If we have the system of surjective ring homomorphisms

$f_{i,i+1}: R_{i+1} \twoheadrightarrow R_i$

for an arbitrary $i \geq 0$ such that all $R_i$ are Gorenstein local ring. Let us put

$R^{\infty} \colon= \varprojlim_{i \geq 0} R_i$.

Question: Assume that $R^{\infty}$ is also a local ring. Then, is $R^{\infty}$ Gorenstein?


This is not true. In fact, large numbers of complete local rings one encounters in "real life" satisfy this condition. Let me give you some background.

Definition: (Hochster) A local ring $(R, \mathfrak{m})$ is called approximately Gorenstein if there is a decreasing sequence of $\mathfrak{m}$-primary ideals $$I_1 \supseteq I_2 \supseteq \ldots$$ that are cofinal with the powers of $\mathfrak{m}$ and such that each $R/I_i$ is Gorenstein.

Remark: It's easy to see that Gorenstein rings are approximately Gorenstein, since you can choose $x_1, \ldots, x_d$ a system of parameters and let $I_i = \langle x_1^i, \ldots, x_d^i \rangle$.

Remark: Any complete approximately Gorenstein ring satisfies the condition you mentioned. Obviously $R = \lim_{i} R/I_i$.

Ok, so now you can ask what rings are approximately Gorenstein. It's easy to see that a local ring is approximately Gorenstein if and only if its completion is.

Theorem: (Hochster) If $(R, \mathfrak{m})$ is complete and reduced, it is approximately Gorenstein. Thus if $(R, \mathfrak{m})$ is excellent and reduced, it is approximately Gorenstein. Furthermore, if $R$ has depth at least $2$, it is approximately Gorenstein.

Hence, many common rings are approximately Gorenstein.

Edit: One source for this information is these notes by Mel Hochster


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