Here is the question I am trying to understand its solution:
I am wondering:
1- Why this proof shows that the ring $\mathbb Z[ \sqrt{2}]$ has exactly $2$ automorphisms? Could anyone explain this to me, please?
2- What are the general steps of finding all automorphisms of a ring and confirming that they are all?
To answer your question no. 1, if $\phi:\mathbb Z[\sqrt2] \longrightarrow \mathbb Z[\sqrt2]$ is an automorphism, then $\phi$ will be an automorphism on $\mathbb Z$ too. From the definition of ring homomorphism, we can write $\phi(m + n \sqrt 2) = \phi(m)+\phi(n)\phi(\sqrt 2)$. Now, find automorphisms on $\mathbb Z$. They are precisely $n \longmapsto n$ (since for any ring homomorphism, $1 \longmapsto 1$). Hence $\phi(m+n\sqrt 2) = m+n\phi(\sqrt 2)$. Your book says clearly why $\phi(\sqrt 2) = \pm\sqrt 2$.
To answer your question no. 2, at first take $1 \longmapsto 1$. It eases things. Usually we come across rings in the form of $R[\xi]$ where $\xi$ is from some bigger ring containing $R$, $R$ is some arbitrary ring. Generally we take $R$ as unital and commutative, hence $R$ contains $\mathbb Z$ as a subring and hence in $R$, $a \longmapsto a$. Now if $\xi$ is algebraic/integral on $R$, take the minimal polynomial of $\xi$ over $R$ and check if you take $\xi$ to its roots, whether they give you automorphisms or not. If $\xi$ is transcendental, then I think we just have one choice, $\xi \longmapsto \xi$, although I'm not quite sure about it.