Second syzygy modules over rings , satisfying $(S_1)$ and $(G_0)$ , are reflexive Let $(R,\mathfrak m,k)$ be a Noetherian local ring  such that $R_P$ is Gorenstein for every minimal prime ideal $P$ of $R$ and $\text{depth }R_P\ge 1$ whenever $ht (P)\ge 1$.
If $0\ne M$ is a finitely generated $R$-module such that for some integers $s,t\ge 1$, there is an exact sequence $0\to M\to R^t\to R^s$, then how to prove that $M$ is reflexive ?
My thoughts: I need to prove $M\cong M^{**}$, or equivalently, $\text{Ext}^i_R(\text{Tr}M,R)=0$ for $i=1,2$, where $\text {Tr}(-)$ denotes Auslander transpose. Firstly, $M$ is a submodule of a finite free module, hence $M$ is torsion-less, so $\text{Ext}^1_R(\text{Tr}M,R)=0$ . Unfortunately, I'm unable to show $\text{Ext}^2_R(\text{Tr}M,R)=0$. Please help.
 A: Let $R$ be a left and right Noetherian ring $R$, and consider a finitely generated left $R$-module $M$. Bass in [Bass 1960, p. 478] states that $M$ is a second syzygy module if and only if it is of the form $\operatorname{Hom}_R(N,R)$ for a finitely generated right $R$-module $N$. (In the commutative case, you can also look at the Lemma and the Remark in [Bass 1963, p. 26]).
Now suppose that $R$ is commutative and Noetherian. Bass in [Bass 1963, Prop. 6.1] proved that $R$ satisfies $S_1$ and $G_0$ (in your notation) if and only if the dual of every finite generated $R$-module is reflexive (see also [Ischebeck 1969, (4.7); Hartshorne 1994, Cor. 1.8]).
Combining these two results, we have:
Proposition. Let $R$ be a commutative Noetherian ring satisfying $S_1$ and $G_0$. Then, an $R$-module is a second syzygy module if and only if it is reflexive.
If you want a single reference for this fact, you can use [Hartshorne 1994, Prop. 1.7].
Remark. Bass's original definition for syzygy modules [Bass 1963, §8] has a shift by 1.
Remark. The notation for $S_1$ and $G_0$ you have in mind I think matches that in [Hartshorne 1994, §1]. As I mentioned in this answer there are other definitions for $G_r$ in the literature.
