Ring isomorphism, very basic question(regarding first isomorphism theorem) Feeling unsure, want to check if I am thinking correctly.
Exercise sounds: Let n>1, square-free integer. Show that $\mathbb{Z}[\sqrt{n}]/\langle \sqrt{n}\rangle\simeq\mathbb{Z}_n$
My take: Lets make suitable homomorphism $\mathbb{Z}[\sqrt{n}]\rightarrow\mathbb{Z}_n$. Let $x=a+\sqrt{n}b$, which is the form of $x\in\mathbb{Z}[\sqrt{n}]$.
Then appropriate homomorphism would be $f(a+\sqrt{n}b)=a+0b\mod{n}(=a\mod{n})$. It's kernel is all elements with $a=0$ and.. and.. I dont know what "and". I think I am done, but feel pretty unsure
Little help apreciated. My 'homomorphism' doesnt look like homomorphism to me.
 A: Your proof's okay.
The map you define is indeed a homomorphism from $(\Bbb Z[\sqrt n],+,*)$ to $(\Bbb Z_n,+_n,*_n)$ since you have,
$$\begin{align}f((a+\sqrt n b)+(c+\sqrt n d))&=f((a+c)+\sqrt n(b+d))\\&=(a+c)~\bmod~n\\&=(a~\bmod~n)+_n(b~\bmod~n)=f(a+\sqrt n b)+_nf(c+\sqrt n d)\end{align}$$
$$\begin{align}f((a+\sqrt n b)(c+\sqrt n d))&=f((ac+nbd)+\sqrt n(bc+ad))\\&=(ac+nbd)~\bmod~n\\&=(ac)~\bmod~n\\&=(a~\bmod~n)*_n (c~\bmod~n)=f(a+\sqrt n b)*_n f(c+\sqrt n d)\end{align}$$
Now, since the kernel of $f$ is given by,
$\begin{align}\textrm{Ker}(f)=\{nd+\sqrt n b\mid d,b\in\Bbb Z\}&=\{(b+\sqrt n d)\sqrt n\mid d,b\in\Bbb Z\}\\&=\{m\sqrt n\mid m\in\Bbb Z[\sqrt n]\}\\&=\langle \sqrt n\rangle\end{align}$
and $f$ is surjective ($\forall~x\in\Bbb Z_n~,~\exists y=x+\sqrt n\in\Bbb Z[\sqrt n]$ such that $f(y)=x$), so $f$ is an epimorphism and by the first isomorphism theorem, we conclude our result.
A: Your morphism  creates the isomorphism between $Z[\sqrt n]/(\sqrt n)$ and $Z/n$. It is obviously surjective. Suppose $f(a+b\sqrt n)=0$. This implies that $a=nc=c\sqrt n\sqrt n$, thus $a+b\sqrt n= \sqrt n(c\sqrt n+b)\in (\sqrt n)$.
