Quick question about solutions to a question concerning quotient ring isomorphism [duplicate]

I have a quick question about the allowed function use to show isomorphism between quotient rings. The question assumes the following exercise:

Let $R=\{a+bi|a,b\in \mathbb{Z}\}$, and $J=\{a+bi| 5|a, 5|b\}.$ Show that $J$ is not a maximal ideal in $R$ (Hint: Consider the principal ideal $K=\{a+bi\}$ in $R$)

The question asks: If $R$ and $J$ are as in Exercise 1, show that $R/J \cong \mathbb{Z_{5}}$

I know that we can define $\psi: R\rightarrow \mathbb{Z_{5}}$ by $\psi(a+bi)=[a-2b]_{5}$, but can I use the function $\psi(a+bi)=[a+2b]_{5}$ instead?

Note: it is question 24 in the attached picture which assumes notations from exercises 21(a) and 22.

Thank you in advance.

Solution 1:

The text doesn't match the screenshot. You're required to prove that if $K=(2+i)$, then $R/K$ is isomorphic to $\mathbb{Z}_5$.

One way is to consider the map $\psi\colon R\to\mathbb{Z}_5$ defined by $\psi(a+bi)=[a-2b]_5$.

Is this a ring homomorphism? One could check this via the definition, but there's a better way. We can consider $\varphi\colon\mathbb{Z}[x]\to\mathbb{Z}_5$ defined by $\varphi(x)=[-2]_5$.

Now we can see that the kernel of this homomorphism contains the ideal $I$ generated by $x^2+1$, because $\varphi(x^2+1)=[-2]_5^2+[1]_5=[0]_5$. Hence this homomorphism induces a ring homomorphism $\psi\mathbb{Z}/I\to\mathbb{Z}_5$ and $\psi(x+I)=[-2]_5$. Now we can “canonically” identify $R$ with $\mathbb{Z}[x]/I$ and we're done, because $$ \psi(a+bi)=\psi(a)+\psi(b)\psi(i)=[a]_5+[b]_5[-2]_5=[a-2b]_5 $$

Can you do the same by mapping $x$ to $[2]_5$? Yes, why not? Repeat the argument and you're done.

Or consider that conjugation is an automorphism of $R$ and if $\psi'(a+bi)=[a+2b]_5$, then $$ \psi'(a+bi)=\psi(\overline{a+bi}) $$ so $\psi'=\psi\circ j$, where $j$ is conjugation.