Since the map is affine, we may assume that $X = \mathrm{Spec}(A)$ and $Y = \mathrm{Spec}(B)$ for some (commutative unital) rings $A$ and $B.$ We denote by $\phi : B \rightarrow A$ the ring map associated to $\pi.$ Write $\mathfrak{q} := y$ for the corresponding prime ideal of $B.$
Without the dimension $1$ hypothesis. We have $$\kappa(y) = B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$$ and $$\Gamma(\pi^{-1}(y), \mathscr{O}_{\pi^{-1}(y)}) = \frac{B_{\mathfrak{q}}}{\mathfrak{q}B_{\mathfrak{q}}} \otimes_{B} A \simeq \frac{B_{\mathfrak{q}}}{\mathfrak{q}B_{\mathfrak{q}}} \otimes_{B_{\mathfrak{q}}} A_{\mathfrak{q}} \simeq \frac{A_{\mathfrak{q}}}{\mathfrak{q}A_{\mathfrak{q}}}.$$ The first isomorphism is quite general, and the map looks like $$b \otimes a \mapsto b \otimes a.$$ For the second one, we consider the following general situation: given an ideal $I$ of a ring $R$ let $M$ be an $R$-module. Consider the exact sequence $$0 \rightarrow IM \rightarrow M \rightarrow M/IM\ \rightarrow 0.$$ We get the exact sequence $$0 = (R/I) \otimes_{R} IM \rightarrow (R/I) \otimes_{R} M \rightarrow (R/I) \otimes_{R} (M/IM)\ \rightarrow 0,$$ which gives $$\begin{align*}(R/I) \otimes_{R} M &\simeq (R/I) \otimes_{R} (M/IM) \\ &\simeq (R/I) \otimes_{R/I} (M/IM) \\ &\simeq M/IM \end{align*}$$ given by $\bar{r} \otimes m \mapsto \overline{rm}.$
Applying this for
- $R = B_{\mathfrak{q}},$
- $I = \mathfrak{q}B_{\mathfrak{q}},$
- $M = A_{\mathfrak{q}} = (A \setminus \phi(\mathfrak{q}))^{-1}A,$
we have $$\frac{B_{\mathfrak{q}}}{\mathfrak{q}B_{\mathfrak{q}}} \otimes_{B_{\mathfrak{q}}} A_{\mathfrak{q}} \simeq \frac{A_{\mathfrak{q}}}{\mathfrak{q}A_{\mathfrak{q}}}$$ given by $$[b/t] \otimes (a/s) \mapsto [\phi(b)a/(\phi(t)s)].$$ Of course, this map is only an $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-linear map. Since the inverse map is a ring map $[a/s] \mapsto 1 \otimes (a/s),$ so is the original map. That is, this isomorphism is an isomorphism of $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-algebra.
With the dimension $1$ hypothesis. The hypothesis that $$\dim_{\kappa(y)}\Gamma(\pi^{-1}(y), \mathscr{O}_{\pi^{-1}(y)}) = \dim_{B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}}(A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}) = 1$$ implies that the map $$B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \rightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}$$ induced by $\phi$ is an isomorphism of rings. Why? First, note that since this is a ring map and $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$ is a field this map must be injective unless $A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} = 0.$ However, it is not, as it is a positive dimensional vector space, so the map must be an inejection. Next, note that this map is $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-linear, and since the target is $1$-dimensional over $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}},$ this injection must be a surjection. This also shows that $$\pi^{-1}(y) = \pi^{-1}(\mathfrak{q}) = \mathrm{Spec}(A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}})$$ has a unique point, which corresponds to a prime ideal $\mathfrak{p}$ of $A.$ What we need to show is that the field extension $$B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \rightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$ induced by $\phi$ is an isomorphism.
Proof. We only need to show the surjectivity. First, note that we have $$A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} \simeq B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \hookrightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$$ and this map literally looks like $a/s \mapsto a/s$ because it is given by $\phi(b)/\phi(t) \mapsto b/t \mapsto \phi(b)/\phi(t).$
Since $\phi^{-1}(\mathfrak{p}) = \mathfrak{q},$ we have $\phi(\mathfrak{q}) \subset \mathfrak{p},$ which implies that $$A \smallsetminus \mathfrak{p} \subset A \smallsetminus \phi(\mathfrak{q}).$$ This induces the ring map $$A_{\mathfrak{p}} \rightarrow A_{\mathfrak{q}}$$ that looks like $a/s \mapsto a/s.$ Denote by $J$ the kernel of the composition $$A_{\mathfrak{p}} \rightarrow A_{\mathfrak{q}} \twoheadrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}},$$ which induces the injection $A_{\mathfrak{p}}/J \hookrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}.$ We can only know that $J$ is a prime ideal, so we have $J \subset \mathfrak{p}A_{\mathfrak{p}}.$ We claim that this inclusion is actually an equality, which will finish the proof. To see this, note that we have $$A_{\mathfrak{p}}/J \hookrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} \hookrightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$$ and this map is given by $(a \mod J) \mapsto (a \mod \mathfrak{p}A_{\mathfrak{p}}).$ Because this is a ring injection, for any $a \in A_{\mathfrak{p}}$ if $a \in \mathfrak{p}A_{\mathfrak{p}},$ then the image of this map is $0,$ so $a$ must be $0$ modulo $J$ as well. This implies that $a \in J,$ so $\mathfrak{p}A_{\mathfrak{p}} \subset J,$ showing that $\mathfrak{p}A_{\mathfrak{p}} = J.$ $\Box$
With the dimension $1$ hypothesis. The hypothesis that $$\dim_{\kappa(y)}\Gamma(\pi^{-1}(y), \mathscr{O}_{\pi^{-1}(y)}) = \dim_{B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}}(A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}) = 1$$ implies that the map $$B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \rightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}$$ induced by $\phi$ is an isomorphism of rings. Why? First, note that since this is a ring map and $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$ is a field this map must be injective unless $A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} = 0.$ However, it is not, as it is a positive dimensional vector space, so the map must be an inejection. Next, note that this map is $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$-linear, and since the target is $1$-dimensional over $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}},$ this injection must be a surjection. This also shows that $$\pi^{-1}(y) = \pi^{-1}(\mathfrak{q}) = \mathrm{Spec}(A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}})$$ has a unique point, which corresponds to a prime ideal $\mathfrak{p}$ of $A.$ What we need to show is that the field extension $$B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \rightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$ induced by $\phi$ is an isomorphism.
Proof. We only need to show the surjectivity. First, note that we have $$A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} \simeq B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}} \hookrightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$$ and this map literally looks like $a/s \mapsto a/s$ because it is given by $\phi(b)/\phi(t) \mapsto b/t \mapsto \phi(b)/\phi(t).$
Since $\phi^{-1}(\mathfrak{p}) = \mathfrak{q},$ we have $\phi(\mathfrak{q}) \subset \mathfrak{p},$ which implies that $$A \smallsetminus \mathfrak{p} \subset A \smallsetminus \phi(\mathfrak{q}).$$ This induces the ring map $$A_{\mathfrak{p}} \rightarrow A_{\mathfrak{q}}$$ that looks like $a/s \mapsto a/s.$ Denote by $J$ the kernel of the composition $$A_{\mathfrak{p}} \rightarrow A_{\mathfrak{q}} \twoheadrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}},$$ which induces the injection $A_{\mathfrak{p}}/J \hookrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}.$ We can only know that $J$ is a prime ideal, so we have $J \subset \mathfrak{p}A_{\mathfrak{p}}.$ We claim that this inclusion is actually an equality, which will finish the proof. To see this, note that we have $$A_{\mathfrak{p}}/J \hookrightarrow A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}} \hookrightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$$ and this map is given by $(a \mod J) \mapsto (a \mod \mathfrak{p}A_{\mathfrak{p}}).$ Because this is a ring injection, for any $a \in A_{\mathfrak{p}}$ if $a \in \mathfrak{p}A_{\mathfrak{p}},$ then the image of this map is $0,$ so $a$ must be $0$ modulo $J$ as well. This implies that $a \in J,$ so $\mathfrak{p}A_{\mathfrak{p}} \subset J,$ showing that $\mathfrak{p}A_{\mathfrak{p}} = J.$ $\Box$
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