# Proper and flat over $\mathbb{P}^1_{\mathbb{Z}}$ implies locally free

Let $$\pi:X\to \mathbb{P}^1_{\mathbb{Z}}$$ be a proper flat morphism with $$X$$ an integral scheme. Is $$\pi_*\mathcal{O}_X$$ necessarily locally free?

• That is not true. Let $X$ be a nonreduced scheme whose underlying reduced scheme is $\mathbb{P}^1_{\mathbb{Z}}\times_{\text{Spec}\ \mathbb{Z}}\mathbb{P}^1_{\mathbb{Z}}$, with $\pi$ equal to projection onto the first factor. Jun 13, 2020 at 17:04
• The OP added the hypothesis that $X$ is an integral scheme after I posted my comment. Jun 13, 2020 at 18:38
• I think I might have a solution when $X_{\mathbb{Q}}$ is geometrically connected, because in that case $\pi_*\mathcal{O}_X$ equals the structure sheaf of $\mathbb{P}^1_{\mathbb{Z}}$ by analyzing the Stein factorization of $\pi$. If you are interested in this more restrictive setting I can write it down.
– Jef
Jun 15, 2020 at 15:06

Like your other question, the answer to this one is related to miracle flatness:

Theorem (Miracle flatness). Let $$f \colon X \to Y$$ be a finite dominant morphism of schemes with $$Y$$ regular. Then $$f$$ is flat if and only if $$X$$ is Cohen–Macaulay.

See for example Tags 00R4 and 00R5.

Now let $$f \colon X \to Y$$ be flat and proper, with $$Y$$ regular of dimension $$2$$. Let $$X \to Y' \to Y$$ be the Stein factorisation, so $$Y'$$ is the normalisation of $$Y$$ in $$X$$ (see Tag 03H0). Here is a positive result:

Claim. If $$X$$ is normal, then $$f_* \mathcal O_X$$ is locally free.

Indeed, if $$X$$ is normal, then so is $$Y'$$, hence it is $$S_2$$ by Serre's criterion (Tag 031S). Since it is finite (dominant) over $$Y$$, it has dimension $$2$$, so $$S_2$$ is equivalent to Cohen–Macaulay. Then miracle flatness says that $$Y' \to X$$ is flat. $$\square$$

Conversely, if we assume a priori that $$f$$ is finite, then $$X = Y'$$, which is $$S_2$$ if and only if it is Cohen–Macaulay if and only if it is flat over $$Y$$. So it suffices to construct a finite map $$X \to \mathbf P^1_{\mathbf Z}$$ with $$X$$ not $$S_2$$.

There are many ways to do this. One construction is to take a double cover $$\mathbf P^1_{\mathbf Z} \to \mathbf P^1_{\mathbf Z}$$ by $$[x:y] \mapsto [x^2:y^2]$$, and glueing together the points $$[1:0]$$ and $$[-1:0]$$ in the fibre above $$3 \in \operatorname{Spec} \mathbf Z$$ of the source $$\mathbf P^1_{\mathbf Z}$$ (which map to the same point in the target). The resulting scheme is regular in codimension $$1$$ but not normal, so cannot be $$S_2$$. (See also this answer for glueing schemes along closed subschemes.)

• but in your construction you are not taking two schemes $X$ and $Y$ and gluing them along a common closed subscheme $Z$; you are somehow glueing closed points on the same scheme. Does that exist in the category of schemes?
– user158636
Jun 16, 2020 at 14:59
• Reverse engineering this answer, if $S$ is an affine piece of $\mathbf P^1_{\mathbf Z}$, you can construct the quotient as the pullback of $S \to S/I \cong \mathbf F_3^2 \leftarrow \mathbf F_3$, i.e. the functions whose values at the two points agree. So geometrically you're glueing $\mathbf P^1_{\mathbf Z}$ and a point $*$ along $* \to * \amalg * \subseteq \mathbf P^1_{\mathbf Z}$, which is a slight generalisation of glueing two schemes along a closed subscheme. See also Tag 0E25. Jun 16, 2020 at 21:25