Definition 29.8.1. A morphism $f : X \to S$ of schemes is called *dominant* if the image of $f$ is a dense subset of $S$.

## 29.8 Dominant morphisms

The definition of a morphism of schemes being dominant is a little different from what you might expect if you are used to the notion of a dominant morphism of varieties.

So for example, if $k$ is an infinite field and $\lambda _1, \lambda _2, \ldots $ is a countable collection of distinct elements of $k$, then the morphism

with $i$th factor mapping to the point $x = \lambda _ i$ is dominant.

Lemma 29.8.2. Let $f : X \to S$ be a morphism of schemes. If every generic point of every irreducible component of $S$ is in the image of $f$, then $f$ is dominant.

**Proof.**
This is a topological fact which follows directly from the fact that the topological space underlying a scheme is sober, see Schemes, Lemma 26.11.1, and that every point of $S$ is contained in an irreducible component of $S$, see Topology, Lemma 5.8.3.
$\square$

The expectation that morphisms are dominant only if generic points of the target are in the image does hold if the morphism is quasi-compact.

Lemma 29.8.3. Let $f : X \to S$ be a quasi-compact morphism of schemes. Then $f$ is dominant (if and) only if for every irreducible component $Z \subset S$ the generic point of $Z$ is in the image of $f$.

**Proof.**
Let $V \subset S$ be an affine open. Because $f$ is quasi-compact we may choose finitely many affine opens $U_ i \subset f^{-1}(V)$, $i = 1, \ldots , n$ covering $f^{-1}(V)$. Consider the morphism of affines

A disjoint union of affines is affine, see Schemes, Lemma 26.6.8. Generic points of irreducible components of $V$ are exactly the generic points of the irreducible components of $S$ that meet $V$. Also, $f$ is dominant if and only if $f'$ is dominant no matter what choices of $V, n, U_ i$ we make above. Thus we have reduced the lemma to the case of a morphism of affine schemes. The affine case is Algebra, Lemma 10.30.6. $\square$

Here is a slightly more useful variant of the lemma above.

Lemma 29.8.4. Let $f : X \to S$ be a quasi-compact morphism of schemes. Let $\eta \in S$ be a generic point of an irreducible component of $S$. If $\eta \not\in f(X)$ then there exists an open neighbourhood $V \subset S$ of $\eta $ such that $f^{-1}(V) = \emptyset $.

**Proof.**
Let $Z \subset S$ be the scheme theoretic image of $f$. We have to show that $\eta \not\in Z$. This follows from Lemma 29.6.5 but can also be seen as follows. By Lemma 29.6.3 the morphism $X \to Z$ is dominant, which by Lemma 29.8.3 means all the generic points of all irreducible components of $Z$ are in the image of $X \to Z$. By assumption we see that $\eta \not\in Z$ since $\eta $ would be the generic point of some irreducible component of $Z$ if it were in $Z$.
$\square$

There is another case where dominant is the same as having all generic points of irreducible components in the image.

Lemma 29.8.5. Let $f : X \to S$ be a morphism of schemes. Suppose that $X$ has finitely many irreducible components. Then $f$ is dominant (if and) only if for every irreducible component $Z \subset S$ the generic point of $Z$ is in the image of $f$. If so, then $S$ has finitely many irreducible components as well.

**Proof.**
Assume $f$ is dominant. Say $X = Z_1 \cup Z_2 \cup \ldots \cup Z_ n$ is the decomposition of $X$ into irreducible components. Let $\xi _ i \in Z_ i$ be its generic point, so $Z_ i = \overline{\{ \xi _ i\} }$. Note that $f(Z_ i)$ is an irreducible subset of $S$. Hence

is a finite union of irreducible subsets whose generic points are in the image of $f$. The lemma follows. $\square$

Lemma 29.8.6. Let $f : X \to Y$ be a morphism of integral schemes. The following are equivalent

$f$ is dominant,

$f$ maps the generic point of $X$ to the generic point of $Y$,

for some nonempty affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is injective,

for all nonempty affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is injective,

for some $x \in X$ with image $y = f(x) \in Y$ the local ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective, and

for all $x \in X$ with image $y = f(x) \in Y$ the local ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective.

**Proof.**
The equivalence of (1) and (2) follows from Lemma 29.8.5. Let $U \subset X$ and $V \subset Y$ be nonempty affine opens with $f(U) \subset V$. Recall that the rings $A = \mathcal{O}_ X(U)$ and $B = \mathcal{O}_ Y(V)$ are integral domains. The morphism $f|_ U : U \to V$ corresponds to a ring map $\varphi : B \to A$. The generic points of $X$ and $Y$ correspond to the prime ideals $(0) \subset A$ and $(0) \subset B$. Thus (2) is equivalent to the condition $(0) = \varphi ^{-1}((0))$, i.e., to the condition that $\varphi $ is injective. In this way we see that (2), (3), and (4) are equivalent. Similarly, given $x$ and $y$ as in (5) the local rings $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ are domains and the prime ideals $(0) \subset \mathcal{O}_{X, x}$ and $(0) \subset \mathcal{O}_{Y, y}$ correspond to the generic points of $X$ and $Y$ (via the identification of the spectrum of the local ring at $x$ with the set of points specializing to $x$, see Schemes, Lemma 26.13.2). Thus we can argue in the exact same manner as above to see that (2), (5), and (6) are equivalent.
$\square$

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