----- - [ ] (Rename when have something better.) > [!proposition] Proposition. ([[checking dimension locally]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]]. Consider the [[multiplicative subset of a ring|multiplicative subset]] $S_{\{ x \}}=\{ x^{n}(1-rx) : n \geq 0, r \in R \}$. Consider the [[localization|ring of fractions]] $R_{\{ x \}}:=S ^{-1}_{\{ x \}}R$. > Then [[Krull dimension|dim]] $R \geq n$ $\iff$ [[Krull dimension|dim]] $R_{\{ x \}} \leq n-1$ $\fa x \in R$. > [!proof]- Proof. ([[checking dimension locally]]) > Three steps. > > **Step 1. Prove that $\mathfrak{m} \cap S_{\{ x \}} \neq \emptyset$ whenever $\mathfrak{m}$ is a [[maximal ideal]] of $R$ and $x \in R$.** > > Let $\mathfrak{m} \in \text{mSpec } R$. First note: $x \in S_{\{ x \}}$, because $x=x^{1}(1-0x)$. So if $x \in \mathfrak{m}$, then $\mathfrak{m} \cap S_{\{ x \}} \neq \emptyset$. Thus assume $x \notin \mathfrak{m}$. $\frac{R}{\mathfrak{m}}$ is a [[field]]; denote $(r+\mathfrak{m}):=(x+\mathfrak{m})^{-1}$. Then $rx+\mathfrak{m}=1_{R/\mathfrak{m}}=1 + \mathfrak{m}$, hence $1-rx \in \mathfrak{m}$, implying $\mathfrak{m} \cap S_{\{ x \}} \neq \emptyset$. > > **Step 2. Prove that $\mathfrak{p} \cap S_{\{ x \}} = \emptyset$ whenever $\mathfrak{p}$ is a non-maximal prime ideal of $R$ and $x \in \mathfrak{m} - \mathfrak{p}$ for some maximal ideal $\mathfrak{m}$ properly containing $\mathfrak{p}$.** > > If $x^{n}(1-rx) \in \mathfrak{p}$, then $1-rx \in \mathfrak{p}$ ($x^{n} \notin \mathfrak{p}$). But then $1-rx+rx=1 \in \mathfrak{m}$, contradicting maximality (properness) of $\mathfrak{m}$. > > **Step 3. Complete the proof.** > > The result in [[extension and contraction under localization]] will be crucial: there is an inclusion-preserving bijection $\{ \mathfrak{p} \in \text{Spec }R : \mathfrak{p} \cap S = \emptyset\} \leftrightarrow \text{Spec }R_{\{ x \}}$ given by [[extension of an ideal|extension]] and [[contraction of an ideal|contraction]] under the [[localization|localization map]]. > > $\implies.$ Suppose $\dim R \leq n$. Consider an arbitrary chain > > $\mathfrak{p}=\mathfrak{p}_{d} \supsetneq \mathfrak{p}_{d-1} \supsetneq \cdots \supsetneq \mathfrak{p}_{0}$ > in $R_{\{ x \}}$. WTS $d \leq n-1$. Under the aforementioned [[bijection]], this corresponds to a chain of prime ideals in $R$ having length $d \leq n$. If $d=n$, then said chain in $R$ witnesses $\dim R=n$, and this means it terminates in a [[maximal ideal]] $\mathfrak{m}$ (otherwise we could append a maximal ideal to the chain and obtain $\dim R=n+1$). By **step 1**, $\mathfrak{m} \cap S_{\{ x \}} \neq \emptyset$, though, hence $\mathfrak{m}$ can't correspond to an element in our original chain. Conclude that $\dim R_{\{ x \}}$ cannot reach $d$. > > $\impliedby$. Suppose $\dim R_{\{ x \}} \leq n-1$ for all $x \in R$. If $\dim R \leq 0$ then certainly $\dim R \leq n$. Otherwise take a maximal chain of (distinct) prime ideals of $R$ of length $\ell$, starting with $\mathfrak{m} \supsetneq \mathfrak{p}$, $\mathfrak{m} \in \text{mSpec }R$. Take $x \in \mathfrak{m}-\mathfrak{p}$. Then $\mathfrak{p} \cap S_{\{ x \}}=\emptyset$ by **step 2**, and so $\mathfrak{p} \supsetneq\cdots$ extends to chain of prime ideals in $R_{\{ x \}}$ of length $\ell-1$. Thus $\ell - 1 \leq n-1$, thus $\ell \leq n$. Hence $\dim R \leq n$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```