Lemma 37.24.1. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta $. If $X_\eta = \emptyset $ then there exists a nonempty open $V \subset Y$ such that $X_ V = V \times _ Y X = \emptyset $.

## 37.24 Generic fibres

Some results on the relationship between generic fibres and nearby fibres.

**Proof.**
Follows immediately from the more general Morphisms, Lemma 29.8.5.
$\square$

Lemma 37.24.2. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta $. If $X_\eta \not= \emptyset $ then there exists a nonempty open $V \subset Y$ such that $X_ V = V \times _ Y X \to V$ is surjective.

**Proof.**
This follows, upon taking affine opens, from Algebra, Lemma 10.30.2. (Of course it also follows from generic flatness.)
$\square$

Lemma 37.24.3. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta $. If $Z \subset X$ is a closed subset with $Z_\eta $ nowhere dense in $X_\eta $, then there exists a nonempty open $V \subset Y$ such that $Z_ y$ is nowhere dense in $X_ y$ for all $y \in V$.

**Proof.**
Let $Y' \subset Y$ be the reduction of $Y$. Set $X' = Y' \times _ Y X$ and $Z' = Y' \times _ Y Z$. As $Y' \to Y$ is a universal homeomorphism by Morphisms, Lemma 29.45.6 we see that it suffices to prove the lemma for $Z' \subset X' \to Y'$. Thus we may assume that $Y$ is integral, see Properties, Lemma 28.3.4. By Morphisms, Proposition 29.27.1 there exists a nonempty affine open $V \subset Y$ such that $X_ V \to V$ and $Z_ V \to V$ are flat and of finite presentation. We claim that $V$ works. Pick $y \in V$. If $Z_ y$ has a nonempty interior, then $Z_ y$ contains a generic point $\xi $ of an irreducible component of $X_ y$. Note that $\eta \leadsto f(\xi )$. Since $Z_ V \to V$ is flat we can choose a specialization $\xi ' \leadsto \xi $, $\xi ' \in Z$ with $f(\xi ') = \eta $, see Morphisms, Lemma 29.25.9. By Lemma 37.22.10 we see that

Hence some irreducible component of $Z_\eta $ passing through $\xi '$ has dimension $\dim _{\xi '}(X_\eta )$ which contradicts the assumption that $Z_\eta $ is nowhere dense in $X_\eta $ and we win. $\square$

Lemma 37.24.4. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta $. Let $U \subset X$ be an open subscheme such that $U_\eta $ is scheme theoretically dense in $X_\eta $. Then there exists a nonempty open $V \subset Y$ such that $U_ y$ is scheme theoretically dense in $X_ y$ for all $y \in V$.

**Proof.**
Let $Y' \subset Y$ be the reduction of $Y$. Let $X' = Y' \times _ Y X$ and $U' = Y' \times _ Y U$. As $Y' \to Y$ induces a bijection on points, and as $U' \to U$ and $X' \to X$ induce isomorphisms of scheme theoretic fibres, we may replace $Y$ by $Y'$ and $X$ by $X'$. Thus we may assume that $Y$ is integral, see Properties, Lemma 28.3.4. We may also replace $Y$ by a nonempty affine open. In other words we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a domain with fraction field $K$.

As $f$ is of finite type we see that $X$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ for some affine opens $X_ i$. By Morphisms, Definition 29.7.1 we see that $U_ i = X_ i \cap U$ is an open subscheme of $X_ i$ such that $U_{i, \eta }$ is scheme theoretically dense in $X_{i, \eta }$. Thus it suffices to prove the result for the pairs $(X_ i, U_ i)$, in other words we may assume that $X$ is affine.

Write $X = \mathop{\mathrm{Spec}}(B)$. Note that $B_ K$ is Noetherian as it is a finite type $K$-algebra. Hence $U_\eta $ is quasi-compact. Thus we can find finitely many $g_1, \ldots , g_ m \in B$ such that $D(g_ j) \subset U$ and such that $U_\eta = D(g_1)_\eta \cup \ldots \cup D(g_ m)_\eta $. The fact that $U_\eta $ is scheme theoretically dense in $X_\eta $ means that $B_ K \to \bigoplus _ j (B_ K)_{g_ j}$ is injective, see Morphisms, Example 29.7.4. By Algebra, Lemma 10.24.4 this is equivalent to the injectivity of $B_ K \to \bigoplus \nolimits _{j = 1, \ldots , m} B_ K$, $b \mapsto (g_1b, \ldots , g_ mb)$. Let $M$ be the cokernel of this map over $A$, i.e., such that we have an exact sequence

After replacing $A$ by $A_ h$ for some nonzero $h$ we may assume that $B$ is a flat, finitely presented $A$-algebra, and that $M$ is flat over $A$, see Algebra, Lemma 10.118.3. The flatness of $B$ over $A$ implies that $B$ is torsion free as an $A$-module, see More on Algebra, Lemma 15.22.9. Hence $B \subset B_ K$. By assumption $I_ K = 0$ which implies that $I = 0$ (as $I \subset B \subset B_ K$ is a subset of $I_ K$). Hence now we have a short exact sequence

with $M$ flat over $A$. Hence for every homomorphism $A \to \kappa $ where $\kappa $ is a field, we obtain a short exact sequence

see Algebra, Lemma 10.39.12. Reversing the arguments above this means that $\bigcup D(g_ j \otimes 1)$ is scheme theoretically dense in $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa )$. As $\bigcup D(g_ j \otimes 1) = \bigcup D(g_ j)_\kappa \subset U_\kappa $ we obtain that $U_\kappa $ is scheme theoretically dense in $X_\kappa $ which is what we wanted to prove. $\square$

Suppose given a morphism of schemes $f : X \to Y$ and a point $y \in Y$. Recall that the fibre $X_ y$ is homeomorphic to the subset $f^{-1}(\{ y\} )$ of $X$ with induced topology, see Schemes, Lemma 26.18.5. Suppose given a closed subset $T(y) \subset X_ y$. Let $T$ be the closure of $T(y)$ in $X$. Endow $T$ with the induced reduced scheme structure. Then $T$ is a closed subscheme of $X$ with the property that $T_ y = T(y)$ set-theoretically. In fact $T$ is the smallest closed subscheme of $X$ with this property. Thus it is “harmless” to denote a closed subset of $X_ y$ by $T_ y$ if we so desire. In the following lemma we apply this to the generic fibre of $f$.

Lemma 37.24.5. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta $. Let $X_\eta = Z_{1, \eta } \cup \ldots \cup Z_{n, \eta }$ be a covering of the generic fibre by closed subsets of $X_\eta $. Let $Z_ i$ be the closure of $Z_{i, \eta }$ in $X$ (see discussion above). Then there exists a nonempty open $V \subset Y$ such that $X_ y = Z_{1, y} \cup \ldots \cup Z_{n, y}$ for all $y \in V$.

**Proof.**
If $Y$ is Noetherian then $U = X \setminus (Z_1 \cup \ldots \cup Z_ n)$ is of finite type over $Y$ and we can directly apply Lemma 37.24.1 to get that $U_ V = \emptyset $ for a nonempty open $V \subset Y$. In general we argue as follows. As the question is topological we may replace $Y$ by its reduction. Thus $Y$ is integral, see Properties, Lemma 28.3.4. After shrinking $Y$ we may assume that $X \to Y$ is flat, see Morphisms, Proposition 29.27.1. In this case every point $x$ in $X_ y$ is a specialization of a point $x' \in X_\eta $ by Morphisms, Lemma 29.25.9. As the $Z_ i$ are closed in $X$ and cover the generic fibre this implies that $X_ y = \bigcup Z_{i, y}$ for $y \in Y$ as desired.
$\square$

The following lemma says that generic fibres of morphisms whose source is reduced are reduced.

Lemma 37.24.6. Let $f : X \to Y$ be a morphism of schemes. Let $\eta \in Y$ be a generic point of an irreducible component of $Y$. Then $(X_\eta )_{red} = (X_{red})_\eta $.

**Proof.**
Choose an affine neighbourhood $\mathop{\mathrm{Spec}}(A) \subset Y$ of $\eta $. Choose an affine open $\mathop{\mathrm{Spec}}(B) \subset X$ mapping into $\mathop{\mathrm{Spec}}(A)$ via the morphism $f$. Let $\mathfrak p \subset A$ be the minimal prime corresponding to $\eta $. Let $B_{red}$ be the quotient of $B$ by the nilradical $\sqrt{(0)}$. The algebraic content of the lemma is that $C = B_{red} \otimes _ A \kappa (\mathfrak p)$ is reduced. Denote $I \subset A$ the nilradical so that $A_{red} = A/I$. Denote $\mathfrak p_{red} = \mathfrak p/I$ which is a minimal prime of $A_{red}$ with $\kappa (\mathfrak p) = \kappa (\mathfrak p_{red})$. Since $A \to B_{red}$ and $A \to \kappa (\mathfrak p)$ both factor through $A \to A_{red}$ we have $C = B_{red} \otimes _{A_{red}} \kappa (\mathfrak p_{red})$. Now $\kappa (\mathfrak p_{red}) = (A_{red})_{\mathfrak p_{red}}$ is a localization by Algebra, Lemma 10.25.1. Hence $C$ is a localization of $B_{red}$ (Algebra, Lemma 10.12.15) and hence reduced.
$\square$

Lemma 37.24.7. Let $f : X \to Y$ be a morphism of schemes. Assume that $Y$ is irreducible and $f$ is of finite type. There exists a diagram

where

$V$ is a nonempty open of $Y$,

$X_ V = V \times _ Y X$,

$g : Y' \to V$ is a finite universal homeomorphism,

$X' = (Y' \times _ Y X)_{red} = (Y' \times _ V X_ V)_{red}$,

$g'$ is a finite universal homeomorphism,

$Y'$ is an integral affine scheme,

$f'$ is flat and of finite presentation, and

the generic fibre of $f'$ is geometrically reduced.

**Proof.**
Let $V = \mathop{\mathrm{Spec}}(A)$ be a nonempty affine open of $Y$. By assumption the Jacobson radical of $A$ is a prime ideal $\mathfrak p$. Let $K = \kappa (\mathfrak p)$. Let $p$ be the characteristic of $K$ if positive and $1$ if the characteristic is zero. By Varieties, Lemma 33.6.11 there exists a finite purely inseparable field extension $K'/K$ such that $(X_{K'})_{red}$ is geometrically reduced over $K'$. Choose elements $x_1, \ldots , x_ n \in K'$ which generate $K'$ over $K$ and such that some $p$-power of $x_ i$ is in $A/\mathfrak p$. Let $A' \subset K'$ be the finite $A$-subalgebra of $K'$ generated by $x_1, \ldots , x_ n$. Note that $A'$ is a domain with fraction field $K'$. By Algebra, Lemma 10.46.7 we see that $A \to A'$ induces a universal homeomorphism on spectra. Set $Y' = \mathop{\mathrm{Spec}}(A')$. Set $X' = (Y' \times _ Y X)_{red}$. The generic fibre of $X' \to Y'$ is $(X_{K'})_{red}$ by Lemma 37.24.6 which is geometrically reduced by construction. Note that $X' \to X_ V$ is a finite universal homeomorphism as the composition of the reduction morphism $X' \to Y' \times _ Y X$ (see Morphisms, Lemma 29.45.6) and the base change of $g$. At this point all of the properties of the lemma hold except for possibly (7). This can be achieved by shrinking $Y'$ and hence $V$, see Morphisms, Proposition 29.27.1.
$\square$

Lemma 37.24.8. Let $f : X \to Y$ be a morphism of schemes. Assume that $Y$ is irreducible and $f$ is of finite type. There exists a diagram

where

$V$ is a nonempty open of $Y$,

$X_ V = V \times _ Y X$,

$g : Y' \to V$ is surjective finite étale,

$X' = Y' \times _ Y X = Y' \times _ V X_ V$,

$g'$ is surjective finite étale,

$Y'$ is an irreducible affine scheme, and

all irreducible components of the generic fibre of $f'$ are geometrically irreducible.

**Proof.**
Let $V = \mathop{\mathrm{Spec}}(A)$ be a nonempty affine open of $Y$. By assumption the Jacobson radical of $A$ is a prime ideal $\mathfrak p$. Let $K = \kappa (\mathfrak p)$. By Varieties, Lemma 33.8.15 there exists a finite separable field extension $K'/K$ such that all irreducible components of $X_{K'}$ are geometrically irreducible over $K'$. Choose an element $\alpha \in K'$ which generates $K'$ over $K$, see Fields, Lemma 9.19.1. Let $P(T) \in K[T]$ be the minimal polynomial for $\alpha $ over $K$. After replacing $\alpha $ by $f \alpha $ for some $f \in A$, $f \not\in \mathfrak p$ we may assume that there exists a monic polynomial $T^ d + a_1T^{d - 1} + \ldots + a_ d \in A[T]$ which maps to $P(T) \in K[T]$ under the map $A[T] \to K[T]$. Set $A' = A[T]/(P)$. Then $A \to A'$ is a finite free ring map such that there exists a unique prime $\mathfrak q$ lying over $\mathfrak p$, such that $K = \kappa (\mathfrak p) \subset \kappa (\mathfrak q) = K'$ is finite separable, and such that $\mathfrak pA'_{\mathfrak q}$ is the maximal ideal of $A'_{\mathfrak q}$. Hence $g : Y' = \mathop{\mathrm{Spec}}(A') \to V = \mathop{\mathrm{Spec}}(A)$ is étale at $\mathfrak q$, see Algebra, Lemma 10.143.7. This means that there exists an open $W \subset \mathop{\mathrm{Spec}}(A')$ such that $g|_ W : W \to \mathop{\mathrm{Spec}}(A)$ is étale. Since $g$ is finite and since $\mathfrak q$ is the only point lying over $\mathfrak p$ we see that $Z = g(Y' \setminus W)$ is a closed subset of $V$ not containing $\mathfrak p$. Hence after replacing $V$ by a principal affine open of $V$ which does not meet $Z$ we obtain that $g$ is finite étale.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)