Codimension of a closed irreducible subset and generic points

Lemma 1 Let $Z$ be an irreducible subset of $X$. Then the closure $\mathrm{cl}_X(Z)$ in $X$, is an irreducible subset of $X$.

Proof) Suppose for contradiction that $\mathrm{cl}_X(Z) = Z_1 \cup Z_2$ where $Z_i$ are closed proper subsets of $\mathrm{cl}_X(Z)$. Then $Z = (Z_1 \cap Z) \cup (Z_2 \cap Z)$ where each $Z_i \cap Z$ is a closed subset of $X$. If either $Z_i \cap Z = Z$, then $Z \subset Z_i \Rightarrow \mathrm{cl}_X(Z) \subset Z_i \subset \mathrm{cl}_X(Z)$ which contradicts our assumption that $Z_i$ are proper subsets of $\mathrm{cl}_X(Z)$. This shows that $Z_i \cap Z$ are proper subset of $Z$.

Lemma 2 Let $Y \subset X$ be a subspace and $Z \subset Y$. Then $\mathrm{cl}_Y(Z) = \mathrm{cl}_X(Z) \cap Y$.

Proof) Since $Z \subset \mathrm{cl}_X(Z) \cap Y$, it follows that $\mathrm{cl}_Y(Z) \subset \mathrm{cl}_X(Z) \cap Y$ as $\mathrm{cl}_X(Z) \cap Y$ is closed in $Y$. Now let $\mathrm{cl}_Y(Z) = W \cap Y$ where $W$ is a closed subset of $X$. Then $Z \subset \mathrm{cl}_Y(Z) = W \cap Y \subset W$ implies that $\mathrm{cl}_X(Z) \subset W$. We conclude that $\mathrm{cl}_X(Z) \cap Y \subset W \cap Y = \mathrm{cl}_Y(Z)$.

Lemma 3 Let $Z_1 \subsetneq Z_2$ be a chain of irreducible subsets of $X$. Then $\mathrm{cl}_X(Z_1) \subsetneq \mathrm{cl}_X(Z_2)$ is a chain of closed irreducible subsets of $X$.

Proof) It follows immediately from Lemma 1 and Lemma 2. If $\mathrm{cl}_X(Z_1) = \mathrm{cl}_X(Z_2)$, then $Z_1 = Z_2$ which contradicts the assumption.



Lemma 4 Let $U$ be a nonempty open subset of an irreducible space $X$. Then $U$ is dense in $X$, i.e. $\mathrm{cl}_X(U) = X$.

Proof) We have both $\mathrm{cl}_X(U)$ and $U^c$ are closed subsets. If $U$ is not dense, then both $\mathrm{cl}_X(U)$ and $U^c$ are proper subsets of $X$. This is a contradiction that $X$ is irreducible.

Lemma 5 Let $Z$ be a irreducible subset of $X$, then $Z \cap U$ is a irreducible subsets of $U$. Equivalently, if $X$ is a irreducible space, then any open subset $U$ of $X$ is irreducible.

Proof) Suppose $U = Z_1' \cup Z_2'$ where $Z_i'$ are proper closed subsets in $U$. Then there exist $Z_i$ closed in $X$ such that $Z_i' = Z_i \cap U$. Then $(Z_1 \cup Z_2) \cap U \subset U$. Hence $U \subset (Z_1 \cup Z_2) \subset X$. As $Z_1 \cup Z_2$ is closed and $U$ is dense by Lemma 4, it follows that $Z_1 \cup Z_2 = Z$. However, if $Z_i = Z$, then $U = Z_i \cap U = Z_i'$. This contradicts our assumption.

Lemma 6 $U \subset X$ be an open subspace. Let $Z$ be an irreducible closed subset of $X$. Then $Z = \mathrm{cl}_X(Z \cap U)$.

Proof) As $Z \cap U \subset Z$, it follows that $\mathrm{cl}_X(Z \cap U) \subset Z$. Suppose $\mathrm{cl}_X(Z \cap U) \neq Z$. Then $(Z \cap U)^c \cup \mathrm{cl}_X(Z \cap U) = Z$. However, it contradicts that $Z$ is an irreducible closed subset.

Lemma 7 Let $Z_1 \subsetneq Z_2$ be a chain of irreducible closed subsets of $X$, then $Z_1 \cap U  \subsetneq Z_2 \cap U$ is a chain of irreducible closed subsets of $U$ where $U$ is an open subset of $X$.

Proof) If $Z_1 \cap U = Z_2 \cap U$, then by Lemma 6, it follows that
$$Z_1 = \mathrm{cl}_X(Z_1 \cap U) = \mathrm{cl}_X(Z_2 \cap U) = Z_2$$



Using Lemma 3 and Lemma 7, we will show that if $X$ is a scheme, $Z$ is a closed irreducible subset of $X$, then for any $U \subset X$ open subscheme, we have

$$\mathrm{codim}_X(Z) = \mathrm{codim}_U(Z \cap U)$$

Suppose that we have a chain of closed irreducible subsets in $U$
$$Z \cap U=Z_l \subsetneq \cdots \subsetneq Z_0$$
Then it induces a chain of closed irreducible subsets in $X$
$$Z = \mathrm{cl}_X(Z_l) \subsetneq \cdots \subsetneq \mathrm{cl}_X(Z_0)$$


Conversely, a chain of closed irreducible subsets in $X$
$$Z = Z_l \subsetneq \cdots \subsetneq Z_0$$
induces a chain of irreducible subsets in $U$
$$Z \cap U = Z_l \cap U \subsetneq \cdots \subsetneq Z_0 \cap U$$


Now assume that $U = \mathrm{Spec}~A$. Then a chain of closed irreducible subsets starting from $Z$ in $U$ corresponds to a chain of prime ideals that is contained in the prime ideal corresponding to the generic point. If $\eta$ is the generic point of $Z$, we conclude that

$$\mathrm{codim}_X(Z) = \mathrm{codim}_U(Z \cap U) = \mathrm{dim}~ \mathcal{O}_{X, \eta}$$