If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, 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), which in particular does not use prime ideals at all. For $x \in R$ let $R_{\{x\}}$ be the localization of $R$ at $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have

$$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \qquad (\ast)$$

It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ You can use this to define the Krull dimension.

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 commutative $K$-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.

Question. Can we use the characterization $(\ast)$ of the Krull dimension by Coquand-Lombardi above to prove $\dim(R[T])=\dim(R)+1$ for Noetherian commutative rings $R$?

Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$.

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 satisfied by Noetherian rings and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$.

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    $\begingroup$ Note: I've asked this on stackexchange math.stackexchange.com/questions/358423 and was encouraged to ask this here. $\endgroup$ Jun 21, 2014 at 8:11
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    $\begingroup$ I would like to know more generally what happens if you take the Coquand-Lombardi characterization as a definition and try to develop the basics of dimension theory from there. $\endgroup$ Jun 26, 2014 at 14:48
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    $\begingroup$ @NeilStrickland: The following example can be found in Hutchins, Examples of Commutative Rings, Example 27: Let $k$ be a field and let $R = k(y)[[x]] \times_{k(y)} k$ the ring of those power series in $x$ with coefficients in $k(y)$ resp. $k$ for the constant term. Then $\dim(R)=1$ and $\dim(R[T])=3$. I've also read that actually every number between $\dim(R)+1$ and $2 \dim(R)+1$ may appear as $\dim(R[T])$. $\endgroup$ Aug 4, 2014 at 14:13
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    $\begingroup$ @Neil: What Martin read (every number between the bound appers) you can also read, namely in A. Seidenberg, On the dimension theory of rings (II), Pacific J. Math. 4 (1954), 603-614. $\endgroup$ Aug 25, 2014 at 21:21
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    $\begingroup$ The elementary characterization is also very useful in a topos-internal context. For example, a scheme $X$ is of dimension $\leq n$ if and only if, from the internal perspective of the little Zariski topos $\mathrm{Sh}(X)$, the (then plain old) ring $\mathcal{O}_X$ is of Krull dimension $\leq n$. See Proposition 3.13 of these sketchy notes. $\endgroup$ Feb 6, 2015 at 15:51

1 Answer 1


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 $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset\mathfrak q_r$ 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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    $\begingroup$ Where do you use the characterization of the Krull dimension by T. Coquand and H. Lombardi? $\endgroup$ Feb 29, 2016 at 16:15
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    $\begingroup$ @MartinBrandenburg: exactly! $\endgroup$
    – darx
    Feb 29, 2016 at 18:20
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    $\begingroup$ If you don't use it, this is not an answer to my question. $\endgroup$ Mar 1, 2016 at 9:54
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    $\begingroup$ @MartinBrandenburg: Of course I agree with you. I could "use" the characterization by just quoting it. But that would make this proof longer! Anyway, my purpose with this answer was twofold: by spelling out what the usual commutative algebra proof amounts to we see that it is quite easy/short and having this here may help the next visitor to do what you asked for. PS: The statement about dimensions of polynomial algebras over fields can be proven very quickly in a course if you do not want to explain more commutative algebra to the students along the way. Best wishes, Darx $\endgroup$
    – darx
    Mar 1, 2016 at 15:59
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    $\begingroup$ You did answer the question "what is a short proof of the dimension formula for polynomial rings", but my question is "what is a proof of the dimension formula using the characterization of the Krull dimension by Coquand and Lombardi". It doesn't matter if you think that this characterization is necessary or not. $\endgroup$ Mar 1, 2016 at 17:08

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