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In the above question, I have mangaged to prove that $r_\varphi$ is a root of $f(x)$, namely, $f(\varphi(\overline{x}))=0$. So that can be used as a known fact. I am now trying to prove that the function mapping $\varphi$ to $r_\varphi$ is a bijection from the set of homomorphisms $\varphi$ to the set of roots of $f(x)$ in $R$. I have managed to prove that this function is injective but do not know how to prove that it is surjective. Can someone help me, please? Thanks so much.

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    What is the source of the question? A book? It is quite interesting, and I want to read more related to this. – 2017-02-25
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    @feralin It is not from a book, and it seems that this question is completely newly designed rather than picked from a book. But in page 9 of this notes, there is a quite related field version of it:http://www.math.columbia.edu/~rf/galoisnotes.pdf – 2017-02-25

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Given a root $r \in R$ of $f$, let $\varphi$ be the homomorphism given by evaluation at $r$, that is, \begin{align*} \varphi: \mathbb{Z}[x] &\to R\\ g &\mapsto g(r) \, . \end{align*} Then $\varphi$ is a ring homomorphism and $\langle f \rangle \subseteq \ker(\varphi)$, so $\varphi$ descends to a well-defined homomorphism $\overline{\varphi}: \mathbb{Z}[x]/\langle f \rangle \to R$ on the quotient. Moreover, $$ \overline{\varphi}(\overline{x}) = \varphi(x) = r $$ so $r = r_\overline{\varphi}$, as desired.

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    It make some sense to me. But we have never been told that $\phi$ is exacely a evaluation map, how can we assume that? Moreover, how can we conclude the bijection relationship then? – 2017-02-25
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    On this track, I find that given a fixed root $r$ of $f(x)$ in $R$, we can always define the map $\varphi$ to be a evaluation map on $r$ and thus a homomorphism. And by using the mapping property of quotient rings we know that there exist a map $\overline{\varphi}$ that sends the root $\overline{x}$ in the quotient ring to the root $r$ in $R$. Thus we can conclude that for all roots in $R$, there exists a map which send a root $\overline{x}$ in the quotient ring to the root $r$ in $R$. Am I correct? – 2017-02-25
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    I don't understand your first comment. To show surjectivity, you need to show that for each root $r \in R$ of $f$, there is some homomorphism $\varphi: \mathbb{Z}[x] \to R$ such that $r = r_\varphi = \varphi(\overline{x})$. I have produced such a map: the map induced by evaluation at $r$. – 2017-02-25
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    Yes, your second comment is a restatement of the argument in my answer. – 2017-02-25