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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