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Let $X$ be a reduced scheme, so, generically regular; you may assume extra conditions like equidimensional and seminormal (though normal is stronger than I'd like, as is Gorenstein).

Is there a reasonable definition of "section of the anticanonical bundle" over $X$? At the very least I'd like to be able to restrict such a "section" to the regular locus and get an actual section. If $X$ is normal, then I'd like any (honest) anticanonical section over $X_{reg}$ to uniquely extend to one of these objects on $X$.

Note: a divisor and a section are not quite the same thing, in that the divisor of zeroes only determines the section up to a global invertible function (which need not even be constant). I really want a section, not just its divisor of zeroes.

The motivation comes from Frobenius splitting. If $X$ is defined over a perfect field of characteristic $p$, and $\sigma$ is a section of the anticanonical over $X_{reg}$ (not vanishing on any component), then away from $\sigma=0$ and the singularities we can define a morphism of sheaves $\varphi:\ F_* \mathcal O_X \to \mathcal O_X$ by $g \mapsto \mathcal C(g/\sigma) \sigma$ (here $F$ is the Frobenius and $\mathcal C$ is the Cartier operator on (top) forms). This extends over $X_{reg}$, and over all $X$ if $X$ is normal. Such maps $\varphi$ make sense even when $X$ is more singular, but I'd rather talk about an anticanonical section than about $\varphi$.

Feel free to add tags.

EDIT: since this hasn't produced an answer "of course you want this standard object well-known to those who know it well", here's more detail about exactly what I want.

Over $X$, we have the sheaf $\mathcal{Hom}(F_* \mathcal O_X, \mathcal O_X)$. If we restrict to $X_{reg}$, where $F$ becomes a (finite) flat morphism, then using Serre duality for such morphisms we can identify this sheaf with $F_* \mathcal{Hom}(\omega^p,\omega) = F_* \omega^{1-p}$ (as in $\S$1.3 of the book [Brion-Kumar]). Now, I'd rather work with $\omega^{-1}$, and take the $p-1$ power (which gets me only very special sections, but that's okay). But I'd like to do that on $X$, where $F$ isn't flat.

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  • $\begingroup$ Are you assuming any good properties in codimension $1$? If so, you may be able to use S2 extension or the det-Div formalism of Knudsen-Mumford and Fogarty to extend across codimension 2 subsets. $\endgroup$ Commented Feb 29, 2016 at 13:16
  • $\begingroup$ Seminormal gets me some codim 1 statements, e.g. that general curves sections can have nodes but no cusps. I'm not sure yet if I'm okay with S2. They're related as follows: S2 + (seminormal in codim 1) $\Leftrightarrow$ S2 + seminormal, rather like Serre's criterion. (Though seminormal doesn't imply S2, e.g. two planes glued at a point.) $\endgroup$ Commented Feb 29, 2016 at 14:23
  • $\begingroup$ As long as your scheme is equidimensional, for an open subset $U\subset X$ whose complement has codimension $\geq 2$ (in every irreducible component), the pushforward from $U$ of a coherent sheaf is still coherent. So, ugly though it may be, you could define your sheaf on $U$ and then extend to all of $X$ by pushforward. By construction, global sections of the pusforward sheaf are the same as sections over $U$. $\endgroup$ Commented Feb 29, 2016 at 19:30
  • $\begingroup$ This may be more than what you want, but "demi-normal"= $S_2$ + $G_1$ + semi-normal (in codim 1) seems the right notion for this. ($G_1$= Gorenstein in codimension $1$) $\endgroup$ Commented Feb 29, 2016 at 22:06
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    $\begingroup$ @Allen I don't think that is Gorenstein. Three lines in the plane is Gorenstein, but $k[x,y,z]/(xy,xz,yz)$ is not. At the same time three lines in the plane going through one point is not semi-normal, while your set of three lines is. $\endgroup$ Commented Mar 1, 2016 at 6:03

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Allen, it seems to me that you are asking what generality you can go to to define canonical/anticanonical (divisorial) sheaves (not bundles!). I think that divisorial sheaves are actually easier to define on a non-normal scheme than divisors and I am guessing this is essentially what Jason is saying. As long as you do not care about actual divisors of zeros than you probably need very little. In fact, it seems to me that you can probably do this without assuming anything additional.

Dualizing complexes exist under fairly general assumptions. If you are dealing with schemes that are finite type over a field, then you are good. Since you are OK with equidimensional schemes, let's assume that $X$ is a reduced scheme of finite type over a field of pure dimension $d$. Then it has a dualizing complex $\omega_X^\bullet$ and define the canonical sheaf of $X$ as the unique cohomology sheaf of $\omega_X^\bullet$ which is not supported on the singular locus of $X$. Usually $\omega_X^\bullet$ is normalized so that this cohomology sheaf is $h^{-d}$. Let us denote this by $\omega_X$ and assume that $\omega_X^\bullet$ is normalized so that $h^{-d}(\omega_X^\bullet)=\omega_X$. In fact, let's assume that $\omega_X^\bullet$ is normalized so that on the regular locus it restricts to $\omega_{X_{\rm reg}}[d]$.

Reduced implies $S_1$, so your $X$ is actually Cohen-Macaulay in codimension $1$. This implies that the support of $h^{i}(\omega_X^\bullet)$ has codimension $2$ for all $i\neq -d$. In other words, there is a closed subset $Z\subseteq X$ of codimension at least $2$ such that $X\setminus Z$ is Cohen-Macaulay and $\omega_{X\setminus Z}^\bullet\simeq \omega_{X\setminus Z}[d]$.

Now here is where you get lucky. In general you'd want to define Weil divisorial sheaves (which is probably what you are looking for) as reflexive sheaves of rank $1$. If $X$ is $S_2$ and $G_1$, then a coherent sheaf is reflexive if and only if it is $S_2$, so you get a relatively easy way to decide what being reflexive means and this is the usual context in which Weil divisorial sheaves (a replacement of divisors as cycles) are considered.

On the other hand, I've just realized that you might not need this much actually. The point is that $\omega_X$ is always $S_2$ in this context even if $X$ might not be. And then you are actually better off with the dual $\omega_X^\vee=\mathscr Hom_X(\omega_X, \mathscr O_X)$, because that is reflexive already if $X$ is $S_1$ and $G_0$ and with $X$ reduced you have more than that (reduced $\Leftrightarrow$ $S_1$ and $R_0$). So, you get that $\omega_X^\vee$ is reflexive and its global sections are the same as those you get on the Cohen-Macaulay locus.

Restricting to the regular locus will give you actual sections. I think that's automatic, no? If $\mathscr F$ is a sheaf and $U\subseteq X$ is an open set, then being a sheaf means that you have a restriction map $$ H^0(X,\mathscr F)\to H^0(U,\mathscr F|_U). $$ So I guess what you are really asking is that the anticanonical sheaf of $X$ would restrict to the usual anticanonical line bundle on the regular locus. This is true for the above definition.

Of course without assuming that $X$ is $R_1$, you cannot expect all sections from the regular locus to extend, but you are wisely only asking that for $X$ normal. If $X$ is normal, then pushing forward the anticanonical sheaf of the regular locus to $X$ will coincide with $\omega_X^\vee$ because it is reflexive, so in particular, sections extend uniquely.

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