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A Noetherian (commutative) ring $A$ is called universally catenary if every $A$-algebra of finite type is catenary. If one wants to know whether $A$ is universally catenary, then this definition suggests checking every $A$-algebra of finite type. Catenarity being preserved under taking quotients it clearly suffices to check every polynomial algebra over $A$ in finitely many indeterminates. This is still quite a bunch of algebras, but fortunately the following result reduces the task to checking only a single algebra.

Theorem (Ratliff): A Noetherian ring $A$ is universally catenary if and only if the polynomial algebra $A[X]$ over $A$ in one indeterminate is catenary.

Again since catenarity is preserved under taking quotients, in the above theorem one can replace $A[X]$ by any polynomial algebra over $A$ in a finite and strictly positive number of indeterminates. One might wonder if there are other "test algebras for universal catenarity", i.e. $A$-algebras $B$ such that $A$ is universally catenary if and only if $B$ is catenary. An interesting and natural candidate would be the Laurent algebra $A[X,X^{-1}]$ over $A$ in one indeterminate. If this is indeed a "test algebra", then it follows readily that every algebra over $A$ of a torsionfree, cancellable, finitely generated monoid different from $0$ has the same property.

So, I would like to pose the following question (equivalent to the above by means of Ratliff's Theorem).

Question: Suppose that the Laurent algebra $A[X,X^{-1}]$ over a Noetherian ring $A$ in one indeterminate is catenary. Is then the polynomial algebra $A[X]$ over $A$ in one indeterminate catenary, too?

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up vote 3 down vote accepted

Yes. Indeed, $S:=\mathrm{Spec}A[X]$ is covered by the open subspaces $U:=\mathrm{Spec}A[X,X^{-1}]$ and $V:=\mathrm{Spec}A[X,(X-1)^{-1}]$. Now $U$ is catenary by assumption, and $V$ is isomorphic to $U$, hence catenary. It follows easily that $S$ is catenary: just observe that if $Y\subset Z$ are irreducible and closed in $S$, then $Y$ has to meet $U$ (say), and then every maximal chain joining $Y$ to $Z$ induces a maximal chain joining $Y\cap U$ to $Z\cap U$.

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Dear Laurent, thank you very much for your quick reply (although it shows that my problem turned out to be embarrassingly easy...). – Fred Rohrer Apr 27 '11 at 10:29

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