# Understanding Theorem 5.32 in the Gortz's Algebraic Geometry book

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?

• 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. May 19, 2023 at 3:35

## 1 Answer

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.