Existence of rational section to a flat projective morphism

Question

Let $X$ be a smooth projective variety over an algebraically closed field $K$ of dimension greater than $1$. Suppose there exists a flat projective morphism $f:X \to \mathbb{P}^n$ for some $n \ge 1$. Suppose there exists a divisor in $X$ which is flat over $\mathbb{P}^n$. Under what condition on $f$ or $X$, does this imply that there exist a rational section to $f$?

Answer 1

This is a follow-up to Mike Roth's important, correct comment (now disappeared). The following is one of a series of examples that I learned of from Tom Graber, but which I guess goes back to the work on "normic forms". Assume that the characteristic is not $3$ (there are similar examples in every characteristic). Let $\mathbb{P}^1$ have homogeneous coordinates $[U,V]$. Let $\mathbb{P}^3$ have homogeneous coordinates $[T_0,T_1,T_2]$. Consider the closed subscheme $Y$ of $\mathbb{P}^1\times \mathbb{P}^2$ with bihomogeneous defining equation $$ F(U,V;T_0,T_1,T_2) = U^2T_0^3 + UVT_1^3 + V^2T_2^3.$$ There is an action of $\mathbb{G}_m$ on $\mathbb{P}^1\times \mathbb{P}^2$ where $\lambda\in \mathbb{G}_m$ acts by $$ \lambda \bullet ([U,V],[T_0,T_1,T_2]) = ([U,\lambda^{-3}V],[T_0,\lambda T_1,\lambda^2 T_2]).$$ The homogeneous polynomial $F$ is invariant for this (bilinearized) action. Thus $Y$ is invariant.

Of course $Y$ is singular at the points $([1,0],[0,0,1])$ and $([0,1],[1,0,0])$, but that changes nothing. There exists a resolution $\nu:X\to Y$ that is projective (just a sequence of blowings up). Since $X$ is a smooth, projective variety of dimension $2$, since $\mathbb{P}^1$ is a smooth, projective variety of dimension $1$, and since the morphism $f := \text{pr}_{\mathbb{P}^1}\circ \nu$ is surjective, automatically $f:X\to \mathbb{P}^1$ is flat and projective.
There is certainly a $\mathbb{P}^1$-flat divisor in $X$: just take the closure in $X$ of any positive degree, effective zero cycle on the generic fiber, e.g., the common zero locus of $F$ and $G=T_0+T_1+T_2$. On the other hand, I claim that there is no rational section of $f$.

If there were a rational section of $f$, then its image in $Y$ would be a rational section of $\text{pr}_{\mathbb{P}^1}:Y\to \mathbb{P}^1$. The Zariski closure of this rational section would be a curve, which then gives a point of the Hilbert scheme parameterizing curves on $Y$. The action of $\mathbb{G}_m$ on $Y$ induces an action of $\mathbb{G}_m$ on the Hilbert scheme. Consider the orbit under $\mathbb{G}_m$ of the specified point of the Hilbert scheme. Since the (connected components of the) Hilbert scheme are projective, the closure of this orbit is proper. In particular, there exists a "limit at infinity" of the original Hilbert point.

This "limit point" parameterizes a curve $C$ in $Y$ that is a limit of rational sections, and that is $\mathbb{G}_m$-invariant. By the valuative criterion of properness applied to $f$, every limit of a one-parameter family of rational sections of $f$ is, again, a rational section of $f$, i.e., there is a unique component of $C$ that dominates $\mathbb{P}^1$, and this component is the image of a rational section of $f$. Since $C$ is $\mathbb{G}_m$-invariant, also the rational section is $\mathbb{G}_m$-equivariant.

However, the only $\mathbb{G}_m$-equivariant rational sections of the projection, $$\text{pr}_{\mathbb{P}^1}:\mathbb{P}^1\times \mathbb{P}^2\to \mathbb{P}^1,$$ are "monomial" sections, i.e., $$ h([U,V]) = ([U,V],[c_0U^{a_0}V^{b_0},c_1U^{a_1}V^{b_1}, c_2U^{a_2}V^{b_2}]), $$ where $c_0$, $c_1$, $c_2$ are elements in the ground field such that $(c_0,c_1,c_2)\neq (0,0,0)$, and where each $a_i$ and $b_i$ is a positive integer such that $a_0+b_0=a_1+b_1=a_2+b_2 = e$ for some positive integer $e$. But then the restriction of the equation $F$ on this section is, $$ F\circ h([U,V]) = c_0^3U^{3a_0+2}V^{3b_0} + c_1^3U^{3a_1+1}V^{3b_1+1} + c_2^3U^{3a_2}V^{3b_2+2}. $$ In particular, the congruence classes modulo $3$ of the exponent vectors of the $3$ terms are $(\overline{2},\overline{0})$, $(\overline{1},\overline{1})$, and $(\overline{0},\overline{2})$. So there can be no cancellation, i.e., the monomials are linearly independent in $k[U,V]$. So the only way that this linear combination of monomials may be zero is if $(c_0^3,c_1^3,c_2^3)$ equals $(0,0,0)$, contradicting that $(c_0,c_1,c_2)\neq (0,0,0)$. This contradiction proves that there is no rational section of $f$.

By the way, this works more generally to show that for every triple of positive integers $(r,d,n)$ with $n+1=d^r$, there exists a degree $d$ hypersurface $Y$ in $\mathbb{P}^r\times \mathbb{P}^n$ such that the projection $Y\to \mathbb{P}^r$ admits no rational section. These are "normic forms" showing that the Tsen-Lang theorem is sharp.

Edit. My equivariance argument above is off, although the conclusion is correct. By the valuative criterion of properness, every equivariant rational section of $\text{pr}_{\mathbb{P}^1}$ extends to an equivariant regular section. By the classification of invertible sheaves on $\mathbb{P}^1$ and the universal property of $\mathbb{P}^2$, every regular section $h$ is of the form, $$ h([U,V]) = ([U,V],[h_0(U,V),h_1(U,V),h_2(U,V)]), $$ where $h_0$, $h_1$ and $h_2$ are homogeneous polynomials in $k[U,V]$ of some common degree $e$. Finally, the equivariance gives that, for some integer $w$ (the "weight"), we have $$ h_0(U,\lambda^3 V) = \lambda^w h_0(U,V), \ h_1(U,\lambda^3 V) = \lambda^{w+1} h_1(U,V),$$ $$ h_2(U,\lambda^3 V) = \lambda^{w+2} h_2(U,V). $$ Now, for $i=0$, $1$ or $2$, if $(a_i,b_i)$ is the exponent vector from a nonzero term in $h_i$, then the equations above give that $$w+i=3b_i.$$ In particular, modulo $3$, $w$ is congruent to $-i$. Thus, if at least two of $h_0$, $h_1$, and $h_2$ is not the zero polynomial, we get a contradiction: $w$ is simultaneously congruent modulo $3$ to two of $0$, $1$ and $2$. However, if only one of the $h_i$ is nonzero, then $F\circ h$ equals $U^{2-i}V^ih_i(U,V)^3$, which is a nonzero polynomial in $k[U,V]$. Since $F\circ h$ is nonzero, this contradicts that $h$ factors through $Y$. So the conclusion is the same: there are no $\mathbb{G}_m$-equivariant rational sections of $f$, hence there are no rational sections at all.