On locally 3-syzygy sheaves

Question

This is a question that came up in the comments section of here. A reflexive sheaf $E$ is called "locally $3$-syzygy" if it fits into an exact sequence $0\rightarrow E \rightarrow F_1\rightarrow F_2 \rightarrow F_3$ where $F_i$s are vector bundles. These reflexive sheaves are vector bundle on the complement of a codimension $4$ subvariety, the question is whether the converse is true i.e. when a reflexive sheaf that is vector bundle on the complement of a closed set of codimension $\geq 4$ is locally 3-syzygy?

I am able to prove through some long method that if the converse is true you get $K_1(k)=0$ (if I'm not making any mistakes in my arguments!). First algebraic $K$-group of fields are trivial but obviously it is wrong. This specially implies that there are counter-examples of the form $\mathbb{A}^n$ with the closed subvariety of the form $\mathbb{A}^k$ where $n-k\geq 4$. So the counter-examples seem to be ubiquitous. My question is, is this almost always wrong or there is some chance of it being true?

Edit: Another relevant question that I cannot answer. Does having a really high codimension wrt the ambient variety guarantee that any reflexive sheaf that is a vector bundle on the complement of the closed variety is locally 3-syzygy? (For example let $Z$ be 3 dimensional variety embedded in $Z\times \mathbb{P}^n$ for a very large $n$)

Answer 1

Locally, the depth increases along syzygies, so there are always counter examples as long as there is a closed point whose local ring has depth at least 3.

For instance, let $R=k[x_1,...,x_n]$ for $n\geq 3$. Let $E$ be the second syzygy of $R/(x_1,...,x_n)$. Then $E$ is reflexive and locally free on $Spec(R)-{m}$ where $m=(x_1,...,x_n)$. But it will not be $3$rd syzygy because the depth of $E_m$ is only 2.

Now some good news: when depth is no more than 2, being Gorenstein would be enough to guarantee. That is because any module of maximal depth on a Gorenstein local ring would be an $n$-syzygy for any $n$.