Suppose X is a pure dimensional projective complex scheme, reducible and non-reduced but without embedded components of lower dimension. Let $X=\cup X_i$ be the decomposition such that $X_i$ is set-theoretically irreducible and $\dim(X_i\cap X_j)<\dim(X_i)$.

Let $F_X$ be a torsion free sheaf of $\mathcal{O}_X$-modules. If $F$ is generically locally free on each $X_i$ then one can define its multi-rank. What to do if $F|_{X_i}$ is not generically locally free?

For example let $X:=(x^p=0)\subset\mathbf P^2$ be the multiple line. Let $F_j$ be the $\mathcal{O}_X$ module generated by $x^j$. For $0 < j < p$ it is not locally free at any point.

My guess: let $pt\in X_i$ be generic enough closed point. Let $p_i$ be the generic multiplicity of $X_i$. At the point $pt\in X$ consider the intersection of $X_i$ with the generic plane of the complementary dimension. This gives the sub-scheme $Y\subset X $ supported on $pt$. Take the fraction $\frac{length(F\otimes\mathcal{O}_Y)}{p_i}$ as the rank on $X_i$.

In the previous example this gives the rank of $F_j$: $\frac{p-j}{p}$.

Is this the commonly accepted definition? References?

  • $\begingroup$ Dear qui-vadis, could you say what ``torsion-free" means? I'm not sure what this means when $X$ is non-reduced. In your example, the sheaf $F_j$ has the property that a non-zero local section can be killed by a non-zero local regular function. $\endgroup$
    – jlk
    Nov 16, 2010 at 6:33
  • $\begingroup$ Torsion-free= every stalk is torsion free= if an element s of the stalk of F is killed by a non-zero divisor in the local ring then s=0. Or, the sheaf does not have a subsheaf supported on a subscheme of smaller dimension $\endgroup$ Nov 16, 2010 at 12:12

1 Answer 1


I don't know about commonly accepted, but here is one definition of rank (not multi-rank) that appears in theory of stable sheaves. Fix an ample line bundle $L$ and consider the Hilbert polynomial $P_{L}(F, t)$, defined by

$$ P_{L}(F, n) = \chi(F \otimes L^{\otimes n}). $$

The leading term of $P_{F, L}(t)$ is of the form $ \frac{r}{d!} t^d$, where $d$ is the dimension of the support. The number $r$ is called the ``rank" (with respect to $L$).

You can try to define multi-rank by doing this component-by-component.

Added: Here is an example. Consider the projective plane $\mathbb{P}^2_{\mathbb{C}}$ with coordinates $X, Y, Z$. A closed subscheme $C$ is defined by $X^p=0$, and the hyperplane class defines a very ample line bundle $\mathcal{L}$ on $C$. What are $\mathcal{L}$-ranks of some sheaves on $C$?

The Hilbert polynomial of the structure sheaf is: $$ P_{\mathcal{L}}(\mathcal{O}_{C}, T) = p T + 1 - \frac{(p-1)(p-2)}{2}. $$

In particular, the $\mathcal{L}$-rank is not $p$ and not $1$! More generally, you can check that every line bundle on $X$ has $\mathcal{L}$-rank $p$. This suggests that the $\mathcal{L}$-rank takes into account both the degree of the polarization and some information about the generators of $F$.

Now let's consider a sheaf that is generically non-zero, but also generically non-free. If $i \colon \mathbb{P}^1 \to C$ is the inclusion of the reduced subscheme, then

$$I := i_{*}(\mathcal{O}_{\mathbb{P}^1})$$

has this property. An application of the projection formula shows that the Hilbert polynomial of this function is: $$ P_{\mathcal{L}}(I, T) = T + 1. $$

In particular, the $\mathcal{L}$-rank is $1$!. The ratio of the $\mathcal{L}$-rank and the degree of $\mathcal{L}$ is equal to $1/p$. This looks closer to the number you are expecting.

  • $\begingroup$ Thanks. (Though I would prefer some standard textbook). It seems his rank coincides with the rank of the sheaf restricted to the reduced locus. (In particular his rank is always integer.) I need the rank of the sheaf on the genuine non-reduced scheme... $\endgroup$ Nov 17, 2010 at 4:13
  • $\begingroup$ @qui-vadis: Sorry, I just noticed this comment. The rank that Simpson uses is always an integer, but it does take into account the non-reduced structure. I added an example to illustrate what happens for non-reduced schemes. $\endgroup$
    – jlk
    Nov 21, 2010 at 5:38

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