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In "The Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn a torsion free sheaf is defined as coherent sheaf $E$ on an integral Noetherian scheme $X$ s.t. for every $x\in X$ and every non-zero germ $s\in O_{X,x}$, multiplication by $s$ $E_x\to E_x$ is injective.

It is then stated that this definition is equivalent to $T(E)=T_{d-1}(E)=0$ where $d=\dim X$ and $T_{d-1}(E)$ is the maximal subsheaf of $E$ of dimension $\leq d-1$.

I am trying to prove this equivalence. I first restricted to the case of $X=\text{Spec} A$ affine and so $E=\tilde{M}$ for $M$ a finitely generated $A$-module. In this case I think I was able to prove that $E$ is torsion free if and only if $\forall m\in M$ we have $\text{Ann}(m)={0}$, which is equivalent to $V(\text{Ann}(m))=X$ (since $A$ is an integral domain) and thus to $\text{Supp}({mM})=X$, which amounts to say $\dim \tilde{mM}=d$ (since $\text{Supp}({mM})$ is closed). Since every submodule of $M$ contains a cyclic submodule of the form $mM$ we should be done.

Is this argument correct? And is it enough restricting to the affine case? Of course the definition of torsion-freeness is local but I am not sure about the vanishing of $T(E)$.

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$T(E)$ is local in the sense that for any $U\subset X$ open, we have $T(E|_U)=T(E)|_U$. Why? $T(E)|_U$ is a subsheaf of $E|_U$ of dimension $\leq d-1$, so it necessarily injects into the maximal such subsheaf, and this injection is an isomorphism because the stalks are the same - we don't notice restriction to an open subset when taking stalks. So your argument should be fine (assuming I didn't miss something else).

Another idea to show this without resorting to the affine case is the following. Suppose $x\in X$ is contained in the support of $T(E)$. Since $T(E)$ is supported on a closed subset of codimension at least one, that means that in $\mathcal{O}_{X,x}$, there exists a germ which vanishes along the support of $T(E)$, and this germ produces a non-injective multiplication map on the stalk of $E$ at $x$.

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