Let $F$ be a coherent sheaf on some scheme $X$. The rank of $F$ at $x$ is the dimension of the $k(x)$-vector space $F_x\otimes_{O_{X,x}} k(x)$ (ie the fibre of $F$ at $x$). It is clear that if $F$ is locally free, then the dual $F^*$ has the same rank as $F$ at every point $x$. I'm wondering if the converse is also true, ie that if a sheaf $F$ is such that the dual has the same rank at every point, then $F$ must be locally free? I tried some simple examples (maybe there is a counterexample), but the ones I have don't work (eg the ideal sheaf $F$ of a point $x\in\mathbf{A}^2$ has rank $1$ at every point in $\mathbf{A}^2-x$, while it has rank $2$ at $x$; on the other hand the dual is a line bundle).
Rank of dual sheaf
algebraic-geometrycommutative-algebra
Related Solutions
Any subsheaf of $\mathcal O_X$-modules $\mathcal F\subset \mathcal O_X$ on a scheme (or even on a ringed space) is an ideal sheaf.
All the other adjectives (rank-one, coherent, smooth, projective, irreducible,...) are irrelevant.
Also, you shouldn't believe that ideal sheaves must be of rank one or quasi- coherent :
On the spectrum $X=\text {Spec} R$ of a discrete valuation ring $R$, consider the ideal sheaf $\mathcal I$ with global sections $\Gamma(X,\mathcal I)=R$ and whose sections over the (open!) generic point are given by $\Gamma(\{\eta\},\mathcal I)=0$.
The sheaf $\mathcal I$ is an ideal sheaf which is not quasi-coherent and which is of rank zero .
There are smooth schemes which don't have this property, usually called resolution property. A simple example is $X=\mathbb{A}^n \cup_{\mathbb{A}^n \setminus \{0\}} \mathbb{A}^n$, the affine $n$-space with a double origin, and $n \geq 2$. A locally free sheaf on $X$ pulls back to two locally free sheaves on $\mathbb{A}^n$ which become isomorphic when restricted to $\mathbb{A}^n \setminus \{0\}$. Since every locally free sheaf on $\mathbb{A}^n$ is free (Quillen-Suslin Theorem), the isomorphism corresponds to an invertible matrix of global sections of $\mathbb{A}^n \setminus \{0\}$, which by Hartog's Lemma coincide with the global sections of $\mathbb{A}^n$ (here we use $n \geq 2$). Hence, the isomorphism of the two free sheaves extends on $\mathbb{A}^n$, which means that the original locally free sheaf on $X$ is free. It follows that not every coherent sheaf on $X$ is a quotient of a locally free i.e. free sheaf, because otherwise $\mathcal{O}_X$ would be ample and therefore $X$ would be separated.
On the other hand, many schemes have the (strong) resolution property, for example divisorial schemes (M. Borelli, Divisorial varieties) - including projective schemes and any separated noetherian locally factorial scheme (SGA 6, Exp. II, Proposition 2.2.7), as well as any separated algebraic surface (P. Gross. The resolution property of algebraic surfaces). For further results see (P. Gross, Vector bundles as generators on schemes and stacks. PhD thesis) and (B. Totaro, The resolution property for schemes and stacks). For algebraic stacks, there is a useful criterion which depends on a presentation (D. Schäppi, A characterization of categories of coherent sheaves of certain algebraic stacks).
It seems to be an open problem if there is a variety (separated scheme of finite type) which doesn't have the resolution property.
Best Answer
Try the example of the sheaf $F$ on $\mathbb{A}^3$ defined by the exact sequence $$ 0 \to F \to \mathcal{O}^{\oplus 3} \stackrel{(x,y,z)}\longrightarrow \mathcal{O} \to \mathcal{O}_o \to 0, $$ where the last term is the structure sheaf of the origin.