Projective dimension of a coherent sheaf in a short exact sequence Let $X$ be a noetherian integral scheme. We define the projective/homological dimension of a torsion free coherent sheaf $E$ to be $\mathrm{dh}(E)= \sup\{dh(E_x)|x\in X\}$, here dh$(E_x)$ denotes the projective dimension of the $\mathcal{O}_x$ module $E_x$. Let $E,F,G$ be torsion free coherent sheaves on $X$, with $F$ locally free. 
Suppose $0\longrightarrow E\longrightarrow F\longrightarrow G\longrightarrow 0$ is a short exact sequence, then I am required to prove that $\mathrm{dh}(E)=\max\{0,\mathrm{dh}(G)-1\}$.
I am not getting this.
I have tried the following. We use the Auslander Buchsbaum formula which says that if $A$ is a regular ring : $\mathrm{dh}(M) + \mathrm{depth}(M) = \dim(A)$.
Since $F$ is locally free, each $F_x$ is free, therefore, $\mathrm{dh}(F_x)=0$, hence $\mathrm{depth}(F_x)=\dim(\mathcal{O}_x)$ for all $x$. Further, since $E_x\longrightarrow F_x$ is injective for all x, any $F_x$ regular element is also an $E_x$ regular element, and hence the $\mathrm{depth}(F_x)\leq\mathrm{depth}(E_x)$ . But  $\mathrm{depth}(F_x)=\dim(\mathcal{O}_x)$, implies that $\mathrm{depth}(E_x)=\mathrm{dim}(\mathcal{O}_x)$ for all $X$, which in turn implies that $\mathrm{dh}(E_x)=0$ for all $x$. Hence $\mathrm{dh}(E)=0$.
Where am I making a mistake? Thanks in advance!
 A: For simplicity, assume $X = \text{Spec}(R)$ for a local integral domain $R$ with maximal ideal $x = {\mathfrak m}$. You're right in that any $F$-regular element is also $E$-regular, but this does not imply that any $F$-regular sequence is also $E$-regular. Namely, if $t\in {\mathfrak m}$ is $F$-regular, then you have an exact sequence $$0\to \text{ker}(G\xrightarrow{\cdot t} G)\to E/tE\to F/tF,$$ so without any assumptions on $G$, the quotient $E/tE$ will not embed canonically into $F/tF$. Luckily, in your situation have assumed that $G$ is torsion free, too, so we indeed get an induced short exact sequence $$0\to E/tE\to F/tF\to G/tG\to 0,$$ and hence any $F/tF$ regular element $s$ will also by $E/tE$-regular. Now, however, we're stuck since $G$ being torsion free does not imply the same for $G/tG$, and we can't proceed to deduce that $E/(s,t)E\to F/(s,t)F$ is injective. Hence, for another $F/(s,t)F$-regular element $r$ we can't tell whether it is also $E/(s,t)E$-regular.
As an example, look at $R = k[[x,y,z]]$, take $\pi: F\twoheadrightarrow G$ to be $\pi: R^{\oplus 3}\xrightarrow{(x\ y\ z)} (x,y,z)R$ and $E := \text{ker}(\pi)$. Then indeed $E,F,G$ are torsion-free, but you only have $\text{depth}(E)=2$ and $\text{depth}(G)=1$. 
Concerning the original statement you wanted to prove, you can restrict to the local case and use the characterization of projective dimension through $\text{Ext}$ to check that for any one-step projective resolution $0\to E\to P\to G\to 0$ with $P$ projective (no assumptions on $E$ and $G$) you have $\text{pdim}(E)=\text{max}\{\text{pdim}(G)-1,0\}$.
