9
$\begingroup$

Let $R$ be a commutative ring with $1$, and let $f \in R$ be an element of $R$ which is neither nilpotent, nor a unit (assuming there exists such an element in $R$). Let $R_f$ be the localization of $R$ with respect to the multiplicative subset $\{ f^n: n \in \mathbb{N} \}$. In this post, the relevant notion of dimension used is Krull dimension.

It is clear that $\dim R_f \leq \dim R$. I have two questions:

1) Are there some sufficient conditions that ensure equality of the dimensions, namely sufficient conditions to have $\dim R_f = \dim R$?

2) Does equality hold in general? If yes, can someone provide a proof please, and if not, can someone provide an example where $\dim R_f < \dim R$?

Edit 1: Following user26857's comment, let $R = \mathbb{C}[x]_{(x)}$. Let $f=x/1$. It is clear that $\dim R = 1$; indeed the only prime ideals of $R$ are $0$ and $(f)$. It follows easily that $\dim R_f = 0$, since the only prime ideal of $R$ which avoids the non-negative powers of $f$ is $0$. This answers my question 2, because in this example, we have $\dim R_f < \dim R$. (Remark: I had written a wrong example in my answer before, instead of this one, for which I apologize.)

Edit 2: Regarding question 1), following Mohan's comments, if $R$ is an integral domain which is a finitely generated algebra over a field or $\mathbb{Z}$, and if $f \in R$ is nonzero, then

$\dim R_f = \dim R$.

We thus have 2 cases to consider in this claim

case A: $R = k[x_1,\cdots,x_n]/P$, where $k$ is a field, and $P$ is a prime ideal in $k[x_1,\cdots,x_n]$.

case B: $R = \mathbb{Z}[x_1,\cdots,x_n]/P$, where $P$ is a prime ideal in $\mathbb{Z}[x_1,\cdots,x_n]$.

I will be happy if someone can post as answer a sketch of a proof of Mohan's claim. I think it amounts to showing that one can find a sequence of chains of prime ideals in R, with all prime ideals involved not containing $f$, and whose lengths go to $\dim R$.

$\endgroup$
5
  • 1
    $\begingroup$ One of the standard places this holds is when $R$ is an integral domain which is a finitely generated algebra over a field or $\mathbb{Z}$. $\endgroup$ – Mohan May 20 '17 at 18:37
  • $\begingroup$ @Mohan thank you! This is basically what I wanted to know. Can one say that the rings you have mentioned are of the form $k[x_1,...,x_n]/P$ or $\mathbb{Z}[x_1,...,x_n]/P$ where $P$ is a prime ideal in the corresponding ring? $\endgroup$ – Malkoun May 20 '17 at 18:47
  • $\begingroup$ Yes, there are other types of rings too, but these are rather important. $\endgroup$ – Mohan May 20 '17 at 19:17
  • $\begingroup$ @Malkoun If $R$ is a Jacobson ring (and both Mohan's examples are) and $f$ is not nilpotent, then there is a maximal ideal $\mathfrak m$ such that $f\notin\mathfrak m$ (since the Jacobson radical of Jacobson rings equals their nilradical). This shows that the height of $\mathfrak m$ remains the same after localizing at $f$. $\endgroup$ – user26857 May 21 '17 at 9:13
  • $\begingroup$ The rings with property $\dim R=\mathrm{ht}(\mathfrak n)$ for every maximal ideal $\mathfrak n$ are called equicodimensional. Now notice that if $R$ is a Jacobson equicodimensional ring, then $\dim R_f=\dim R$. Once again, both examples of Mohan's satisfy this condition (see this answer). $\endgroup$ – user26857 May 21 '17 at 9:14
5
$\begingroup$

I thank user26857 for his comments, which I am using to provide an answer to this post, in the hope it will be useful for others. I have learned a lot. Credit goes to user26857, and mistakes in this answer (if any) are my sole responsibility!

Definition 1. A commutative ring with 1 is said to be a Jacobson ring if every prime ideal is the intersection of the maximal ideals containing it.

Definition 2. A commutative ring with 1 $R$ is said to be equicodimensional if for every maximal ideal $\mathfrak{m}$ of $R$, $\dim R = \operatorname{ht}(\mathfrak{m})$.

Proposition. Let $R$ be an equicodimensional Jacobson commutative ring with 1, and let $f \in R$ be a non-nilpotent element. Then $\dim R_f = \dim R$.

Note: Since the examples mentioned by Mohan are equicodimensional Jacobson, namely if we have an integral domain which is a finitely generated algebra over a field or $\mathbb{Z}$, it follows that $\dim R_f = \dim R$ for these very important rings in particular.

Proof. I will follow user26857's sketch of a proof. Since $f$ is not nilpotent, then $f$ does not belong to the nilradical of $R$. Since $R$ is Jacobson, then its nilradical equals its Jacobson radical. Hence there exists some maximal ideal $\mathfrak{m}$ of $R$ such that $f \notin \mathfrak{m}$. Hence $\operatorname{ht}(\mathfrak{m}) = \operatorname{ht}(\mathfrak{m}_f)$, where $\mathfrak{m}_f$ denotes the extension of $\mathfrak{m}$ in $R_f$.

Since $R$ is equicodimensional, we also have that $\dim R = \operatorname{ht}(\mathfrak{m})$. But $\operatorname{ht}(\mathfrak{m}_f) \leq \dim R_f$, so that $\dim R \leq \dim R_f$, and thus $\dim R = \dim R_f$, as required.

$\endgroup$
3
  • 1
    $\begingroup$ In general, Jacobson rings are not necessarily equicodimensional as we learn from this paper. $\endgroup$ – user26857 May 21 '17 at 15:28
  • 1
    $\begingroup$ thank you for your interesting information once more. $\endgroup$ – Malkoun May 21 '17 at 15:31
  • $\begingroup$ @user26857 It is a nice example in the paper you linked to, providing a nice mix of real inequalities and complex geometry. $\endgroup$ – Malkoun May 22 '17 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.