Hartshorne proposition III.9.7

Solution 1:

Are you asking why $f(x')=y$? Suppose $f:U=\operatorname{Spec} B \to V=\operatorname{Spec}A $ where $U\subset X, V\subset Y,A$ is an integral domain and $y\in V$. Since $y$ is a closed point, it corresponds to a maximal ideal $m$ of $A$. Since $dimA=1$ and $A$ is an integral domain, $(0)\subsetneq m$. We have the following commutative diagram: $$\require{AMScd} \begin{CD} A @>{\phi} >> B;\\ @VVV @VVV \\ A_{m} @>{f^{\#}}>> B_{n}; \end{CD}$$

Suppose $nB_{n}$ is not torsion free, then by definition $f^{\#}t$ is a zero divisor where $mA_{m}=(t)$ . By prop 1.11 and the remark below prop 4.7 of Atiyah and MacDonald's Commutative Algebra, $f^{\#}t$ is contained in some associated prime ideal $\tilde{\mathfrak{p}}$ of $B_{n}$. Let $\mathfrak{p}=\tilde{\mathfrak{p}}\bigcap B$, then $\phi^{-1}(\mathfrak{p})=m$ since $f^{\# -1}(\tilde{\mathfrak{p}})=(t)$. Hence $f(x')=y$, where $x'$ is the point corresponding to $\mathfrak{p}$. Contradiction.