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[parent] bound on the Krull dimension of polynomial rings (Theorem)

If $ A$ is a commutative ring, and $ \operatorname{dim}$ denotes Krull dimension, then

$\displaystyle \operatorname{dim}(A)+1\leq \operatorname{dim}(A[x])\leq 2\operatorname{dim}(A)+1.$    

It is known (see [Seid],[Seid2]) that for any $ k\geq 0$ and $ n$ with $ k\leq n\leq 2k+1$, there exists a ring $ A$ such that $ \dim A=k$ and $ \dim A[x]=n$.

Bibliography

Seid
A. Seidenberg, A note on the dimension theory of rings. Pacific J. of Mathematics, Volume 3 (1953), 505-512.
Seid2
A. Seidenberg, On the dimension theory of rings (II). Pacific J. of Mathematics, Volume 4 (1954), 603-614.



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Cross-references: ring, Krull dimension, commutative ring
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This is version 5 of bound on the Krull dimension of polynomial rings, born on 2005-06-29, modified 2006-12-12.
Object id is 7196, canonical name is BoundOnKrullDomainsOfPolynomialRings.
Accessed 1374 times total.

Classification:
AMS MSC13C15 (Commutative rings and algebras :: Theory of modules and ideals :: Dimension theory, depth, related rings )

Pending Errata and Addenda
1. k < n by gel on 2008-11-25 00:09:09
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intuition and examples by rspuzio on 2005-07-01 01:03:54
Could someone give examples in which these bounds are satisfied? Is there an intuitive explanation for the upper bound? A rigorous proof would, of course, also be welceme. Depending on the nature of the proof, it might explain the bound but, if the proof is obscure and technical, an intuitive explanation would also be welcome even if it is not striclty correct.
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