# A proof for $\dim(R[T])=\dim(R)+1$ without prime ideals?

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 $$R$$.

• Note: I've asked this on stackexchange math.stackexchange.com/questions/358423 and was encouraged to ask this here. Jun 21, 2014 at 8:11
• 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. Jun 26, 2014 at 14:48
• @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 $\dim(R[T])$. Aug 4, 2014 at 14:13
• @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. Aug 25, 2014 at 21:21
• 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. Feb 6, 2015 at 15:51

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!

• Where do you use the characterization of the Krull dimension by T. Coquand and H. Lombardi? Feb 29, 2016 at 16:15
• @MartinBrandenburg: exactly!
– darx
Feb 29, 2016 at 18:20
• If you don't use it, this is not an answer to my question. Mar 1, 2016 at 9:54
• @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
– darx
Mar 1, 2016 at 15:59
• 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. Mar 1, 2016 at 17:08