bi+\langle 2-i \rangle$, with $a,b$ integers. Now, I read on this example in my book and see the following:
$$ 5(1+\langle 2-i\rangle) =5+\langle 2-i\rangle = 0+\langle 2-i\rangle$$
But how can $5\langle 2-i\rangle = \langle 2-i\rangle$? That is, how do we know that $\langle 10-5i\rangle$ is the same generator as $\langle 2-i\rangle$?
Thank you for clarifying this.
 A: It depends on how you define $n\cdot a$ when $n\in \Bbb Z$ (and not $R$!) and $a\in R$ in the first place.
For any abelian group $A$, we have an action of $\Bbb Z$ on $A$, i.e., a ring homomorphism $\Bbb Z\to\operatorname{End}(A)$ that maps $1\in \Bbb Z$ to $\operatorname{ id}_A\in\operatorname{End}(A)$. This is commonly denoted with the multiplication sign. In other words, $$n\cdot a=\underbrace{a+a+\ldots +a}_n$$ (with inverses used for negative $n$). If instead of an abelian group $A$ we deal with a ring $R$, recall that the addition in $R$ makes it an abelian group. We use the same notation for a corresponding action on $R$. In other words, 
$$\begin{align}5\cdot (1+\langle 2-i\rangle)&= (1+\langle 2-i\rangle)+\ldots+(1+\langle 2-i\rangle)\\&=1+1+1+1+1+\langle 2-i\rangle\\&=5+\langle 2-i\rangle\end{align}$$
So there is no step $5\cdot\langle 2-i\rangle=\langle10-5i\rangle$ involved after all.

For the final step, that the result is in fact $0+\langle 2-i\rangle$, we need not observe that in fact $5\in\langle 2-i\rangle$, i.e., that  we have $5=\alpha\cdot(2-i)$ for suitable $\alpha\in\Bbb Z[i]$; we see that $\alpha=2+i$ does the trick.
A: In $\mathbb{Z}[i]$, we have that $5 \in \langle 2-i \rangle$ since $5 = (2+i)(2-i).$
More explicitly, in $R$, we have: $$ 5 + \langle 2-i \rangle= (2+i)(2-i) + \langle 2-i \rangle = 0 + \langle 2-i \rangle. $$
