We take the following definition from [IV, Definition 11.10.2, EGA].
Definition 29.7.1. Let X be a scheme. Let U \subset X be an open subscheme.
The scheme theoretic image of the morphism U \to X is called the scheme theoretic closure of U in X.
We say U is scheme theoretically dense in X if for every open V \subset X the scheme theoretic closure of U \cap V in V is equal to V.
With this definition it is not the case that U is scheme theoretically dense in X if and only if the scheme theoretic closure of U is X, see Example 29.7.2. This is somewhat inelegant; but see Lemmas 29.7.3 and 29.7.8 below. On the other hand, with this definition U is scheme theoretically dense in X if and only if for every V \subset X open the ring map \mathcal{O}_ X(V) \to \mathcal{O}_ X(U \cap V) is injective, see Lemma 29.7.5 below. In particular we see that scheme theoretically dense implies dense which is pleasing.
Example 29.7.2. Here is an example where scheme theoretic closure being X does not imply dense for the underlying topological spaces. Let k be a field. Set A = k[x, z_1, z_2, \ldots ]/(x^ n z_ n) Set I = (z_1, z_2, \ldots ) \subset A. Consider the affine scheme X = \mathop{\mathrm{Spec}}(A) and the open subscheme U = X \setminus V(I). Since A \to \prod _ n A_{z_ n} is injective we see that the scheme theoretic closure of U is X. Consider the morphism X \to \mathop{\mathrm{Spec}}(k[x]). This morphism is surjective (set all z_ n = 0 to see this). But the restriction of this morphism to U is not surjective because it maps to the point x = 0. Hence U cannot be topologically dense in X.
Lemma 29.7.3. Let X be a scheme. Let U \subset X be an open subscheme. If the inclusion morphism U \to X is quasi-compact, then U is scheme theoretically dense in X if and only if the scheme theoretic closure of U in X is X.
Proof. Follows from Lemma 29.6.3 part (3). \square
Example 29.7.4. Let A be a ring and X = \mathop{\mathrm{Spec}}(A). Let f_1, \ldots , f_ n \in A and let U = D(f_1) \cup \ldots \cup D(f_ n). Let I = \mathop{\mathrm{Ker}}(A \to \prod A_{f_ i}). Then the scheme theoretic closure of U in X is the closed subscheme \mathop{\mathrm{Spec}}(A/I) of X. Note that U \to X is quasi-compact. Hence by Lemma 29.7.3 we see U is scheme theoretically dense in X if and only if I = 0.
Lemma 29.7.5. Let j : U \to X be an open immersion of schemes. Then U is scheme theoretically dense in X if and only if \mathcal{O}_ X \to j_*\mathcal{O}_ U is injective.
Proof. If \mathcal{O}_ X \to j_*\mathcal{O}_ U is injective, then the same is true when restricted to any open V of X. Hence the scheme theoretic closure of U \cap V in V is equal to V, see proof of Lemma 29.6.1. Conversely, suppose that the scheme theoretic closure of U \cap V is equal to V for all opens V. Suppose that \mathcal{O}_ X \to j_*\mathcal{O}_ U is not injective. Then we can find an affine open, say \mathop{\mathrm{Spec}}(A) = V \subset X and a nonzero element f \in A such that f maps to zero in \Gamma (V \cap U, \mathcal{O}_ X). In this case the scheme theoretic closure of V \cap U in V is clearly contained in \mathop{\mathrm{Spec}}(A/(f)) a contradiction. \square
Lemma 29.7.6. Let X be a scheme. If U, V are scheme theoretically dense open subschemes of X, then so is U \cap V.
Proof. Let W \subset X be any open. Consider the map \mathcal{O}_ X(W) \to \mathcal{O}_ X(W \cap V) \to \mathcal{O}_ X(W \cap V \cap U). By Lemma 29.7.5 both maps are injective. Hence the composite is injective. Hence by Lemma 29.7.5 U \cap V is scheme theoretically dense in X. \square
Lemma 29.7.7. Let h : Z \to X be an immersion. Assume either h is quasi-compact or Z is reduced. Let \overline{Z} \subset X be the scheme theoretic image of h. Then the morphism Z \to \overline{Z} is an open immersion which identifies Z with a scheme theoretically dense open subscheme of \overline{Z}. Moreover, Z is topologically dense in \overline{Z}.
Proof. By Lemma 29.3.2 or Lemma 29.3.3 we can factor Z \to X as Z \to \overline{Z}_1 \to X with Z \to \overline{Z}_1 open and \overline{Z}_1 \to X closed. On the other hand, let Z \to \overline{Z} \subset X be the scheme theoretic closure of Z \to X. We conclude that \overline{Z} \subset \overline{Z}_1. Since Z is an open subscheme of \overline{Z}_1 it follows that Z is an open subscheme of \overline{Z} as well. In the case that Z is reduced we know that Z \subset \overline{Z}_1 is topologically dense by the construction of \overline{Z}_1 in the proof of Lemma 29.3.3. Hence \overline{Z}_1 and \overline{Z} have the same underlying topological spaces. Thus \overline{Z} \subset \overline{Z}_1 is a closed immersion into a reduced scheme which induces a bijection on underlying topological spaces, and hence it is an isomorphism. In the case that Z \to X is quasi-compact we argue as follows: The assertion that Z is scheme theoretically dense in \overline{Z} follows from Lemma 29.6.3 part (3). The last assertion follows from Lemma 29.6.3 part (4). \square
Lemma 29.7.8. Let X be a reduced scheme and let U \subset X be an open subscheme. Then the following are equivalent
U is topologically dense in X,
the scheme theoretic closure of U in X is X, and
U is scheme theoretically dense in X.
Proof. This follows from Lemma 29.7.7 and the fact that a closed subscheme Z of X whose underlying topological space equals X must be equal to X as a scheme. \square
Lemma 29.7.9. Let X be a scheme and let U \subset X be a reduced open subscheme. Then the following are equivalent
the scheme theoretic closure of U in X is X, and
U is scheme theoretically dense in X.
If this holds then X is a reduced scheme.
Proof. This follows from Lemma 29.7.7 and the fact that the scheme theoretic closure of U in X is reduced by Lemma 29.6.7. \square
Lemma 29.7.10. Let S be a scheme. Let X, Y be schemes over S. Let f, g : X \to Y be morphisms of schemes over S. Let U \subset X be an open subscheme such that f|_ U = g|_ U. If the scheme theoretic closure of U in X is X and Y \to S is separated, then f = g.
Proof. Follows from the definitions and Schemes, Lemma 26.21.5. \square
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