A non-zero section of an invertible sheaf on a geometrically integral (smooth?) projective $k$-scheme is regular?

Question

Let $X$ be a projective, geometrically integral $k$-scheme $X$ for a field $k$ (possibly we also need $X$ smooth, i.e. $X_{\overline{k}}$ regular). It seems implicit in Hartshorne's discussion (with the smoothness hypothesis) of linear systems (beginning about half way down page 156, continuing through page 157) that if $\mathscr{L}$ is an invertible $\mathscr{O}_X$-module and $s\in\Gamma(X,\mathscr{L})$ is non-zero, then in fact $s$ is regular in the sense that the corresponding $\mathscr{O}_X$-module map $\mathscr{O}_X\to\mathscr{L}$ is injective (explicitly: if $U\subseteq X$ is open and $f\in\Gamma(U,\mathscr{O}_X)$ is non-zero, then $f\cdot(s\vert_U)$ is non-zero in $\Gamma(U,\mathscr{L})$). This would ensure that the so-called divisor of zeros of $f$ is actually a Cartier divisor. Minimally, I think one needs a trivialization of $\mathscr{L}$ with the property that $s$ does not restrict to zero on any member of the trivialization (so regularity is sufficient at least).

On page 395 of the book of Görtz-Wedhorn, they work with $X$ projective and satisfying $\Gamma(X,\mathscr{O}_X)=k$ (which implies in particular that $X$ is geometrically connected) in their discussion of linear systems and seem to take the same fact about sections of invertible modules for granted. I can't see why this is true.

Question: If $s\in\Gamma(X,\mathscr{L})$ is non-zero, is $s$ necessarily regular in the sense described above?

Answer 1

An alternative way to see this is by using the following lemma.

Lemma. Let $X$ be integral with generic point $\eta$, and let $\mathscr E$ be locally free. Then the map $$\Gamma(X, \mathscr E) \to \mathscr E_\eta$$ is injective.

Proof. Suppose $s \in \Gamma(X,\mathscr E)$ is nonzero. Cover $X$ by affine opens $U$ on which $\mathscr E|_U \cong \mathcal O_U^n$. By the sheaf condition, we have $s|_U \neq 0$ for one of these $U$. Say $U = \operatorname{Spec} A$, and let $K = \kappa(\eta)$ be the fraction field of $A$. Then we are considering the map $A^n \to K^n$, which is clearly injective. $\square$

Remark. The result is of course not true for torsion sheaves, e.g. because $\mathscr E_\eta$ can be zero even if $\mathscr E \neq 0$. (Example: $\mathbb Z/n \mathbb Z$ on $\operatorname{Spec} \mathbb Z$.)

Corollary. Let $X$ be integral. Then a map $\phi \colon \mathscr E \to \mathscr F$ with $\mathscr E$ locally free and $\mathscr F$ quasicoherent is injective if and only if the map $\phi_\eta \colon \mathscr E_\eta \to \mathscr F_\eta$ is injective.

Proof. Clearly if $\phi$ is injective, then so is $\phi_\eta$; this follows from exactness of filtered colimits. Conversely, assume $\phi_\eta$ is injective. We need to prove that $\phi(U)$ is injective for all $U \subseteq X$ open; wlog we may take $U = X$. Consider the commutative diagram $$\begin{array}{ccc}\Gamma(X,\mathscr E) & \stackrel \phi \to & \Gamma(X,\mathscr F) \\ \downarrow & & \downarrow \\ \mathscr E_\eta & \stackrel{\phi_\eta}\to & \mathscr F_\eta.\end{array}$$ The bottom map is injective by assumption, and the left vertical map by the lemma. Hence, the top map is injective as well. $\square$

Remark. To come back to your question: you know that $s \neq 0$, hence $s_\eta$ is nonzero, so multiplication by $s$ is injective on the $1$-dimensional vector space $\mathcal O_\eta = \kappa(\eta)$.

Remark. There is probably a version of the results above for arbitrary quasicoherent sheaves where we have to check injectivity at the associated primes of $\mathscr E$ (a notion to be made precise). In the affine case see Eisenbud, Corollary 3.5. If $\mathscr E$ is coherent, the set of associated primes should be (locally?) finite, and if $\mathscr E$ is torsion-free, it's only the generic point.

Answer 2

All you need for this is for $X$ to be integral. Injectivity of a map $\mathscr{O}_X\to\mathscr{L}$ can be checked locally. Locally, $\mathscr{L}\cong\mathscr{O}_X$, and so the map $\mathscr{O}_X\to\mathscr{L}\cong\mathscr{O}_X$ is multiplication by some section of $\mathscr{O}_X$, which is injective as long as the section is nonzero since $X$ is integral. This says that if $s$ is a section of $\mathscr{L}$ which is not regular, then there must be some nonempty open set $U$ such that the restriction of $s$ to $U$ is the zero section. But the set of points where $s$ vanishes (in the sense of vanishing in the fiber) is closed, so by irreducibility $s$ vanishes at every point of $X$. Since $\mathscr{L}\cong\mathscr{O}_X$ locally, a section which vanishes everywhere looks locally like an element of the ring which is in every prime ideal, which must be $0$ because $X$ is reduced. Thus if $s$ is not regular, it must be the $0$ section.

The assumption that $X$ is projective and $\Gamma(X,\mathscr{O}_X)=k$ is not sufficient for this. For instance, if $X$ is the union of two lines in $\mathbb{P}^2$ and $\mathscr{L}=\mathscr{O}(1)$, then the equation of one of the two lines gives a nonzero section of $\mathscr{L}$ which is not regular.