I am reading the Liu's Algebraic Geometry and arithmetic curves, p.348, Example 3.2. and stuck at some point.
Definition 3.1 ( In his book p.347 ). Let $S$ be a Dedekind scheme. We call an integral, projective, flat $S$-scheme $\pi : X \to S$ of dimension $2$ a fibered surface over $S$. The generic point of $S$ will be denoted by $\eta$. We call $X_{\eta}$ the generic fiber of $X$. A fiber $X_s$ with $s\in S$ closed is called a closed fiber. When $\operatorname{dim}S=1$, $X$ is called a projective flat $S$-curve (see Lemma 3.3). Note that the flatness of $\pi$ is equivalent to the surjectivity of $\pi$. (C.f. Fibered surface is flat if and only if it surjects onto the base ) We will say that $X$ is normal (resp. regular) fibered surface if $X$ is normal (resp. regular).
Example 3.2. Let $S:= \operatorname{Spec}\mathbb{Z}$ and $X:=\operatorname{Proj}\mathbb{Z}[x,y,z]/(y^2z+yz^2-x^3+xz^2)$. Let us show that $X$ is a normal fibered surface. The only point needing verification is that $X$ is normal. ( Next argument for showing normality is omitted ).
I don't understand why the bold statement is true. $X$ is integral, projective scheme (over $S$).
My question is, for $X$ to become fibered surface over $S$,
- Q.1. Why $X$ has dimension $2$?
EDIT : First attempt to the first question. Note that by the Liu's book, p.53, Lemma 3.41, letting $I :=(f(x,y,z):=y^2z+yz^2-x^3+xz^2)$, $ \operatorname{Proj}\mathbb{Z}[x,y,z]/I$ is isomorphic to a closed subscheme of $\mathbb{P}^{2}_{\mathbb{Z}}$, with support ( i.e., with underlying topological space) $V_{+}(I)$.
So, $\dim \operatorname{Proj}\mathbb{Z}[x,y,z]/I = \dim V_{+}(I)$ and it suffices to show that $\dim V_{+}(I)=2$. First note that $V_{+}(I)$ is a proper closed subset of $\mathbb{P}^{2}_{\mathbb{Z}}$ (True?). So,
$$ \dim V_{+}(I) < \dim \mathbb{P}^{2}_{\mathbb{Z}} = \dim Z + 2 = 3 .$$
so that $\dim V_{+}(I) \le 2$. So it suffices to show that $ 2 \le \dim V_{+}(I)$.
Now I arrange some preliminary definitions for showing this.
We set $A_{+} := \oplus_{d \ge 1} A_d$. This is a graded ideal. More generally, if $I \subseteq A$ is a homogeneous ideal, we set $I_{+} := I \cap A_{+}$. This is again a homogeneous ideal. A homogeneous prime ideal $\mathfrak{p} \subset A$ is called relevant if it does not contain $A_{+}$, i.e., if $\mathfrak{p}_{+} \subsetneq A_{+}$.
Proposition 13.4. (Gortz's Algebraic Geometry, p.369) Let $A$ be a graded ring. For a subset $Y \subseteq \operatorname{Proj}A$ define
$$ I_{+}(Y) := (\cap_{\mathfrak{p} \in Y} \mathfrak{p}) \cap A_{+}.$$
(2) The maps $Y \mapsto I_{+}(Y)$ and $I \mapsto V_{+}(I)$ define mutually inverse, inclusion reversing bijections between the set of homogeneous ideals $I \subseteq A_{+}$ such that $I=\operatorname{rad}(I)_{+}$ and the set of closed subsets of $\operatorname{Proj}A$.
Via this bijection, the closed irreducible subsets correspond to ideals of the form $\mathfrak{p}_{+}$, where $\mathfrak{p}$ is relevant prime ideal.
Note that $\mathbb{Z}[x,y,z]_{+} = (x,y,z)$ (True?). And note that the above $I$ is homogeneous relevant prime ( In fact $I$ is a prime ideal since $f(x,y,z):=y^2z+yz^2-x^3+xz^2$ can be viewed as $-x^3 + z^2 x + (y^2z +yz^2) \in \mathbb{Z}[y,z][x]$ and applying the Eisenstein's Criterion ) ideal of $\mathbb{Z}[x,y,z]$ such that $I = I \cap \mathbb{Z}[x,y,z]_{+} =: I_{+}$.
So we have $I = I_{+}(V_{+}(I))$. Note that if we can find two relevant prime ideal $\mathfrak{p}_1$, $\mathfrak{p}_2$ such that
$$I = I_{+}(V_{+}(I)) \subsetneq \mathfrak{p}_{1, +} \subsetneq \mathfrak{p}_{2, +}$$, then by the above inclusion reversing bijection, we obtain chain of closed irreducible subsets
$$ V_{+}(\mathfrak{p}_{2, +}) \subsetneq V_{+}(\mathfrak{p}_{1, +}) \subsetneq V_{+}(I) .$$
( $\divideontimes$ If $C \subseteq X$ is closed and $F \subseteq X$ is closed irreducible subset in $X$ such that $F \subseteq C$, then $F$ is closed irreducible in $C$. )
And so we have $ 2 \le \dim V_{+}(I)$. And we really can find such $\mathfrak{p}_1$, $\mathfrak{p}_2$ ?
Is it possible?
EDIT : Yes. It seems possible. Let $\mathfrak{p}_1 := (f, x,z)$ and $\mathfrak{p}_2:= (f,x,z,7y,7)$ where $f(x,y,z):=y^2z+yz^2-x^3+xz^2$. Such ideals may works. C.f. Are there homogeneous prime ideals $p_1 , p_2 \nsupseteq (x,y,z)$ such that $I:=(y^2z+yz^2-x^3+xz^2) \subsetneq p_{1,+} \subsetneq p_{2,+}$? .
- Q.2. why $X$ is flat over $S$? Or equivalently (as the bold statement in the above definition 3.1.), why $X$ is surjective over $S$?
Here I think that the morphism $\pi : X\to S$ is as : What is explicit description of the structure morphism $\operatorname{Proj}A \to \operatorname{Spec}R$?.
EDIT : First attempt for the question 2 : I tried to show the flatness of $X$ over $S$ directly. Note https://stacks.math.columbia.edu/tag/01U5 and https://stacks.math.columbia.edu/tag/0AUW. From these, it sufficient to show that there is an open cover $X = \pi^{-1}(S) = \cup_{ i \in I} U_i$ such that $\mathbb{Z} = \mathcal{O}_S(S) \to \mathcal{O}_X(U_i)$ is torsion free for all $i\in I$. In particular, note that $X = \cup _{i \in I} D_{+}(f_i)$ where $f_i$ is homogeneous elements of $A_{+}$ where $A:= \mathbb{Z}[x,y,z]/(y^2z+yz^2-x^3+xz^2)$.
My question is, the induced ring maps $\mathbb{Z} = \mathcal{O}_S(S) \to \mathcal{O}_X(D_{+}(f_i)) = A_{(f_i)}$ is torsion free ? If so we are done.
Second attempt for the question 2 : I tried to show the surjectivity of the $\pi : X \to S$ which is equivalent to the flatness, as I mentioned above. As I linked What is explicit description of the structure morphism $\operatorname{Proj}A \to \operatorname{Spec}R$?, the $\pi : X \to S$ is constructed from glueing next morphisms ( $f\in A_{+}$ homogeneous elements )
$$\psi_{f} : D_{+}(f) \xrightarrow{\cong} \operatorname{Spec}A_{(f)} \xrightarrow{^al_f} \operatorname{Spec}A_{0}= \operatorname{Spec}\mathbb{Z}$$
, where $l_f : A_0 \to A_{(f)}$ be the natural map $a \mapsto \frac{a}{1}$. Note that $A_0 = \mathbb{Z}/(I\cap \mathbb{Z}) = \mathbb{Z}$.
To show the surjectivity of $\pi$, let $(p)$ be a prime ideal of $\mathbb{Z}$. Then, my question is,
" does there exists homogeneous element $ f\in A_{+}$ and there exists $\mathfrak{q} \in \operatorname{Spec}A_{(f)}$ such that $^a l_f (\mathfrak{q}) = l_f^{-1}(\mathfrak{q}) = \{n \in A_0=\mathbb{Z} : \frac{n}{1} \in \mathfrak{q} \} = (p)$. " ?
Assume that $\{n \in A_0=\mathbb{Z} : \frac{n}{1} \in \mathfrak{q} \} = (p)$ is true for some $f$ and $\mathfrak{q}$. Then
- If $np \in (p)$, then $\frac{np}{1} \in \mathfrak{q}$.
- If $n\in l_f^{-1}(\mathfrak{q})$ so that $\frac{n}{1}\in \mathfrak{q}$, then $n \in (p)$
From these, can we guess such $f \in A_{+}$ and $\mathfrak{q}$?
Can any one provide such $f \in A_{+}$ and $\mathfrak{q}$?
Third attempt for the question 2 : Now I'm trying to show the sujectivity of $\pi :X \to S$ in other direction, helped from below hint of Aphelli.
Let $s= (p) \in S:= \operatorname{Spec}\mathbb{Z}$. It suffices to show that $\pi^{-1}(s) \neq \varnothing$. Note that by the Liu's Algebraic geometry book, p.82, Proposition 3.1.9., we have ( $I:= (y^2z+yz^2−x^3+xz^2)$ )
$$ \pi^{-1}(s) \cong \operatorname{Proj}(\mathbb{Z}[x,y,z]/I) \times _ {\operatorname{Spec}\mathbb{Z}}\operatorname{Spec}k(s) \cong \\ \operatorname{Proj}(\mathbb{Z}[x,y,z]/I \otimes_{\mathbb{Z}} k(s)) \cong \operatorname{Proj}( k(s)[x,y,z]/Ik(s)[x,y,z]) \cong V_{+}(Ik(s)[x,y,z]) \subseteq \mathbb{P}^{2}_{k(s)}$$
( C.f. The final isomorphism is true by the Liu's Lemma 2.3.41 (p.53). For the third isomorphism, see Quotients and extensions of scalars in polynomial algebras. True? )
Note that $k(s) = \mathbb{Q}$ when $p=0$ and $k(s)=\mathbb{Z}/(p)$ when $p$ is prime number. So we can view the $Ik(s)[x,y,z]$ as the principal ideal generated by $f(x,y,z):= y^2z+yz^2-x^3+xz^2$ whose coefficients are viewed as in $\mathbb{Q}$ or $\mathbb{Z}/(p)$. If this polynomial is irreducible over $\mathbb{Q}$ and $\mathbb{Z}/(p)$, then $V_{+}(Ik(s)[x,y,z]) \subseteq \mathbb{P}^{2}_{k(s)}$ is of codimension $1$ by the following thorem (Gortz's Algebraic Geometry, Corollary 5.42 ) so that it is nonempty so that $\pi^{-1}(s)$ is nonempty and we are done
Corollary 5.42. Let $Z\subseteq \mathbb{P}^n_{k}$ be an integral closed subscheme. Then $Z$ is of codimension $1$ if and only if $Z=V_{+}(f)$ for an 'irreducible' homogeneous polynomial $f$.
Q. And $f(x,y,z)$ is irreducible over $\mathbb{Q}$ and $\mathbb{Z}/(p)$ ?
Fourth attempt for the question 2 : Through communicating with Liu, I noticed that we may use next theorem ( Corollary 4.3.10 in Liu's book ) :
Corollary 4.3.10. Let $Y$ be a Dedekind scheme, and $f:X\to Y$ be a non-constant morhpism ( i.e., $f(X)$ is not reduced to a point ) with $X$ integral. Then $f$ is flat.
So, my question is, the above $\pi : X \to S$ is non-constant? Is there any criteria for the non-constancy of a morphism ?
P.s. I am struggling with this issue for few hours. It will be appreciate for someone who can give a any crucial hint. Can anyone helps?
