Let $X$ be a curve (scheme of dimension 1, say noetherian, projective, Cohen-Macaulay all of whose irreducible components have dimension 1) and let $\mathcal{F}$ be a coherent, torsion-free $\mathcal{O}_X$-module (i.e. locally on open subsets $U \subseteq X$ we know that only zero-divisors of $\mathcal{O}_X(U)$ can kill elements of $\mathcal{F}$).
Does $\mathcal{F}$ have dimension 1?
On affine schemes $\operatorname{Spec}(A)$ this is true:
Then $\mathcal{F}$ is given by some $A$-module $M$ and we have $$\dim M = \dim \operatorname{Supp}(M) = \dim V(\operatorname{Ann}(M)).$$ Now that $M$ is torsion-free, the annihilator of $M$ is contained in the set of zero-divisors of $A$, which is $\bigcup_{P \in \operatorname{Spec}(A),\,P \text{ minimal}} P$.
By prime avoidance we obtain that $\operatorname{Ann}(M) \subseteq P$ for $P$ minimal prime. And hence $$V(P) \subseteq V(\operatorname{Ann}(M))$$ which now provides $\dim V(\operatorname{Ann}(M)) = 1$ since the same is true for the irreducible component $V(P)$ of $\operatorname{Spec}(A)$, by assumption.
But I don't know how to go to the global case.