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Let $R$ be a commutative ring of finite Krull dimension $n_0$. Let $\dim(R[X_1,\dots,X_m])=n_m$. The sequence $\{n_i\}_{i=0}^\infty$ is called the dimension sequence of $R$. Let $d_i=n_i-n_{i-1}$. The sequence $\{d_i\}_{i=1}^\infty$ is called the difference sequence of R. The sequences of nonnegative integers that can be realized as a dimension sequence of a ring have been determined by J. Arnold and R. Gilmer in the paper "The dimension sequence of a commutative ring."

A simple example of such a sequence is any sequence satisfying the following two conditions: $n_0+1\geq d_1\geq d_2\geq\cdots$ and for some positive integer $k$, $1\leq d_k=d_{k+1}=\cdots$

In the paper "A number theoretic characterization of dimension" T. Parker proves the following: In order that a sequence be a dimension sequence it is necessary and sufficient that the following inequality is satisfied for every $n>0$: $na_n\leq (n+1)a_{n-1}+1$.

Let $m>1$ , $n_0=m-1$ and $n_i=mi$. Then $\{n_i\}$ is a dimension sequence but does not satisfy Parker's criterion. I am assuming that I made a mistake on this simple algebra, but I simply cannot see it. Does anyone see where the problem lies?

Andrew Chiriac
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The problem is the $0^{th}$ term, $n_0 = m-1, n_1 = m$. Thus $d_1 = 1$, $d_i = m$ for $i \geq 2$. For this sequence to be an example of the kind you describe, $d_1$ should bound all the other $d_i$, so we must have $m = 1$. The sequence also satisfies the condition of Parker if and only if $m = 1$, so there is no contradiction.

Tom Oldfield
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