I am reading the Gortz's Algebraic Geometry, proof of the Theorem 5.32 and stuck at understanding some statement :

Theorem 5.32. Let $X$ be an integral $k$-scheme of finte type, and let $f\in \Gamma(X, \mathcal{O}_X)$ be a non-unit, and different from $0$ ( i.e., $\varnothing \subsetneq V(f) \subsetneq X$ ). Then $V(f)$ is equi-dimensional of codimension $1$ in $X$.

Here ( as Lukas Heger below commented ) $V(f) := \{ x\in X : f_x \in \mathfrak{m}_x \mathcal{O}_{X,x}\}$ ( c.f. Liu's Algebraic Geometry book, p.74 ) Note that $V(f)$ is the complement of $X_f$ and for any affine open subset $U$ of $X$, $V(f) \cap U$ is the principal closed subset $V(f|_U)$ in the affine scheme $U$.

Proof. Since $V(f)$ has only finitely many irreducible components $Z_1, \dots, Z_r$, there exists for each $i=1, \dots, r$ an open affine neighborhood $U_i = \operatorname{Spec}A_i$ of the generic point $\eta_i$ of $Z_i$ such that $U_i \cap Z_j = \varnothing$ for $j\neq i$. ( This is possible by considering open subschemes $W_i := X - (Z_1 \cup \cdots \cup \hat{U_i} \cup \cdots \cup U_r )$, each of which containing the generic point $\eta_i$ ). By Theorem 5.22 (3) we have $\dim X = \dim U_i$. Replacing $X$ by $U_i$ and $f$ by $f|_{U_i}$, we therefore may assume that $X= \operatorname{Spec}A$ is affine and that $V(f)$ is irreducible. Let..(Omitted)

I can't understand the bold statement. Consider next theorem ( his book Proposition 5.30 )

Proposition 5.30. Let $X$ be an irreducible scheme finite type over a field $k$. Then for all closed subsets $Y$ of $X$ we have $ \dim Y + \operatorname{codim}_XY = \dim X$.

By the reduction assumption, $V(f|_{U_i}) = V(f) \cap U_i$ is equi-codimensional of codimension $1$ of $U_i$ ; i.e., its irreducible components has codimension $1$ in $U_i$. From this and $U_i \cap Z_j = \varnothing$ for $j\neq i$, can we show that $\operatorname{codim}_X Z_1, \dots, \operatorname{codim}_X Z_r =1 $? To show this, by the Proposition 5.30. , it suffices to show that $\dim Z_1 , \dots , \dim Z_r = \dim X-1$. Can anyone helps?

  • 2
    $\begingroup$ You can define $V(f)$ as the set of points $x$ such that the image of $f$ under $\mathcal O_X \to \mathcal O_{X,x}$ is contained in the maximal ideal of $\mathcal O_{X,x}$. This agrees with the usual notion of $V(f)$ in the affine case. $\endgroup$ May 19, 2023 at 3:35

1 Answer 1


O.K. I got it. I will use theorem 5.22-(3) of the Gortz's book :

Theorem 5.22-(3) . Let $X$ be an irreducible $k$-scheme of finite type with generic point $\eta$. Then for any non-empty open subscheme $U$ of $X$, we have $\dim U= \dim X$.

Fix $i$ ( $ 1 \le i \le r$ ).

Since $U_i \cap Z_i $ is non-empty ( containing the generic point of $Z_i$ ), by the Theorem 5.22-(3), $\dim Z_i = \dim (U_i \cap Z_i)$.

Note that $V(f) \cap U_i = (Z_1 \cup \cdots \cup Z_r ) \cap U_i = Z_i \cap U_i$ since $ U_i \cap Z_j = \varnothing$ for $j \neq i$. So, $\dim Z_i = \dim (U_i \cap Z_i) = \dim (V(f) \cap U_i) = \dim V(f|_{U_i})$.

Since by the reduction assumption $V(f|_{U_i})$ is equi-codimensional of codimension $1$ in $U_i$ so that $\dim Z_{i,1} , \dots \dim Z_{i,n} = \dim U_i -1 = \dim X -1$ ( where $Z_{i,j}$ are the irreducible components of $V(f|_{U_i})$ and we used the Proposition 5.30 above in the question ), we have $\dim V(f|_{U_i}) = \dim X -1$ so that $\dim Z_i = \dim X -1 $ and we are done.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .