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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