If \(f: X \to Y\) is a morphism and \(y \in Y\) a point, show that \(\operatorname{sp}(X_y)\) is homeomorphic to \(\inverseof{f}(y)\) with the induced topology.
Let \(X = \Spec k[s,t]/(s - t^2)\), let \(Y = \Spec k[s]\), and let \(f: X \to Y\) be the morphism defined by sending \(s \mapsto s\). Assume \(k\) algebraically closed.
If \(y \in Y\) is the point \(a \in k\) with \(a \neq 0\), show that the fibre \(X_y\) consists of two points, with residue field \(k\).
If \(y \in Y\) corresponds to \(0 \in k\), show that the fibre \(X_y\) is a nonreduced one-point scheme.
If \(\eta\) is the generic point of \(Y\), show that \(X_\eta\) is a one-point scheme whose residue field is an extension of degree two of the residue field of \(\eta\).
(Erratum)
The first bullet in part (b) needs the hypothesis \(\operatorname{char}k\ne2\). If \(\operatorname{char}k=2\) and \(a\ne0\), then for the unique \(r\in k\) with \(r^2=a\), \[ t^2-a=(t-r)^2, \] so the fibre is a nonreduced one-point scheme rather than two points. The fibre over \(0\) and the generic degree-two fibre statements are valid in every characteristic.
Solution
1. It is enough to identify the topology of a fibre affine-locally on \(X\) and \(Y\).
Choose an affine neighborhood \[ V=\Spec B\subseteq Y \] of \(y\), corresponding to a prime \(\mathfrak p\subseteq B\), and cover \(f^{-1}(V)\) by affine opens \[ U=\Spec A. \] The fibre restricted to \(U\) is \[ U_y = U\times_V\Spec\kappa(y) \cong \Spec\bigl(A\otimes_B\kappa(\mathfrak p)\bigr). \] These affine pieces cover \(X_y\), so it suffices to compare each \(|U_y|\) with \(U\cap f^{-1}(y)\).
2. There is a canonical ring isomorphism\[ A\otimes_B\kappa(\mathfrak p) \cong S^{-1}(A/\mathfrak p A), \qquad S=B\setminus\mathfrak p. \]
Since \[ \kappa(\mathfrak p) = \operatorname{Frac}(B/\mathfrak p) = S^{-1}(B/\mathfrak p), \] tensoring first with \(B/\mathfrak p\) and then localizing gives \[ A\otimes_B\kappa(\mathfrak p) \cong S^{-1}(A\otimes_B B/\mathfrak p) \cong S^{-1}(A/\mathfrak p A). \]
3. The points of \(U_y\) correspond bijectively to the primes\[ \mathfrak q\in\Spec A \]such that\[ \mathfrak q\cap B=\mathfrak p. \]
By 2, primes of the fibre ring correspond to primes \(\mathfrak q\subseteq A\) which contain \(\mathfrak pA\) and are disjoint from the image of \(S\).
The first condition gives \[ \mathfrak p\subseteq\mathfrak q\cap B, \] while disjointness from \(B\setminus\mathfrak p\) gives the reverse inclusion. Thus \[ \mathfrak q\cap B=\mathfrak p. \] These are exactly the points of \(U\) mapping to \(y\).
4. The bijection in 3 is a homeomorphism from \(|U_y|\) onto \(U\cap f^{-1}(y)\) with the induced topology.
Let \(R=S^{-1}(A/\mathfrak pA)\) be the fibre ring. Any closed subset of \(\Spec R\) is \(V_R(J)\) for an ideal \(J\subseteq R\). Let \(I\subseteq A\) be the inverse image in \(A\) of the contraction of \(J\) to \(A/\mathfrak pA\). Under the prime correspondence of 3, \[ V_R(J) \] corresponds exactly to \[ V_A(I)\cap f^{-1}(y)\cap U. \] Thus closed subsets on the fibre side are precisely intersections with closed subsets of \(U\), which is the induced topology.
5. Therefore\[ \boxed{|X_y|\xrightarrow{\sim}f^{-1}(y)} \]is a homeomorphism onto the set-theoretic fibre with its subspace topology.
The affine homeomorphisms from 4 are compatible on overlaps because they all arise from the projection \(X_y\to X\). They glue to the claimed global homeomorphism.
6. In part (b), the coordinate ring of \(X\) is canonically isomorphic to \(k[t]\), with the structural map\[ k[s]\longrightarrow k[t], \qquad s\longmapsto t^2. \]
The quotient relation \(s-t^2=0\) eliminates \(s\): \[ k[s,t]/(s-t^2)\cong k[t]. \] Under this isomorphism, the inclusion of the base coordinate ring is exactly \(s\mapsto t^2\).
7. For a closed point \(y=(s-a)\) of \(Y\), the fibre is\[ \boxed{ X_y \cong \Spec k[t]/(t^2-a).} \]
Since \(k\) is algebraically closed, \[ \kappa(y)=k. \] Hence \[ X_y \cong \Spec\bigl(k[t]\otimes_{k[s]}k[s]/(s-a)\bigr) \cong \Spec k[t]/(t^2-a). \]
8. Assume \(a\ne0\) and \(\operatorname{char}k\ne2\). Then \(X_y\) consists of two reduced points, each with residue field \(k\).
Choose \(r\in k\) with \(r^2=a\). Since \(a\ne0\), we have \(r\ne0\), and since \(2\ne0\) in \(k\), \[ r\ne-r. \] Thus \[ t^2-a=(t-r)(t+r) \] has two distinct linear factors. The Chinese remainder theorem gives \[ k[t]/(t^2-a) \cong k[t]/(t-r)\times k[t]/(t+r) \cong k\times k. \] Therefore the fibre is the disjoint union of two reduced \(k\)-points.
9. If \(a=0\), then the fibre is a nonreduced one-point scheme.
By 7, \[ X_0=\Spec k[t]/(t^2). \] The quotient has the unique prime ideal \((t)\), so the spectrum has one point. The class of \(t\) is nonzero and nilpotent, so the scheme is not reduced.
10. If \(a\ne0\) and \(\operatorname{char}k=2\), then the fibre is likewise a nonreduced one-point scheme.
Choose \(r\in k\) with \(r^2=a\). In characteristic \(2\), \[ t^2-a=t^2-r^2=(t-r)^2. \] Hence \[ X_y\cong\Spec k[t]/(t-r)^2, \] which has one point and a nonzero nilpotent. This is the correction recorded in the erratum.
11. Let \(\eta\) be the generic point of \(Y\). Then\[ \kappa(\eta)=k(s) \]and\[ X_\eta \cong \Spec\bigl(k(s)[t]/(t^2-s)\bigr). \]
The generic point corresponds to the zero prime of \(k[s]\), whose residue field is its fraction field \(k(s)\). Base change along \[ k[s]\hookrightarrow k(s) \] gives \[ k[t]\otimes_{k[s]}k(s) \cong k(s)[t]/(t^2-s). \]
12. The polynomial\[ t^2-s\in k(s)[t] \]is irreducible in every characteristic.
A quadratic polynomial of the form \(t^2-s\) is reducible over the field \(k(s)\) exactly when \(s\) is a square in \(k(s)\).
But the discrete valuation \(v_s\) on \(k(s)\) associated to the prime \((s)\) satisfies \[ v_s(s)=1. \] Every square has even \(v_s\)-valuation. Therefore \(s\) is not a square in \(k(s)\), so \(t^2-s\) is irreducible.
13. The generic fibre is a one-point scheme whose residue field has degree two over \(k(s)\):\[ \boxed{ X_\eta = \Spec k(s)(\sqrt{s}), \qquad [\kappa(X_\eta):k(s)]=2.} \]
By 12, the quotient \[ k(s)[t]/(t^2-s) \] is a field and has vector-space dimension \(2\) over \(k(s)\). The spectrum of a field is one point.
If \(\operatorname{char}k\ne2\), the extension is separable. If \(\operatorname{char}k=2\), it is purely inseparable. Its degree is \(2\) in either case.
14. Q.E.D.
Steps <1>1–<1>5 prove part (a). Steps <1>6–<1>9 prove the source’s intended characteristic-\(\ne2\) closed-fibre statements, 10 records the characteristic-\(2\) correction, and <1>11–<1>13 prove the generic-fibre statement.