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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.

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    Note that $\langle 2-i\rangle$ differs from $\langle 10-5i\rangle$ – 2017-01-30
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    I think the confusion is coming from misunderstanding when cosets are equal. $a + \langle 2-i \rangle = b + \langle 2-i \rangle$ preciesly when $(a-b) \in \langle 2-i \rangle$. Therefore to show that $5 + \langle 2-i \rangle = 0 + \langle 2-i \rangle$, you can show that $5 \in \langle 2-i \rangle$. – 2017-01-30

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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.

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    But how is $2+i \in \langle 2-i\rangle$? – 2017-01-30
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    @sequence It isn't. The ideal contains $\textit{all}$ ring multiples of $(2-i)$. – 2017-01-30
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    So $\langle 2-i\rangle$ is an ideal, not a generator? I guess I got confused in the notation. – 2017-01-30
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    @sequence Yes, $\langle 2-i \rangle$ is an ideal. Its generator is $2-i$. – 2017-01-30
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    @KenDuna How can you generate this ideal with $2-i$? Say, since $5$ is in this ideal, how can you generate $5$ with $2-i$? – 2017-01-30
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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: $$ 5 + \langle 2-i \rangle= (2+i)(2-i) + \langle 2-i \rangle = 0 + \langle 2-i \rangle. $$

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    But how is $(2+i)(2-i)\in \langle 2-i\rangle$ if $\langle 2-i\rangle$ means only products and sums of $(2-i)$? – 2017-01-30
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    $\langle 2-i \rangle = \{(a+bi)(2-i) \ | \ a+bi \in \mathbb{Z}[i] \}$. – 2017-01-30