The exercise 7.14 of Introduction to Commutative Algebra (by Atiyah and Macdonald) states:
Let $k$ be an algebraically closed field, let $A$ denote the polynomial ring $k[t_1,\ldots,t_n]$ and let $\mathfrak{a}$ be an ideal in $A$. Let $V$ be the variety in $k^n$ defined by the ideal $\mathfrak{a}$, so that $V$ is the set of all $x = (x_1,\ldots,x_n) \in k^n$ such that $f(x)=0$ for all $f \in \mathfrak{a}$. Let $I(V)$ be the ideal of $V$, i.e. the ideal of all polynomials $g \in A$ such that $g(x) = 0$ for all $x \in V$. Then $I(V) = r(\mathfrak{a}).$
and the hint (almost the sketch of proof) says:
It is clear that $r(\mathfrak{a}) \subseteq I(V).$ Conversrly, let $f \notin r (\mathfrak{a}),$ then there is a prime ideal $\mathfrak{p}$ containing $\mathfrak{a}$ such that $f \notin \mathfrak{p}$. Let $\overline{f}$ be the image of $f$ in $B = A/\mathfrak{p}$, let $C = B_f = B[1/\overline{f}],$ and let $\mathfrak{m}$ be a maximal ideal of $C$. Since $C$ is a finitely generated $k$-algebra we have $C/\mathfrak{m} \cong k$, by (7.9). The images $x_i$ in $C/\mathfrak{m}$ of the generators $t_i$ of $A$ thus define a point $x=(x_1,\ldots,x_n) \in k^n$, and the construction shows that $x \in V$ and $f(x) \neq 0$.
I'm not quite understand why we can deduce the result $x \in V$ in the last sentence (The bold part). If I'm right, then the way we construct $x_i$ is to consider the map. \begin{alignat*}{0} A&\longrightarrow&A/\mathfrak{p} = B&\longrightarrow&B[1/\overline{f}] = C&\longrightarrow&C/\mathfrak{m} \cong k \ \\ f&\longmapsto & \overline{f} &\longmapsto & \overline{f}&\longmapsto & \overline{f} + \mathfrak{m} \\ t_i&&&\longmapsto &&& x_i \end{alignat*} Since $1/\overline{f} \in C, \overline{f} \notin \mathfrak{m},$ we get $f(x) = f(x_1,\ldots,x_n) = \overline{f} + \mathfrak{m} \neq 0.$
But I don't see why $x \in V$. I know that it suffices to show: for any $g \in \mathfrak{a}, g(x) = 0$, while the information about $\mathfrak{a}$ in this map is not much. Did I miss something? Any hint would help.