If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.

Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension, see here. It states:

For $l \in \mathbb{N}$ we have $\dim(R) \leq l$ if and only if for all $x_0,\dotsc,x_l \in R$ there are $a_0,\dotsc,a_l \in R$ and $m_0,\ldots,m_l \in \mathbb{N}$ such that $x_0^{m_0} (\cdots ( x_l^{m_l} (1+a_l x_l)+\cdots)+a_0 x_0)=0$.

A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated $K$- algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.

Can we use this elementary characterization of the Krull dimension to give a new short proof of $\dim(R[T])=\dim(R)+1$ for noetherian commutative rings $R$?

Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is equivalent or even weaker than Noetherian and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ (see here) for every $K$-algebra $R$.

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    $\begingroup$ At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows? $\endgroup$ – user26857 Apr 12 '13 at 8:11
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    $\begingroup$ Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know. $\endgroup$ – Martin Brandenburg Apr 12 '13 at 9:11
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    $\begingroup$ Shall I ask this on mathoverflow? $\endgroup$ – Martin Brandenburg Aug 26 '13 at 12:43
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    $\begingroup$ MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1 $\endgroup$ – Martin Sleziak Jun 21 '14 at 19:55
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    $\begingroup$ Since the question is on MO now, should it be closed here? $\endgroup$ – Martin Brandenburg Jun 27 '14 at 18:59

User "darx" posted the following at mathoverflow:

There are two kinds of primes $\mathfrak q \subset R[T]$. The first possibility is that $\mathfrak q = \mathfrak p[T]$ with $\mathfrak p = R \cap \mathfrak q$. Let us call such a prime "small". The second possibility is that the inclusion $\mathfrak p[T] \subset \mathfrak q$ is strict. Let us call such a prime "big".

Suppose we have a sequence of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset q_r$ in $R[T]$. Then we get a sequence of big/small. If the sequence has only one switch from small to big, then of course $r \leq \dim(R) + 1$. The problem comes from a sequence with multiple switches. But thinking about it for a moment we see that it suffices to prove the following.

Scholium: If $\mathfrak q_0 \subset \mathfrak q_1$ in $R[T]$ lies over $\mathfrak p_0 \subset \mathfrak p_1$ in $R$ and if $\mathfrak q_0$ is big and $\mathfrak q_1$ is small, then there is a prime strictly in between $\mathfrak p_0$ and $\mathfrak p_1$.

To prove this we argue by contradiction and assume there is no prime strictly in between. Observe that in any case $\mathfrak p_0 \not = \mathfrak p_1$ by our definition of big and small. After replacing $R$ by $(R/\mathfrak p_0)_{\mathfrak p_1}$ we reach the situation where $R$ is a local Noetherian domain of dimension $1$. Then $\mathfrak p_0 = (0)$ and $\mathfrak p_1 = \mathfrak m$ is the maximal ideal.

Translating we have to derive a contradiction from the following: we have a nonzero prime $\mathfrak q \subset \mathfrak m[T]$ with $\mathfrak q \not = \mathfrak m[T]$.

Let $K$ be the fraction field of $R$. Let $\mathfrak q_K \subset K[T]$ be the ideal generated by $\mathfrak q$ in $K[T]$. Then $\mathfrak q = \mathfrak q_K \cap R[T]$.

For every $n \geq 0$ let $R[T]_{\leq n}$ be the polynomials of degree $\leq n$. Let $M_n = \mathfrak q \cap R[T]_{\leq n}$ and $Q_n = R[T]_{\leq n}/M_n$ so that we have a short exact sequence $$ 0 \to M_n \to R[T]_{\leq n} \to Q_n \to 0 $$ Now observe that $Q_n$ is a finite $R$-module, is torsion free, and has rank bounded independently of $n$. Namely, over $K$ we know that $\mathfrak q_K$ is generated by a polynomial of degree $d$ and we see that $Q_n \otimes_R K$ has dimension over $K$ at most $d$.

Pick $a \in \mathfrak m$ nonzero. Then (1) $R/aR$ has finite length $c$, (2) for any finite torsion free module $Q$ of rank $r$ the length of $Q/aQ$ is $rc$, and (3) a module $Q$ with length $Q/aQ$ bounded by $rc$ is generated by $\leq rc$ elements. [Hints for elementary proofs: To prove (1) you show for any $b \in \mathfrak m$ some power of $b$ is in $aR$ otherwise $R/aR$ would have a second prime. To prove (2) you choose $R^{\oplus r} \subset Q$ and you use the snake lemma for multiplication by a on the corresponding ses. To prove (3) use Nakayama and that a finite length module is generated by at most its length number of elements.]

Take $n > dc$ where $d$ is the upper bound for the ranks of all $Q_n$ found above. Then we conclude that there exists an element in $M_n$ which is not in $\mathfrak m(R[T]_{\leq n})$ because we have seen above that $Q_n$ can be generated by $\leq dc$ elements. Small standard argument omitted.

This is the desired contradiction because we assumed $\mathfrak q \subset \mathfrak m[T]$. QED

This answer shows that with usual commutative algebra there is a very short proof. Enjoy!

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    $\begingroup$ And it still does not answer Martins question as the given characterization is not used. $\endgroup$ – TMO Oct 9 '19 at 4:59
  • $\begingroup$ Yes, and this is why I downvote, sorry. My question is not "find some proof of the dimension formula". $\endgroup$ – Martin Brandenburg Dec 22 '19 at 14:25

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