5
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.