Proof $\mathbb{Z}[i]/(2- i) \cong \mathbb{Z_5}$

Question

My worked example sheet states

Prove $\mathbb{Z}[i]/(2- i) \cong \mathbb{Z_5}$ using the following steps:

Let $G = \mathbb{Z}[i]/(2- i)$

• $G = \{ \bar{x} + \bar{y}\bar{i} \mid x,y \in \mathbb{Z} \}$
• $G = \{ \bar{x} \mid x \in \mathbb{Z} \}$, because $\bar{2} = \bar{i}$
• Show that $G = \{ \bar{0},\ldots,\bar{4}\}$
• Finally, prove $G \cong \mathbb{Z_5}$

The proof is clear to me but I do not know how to rigorously show $G = \{ \bar{x} + \bar{y}\bar{i} \mid x,y \in \mathbb{Z} \}$ and $G = \{ \bar{0},\ldots,\bar{4}\}$.

For the first one, I think this is just the definition of a quotient ring but is that enough for rigour? While for the second one, we have the relation $\bar{5} = \bar{0}$ in $R$ but how can I show ( rather write ) $G$ has the mentioned elements.

I guess, I am asking for a formal way of writing this.

PS: The proof for similar problem is available on the forum, but my problem is how much detail is enough. ( For me the context is an examination. ) Pardon my elementary question.

Answer 1

Following your argumentation:

• $G = \{\bar x + \bar y \bar i\,|\, x,y\in\Bbb Z\}$ is precisely because there is canonical ring epimorphism $\Bbb Z[i]\twoheadrightarrow G$ which is both additive and multiplicative, thus $\overline{x+yi} = \bar x + \bar y \bar i$.

• $G = \{\bar x\,|\, x\in\Bbb Z\}$ because $\bar 2=\bar i$ should be good enough, but perhaps more detailed argument could be $\overline{x+yi} = \overline{x+2y}$ because $\bar 2=\bar i$. What this actually tells us is that there is epimorphism $\varphi\colon\Bbb Z\twoheadrightarrow G$ given by $x\mapsto \bar x$.

• $G = \{\bar 0,\ldots,\bar 4\}$ will follow from $\bar 0 = \bar 5$ (since $5 = (2-i)(2+i)$) because this shows us that $5\Bbb Z\subseteq \ker \varphi$, and thus $\varphi$ induces epimorphism $\bar\varphi\colon \Bbb Z_5\twoheadrightarrow G$ given by $\bar x\mapsto \bar x$.

We are almost done at this point. All we have to show now is that $\bar\varphi$ is actually isomorphism. This amounts to saying that $\{\bar 0,\ldots,\bar 4\}$ really are different elements in $G$, and probably the easiest way to show this is to explicitly give inverse for $\bar\varphi$. For this, we should use the point we made before: $\overline{x+yi} = \overline{x+2y}$ in $G$. Hence, let us define $\psi\colon\Bbb Z[i]\to \Bbb Z_5$ with $x+yi\mapsto \overline{x+2y}$. I will leave it as exercise to show that this induces epimorphism $\bar\psi\colon G\to\Bbb Z_5$ which is inverse of $\bar\varphi$. Of course, for this to be rigourous, one has to show that $\psi$ is homomorphism in the first place. This can be done either directly or using universal property of $\Bbb Z[i]$ (we have $\psi(i)^2 = -\bar 1$).

Answer 2

For part 1, I think it is indeed sufficiently rigorous to say something like "this is the definition of a quotient ring".

For the second, you should state something like $$ \overline{2 - i} = 0 \implies \bar 2 = \bar i $$ For the third, if you have already shown that $\bar 5 = \bar 0$, then you should use the previous statements to reach the desired conclusion. In particular:

We already know that $G = \{\bar x : x \in \Bbb Z\}$. So, consider any element $\bar x \in G$ (where $x \in \Bbb Z$). Note that by the division algorithm, there exists a $q \in \Bbb Z$ and $r \in \{0,1,2,3,4\}$ such that $$ \bar x = \overline{q\cdot 5 + r} = \overline{r} $$ So, $\bar x \in \{\bar 0,\dots,\bar 4\}$. So, $G = \{\bar 0,\dots,\bar 4\}$.

For the last bit, there is an obvious isomorphism between these two groups. In particular, we simply take $\bar x \mapsto \bar x$ for $x \in \Bbb Z$. It suffices to state that this map has all the required properties of an isomorphism.