---- > [!theorem] Theorem. ([[dimension theorem for Noetherian local rings]]) > Let $(A, \mathfrak{m})$ be a [[Noetherian ring|Noetherian]] [[local ring|local]] [[ring]]. Here are three numbers we can extract from $A$: > 1. $\text{dim }A$, the [[Krull dimension]] of $A$; > 2. $\delta(A)=\min\{ \delta(\mathfrak{q}) : \mathfrak{q} \text{ is an } \mathfrak{m}\text{-primary ideal of }A \}$, where $\delta(\mathfrak{q})$ denotes the minimal number of [[submodule generated by a subset|generators]] for the [[ideal]] $\mathfrak{q}$ > 3. $d\big( G_{\mathfrak{m}}(A) \big)$, the order of the pole at $T=1$ of the [[rational function]][^1] $P\big( G_{\mathfrak{m}}(A), T \big)=\sum_{n=0}^{\infty} \ell(\mathfrak{m}^{n} / \mathfrak{m}^{n+1} ) \cdot T^{n}$. > > The claim is that $\delta(A) \leq d(G_{\mathfrak{m}}(A)) \leq \text{dim } A \leq \delta(A)$, hence $\text{dim }A=\delta(A)=d\big( G_{\mathfrak{m}}(A) \big).$ > [!proposition] Corollary. (Krull's Height Theorem) > Let $\mathfrak{a}=\langle x_{1},\dots,x_{r} \rangle$ a an [[ideal]] of $R$ a [[Noetherian ring|Noetherian]] (not assumed local). [[height of an ideal|Then]] $\text{ht }\mathfrak{p} \leq r$ for every [[minimal prime ideal]] $\mathfrak{p}$ over $\mathfrak{a}$. > > > > [!proof] Proof. > > > > Write $\mathfrak{a}^{e}=\mathfrak{a}R_{\mathfrak{p}}$ for the [[extension of an ideal|extension]] of $\mathfrak{a}$ under the [[localization|localization map]] $R \to R_{\mathfrak{p}}$. If we can show $\mathfrak{a}R_{\mathfrak{p}}$ is $\mathfrak{p}R_{\mathfrak{p}}$-[[primary ideal|primary]], then we're done, for then $\text{ht }\mathfrak{p}=\text{dim }R_{\mathfrak{p}}= \min_{\mathfrak{q} \subset R_{\mathfrak{p}} \text{ }\mathfrak{m}\text{-primary}} \delta(\mathfrak{q}) \leq \delta(\mathfrak{a}R_{\mathfrak{p}}) \leq r.$ > > > > Recall in general $\sqrt{ I }$ [[maximal ideal|maximal]] $\implies I$ is $\sqrt{ I }$-primary. So we just need to show $\sqrt{ \mathfrak{a }R_{\mathfrak{p}} }=\mathfrak{p}R_{\mathfrak{p}}$. > > > > *Subclaim. $\sqrt{ \mathfrak{a}^{e}=\mathfrak{a}R_{\mathfrak{p}} }=\mathfrak{p}R_{\mathfrak{p}}$.* > > *Proof of subclaim.* [[nilradical equals intersection of all prime ideals|Since]] $\sqrt{ \mathfrak{a}R_{\mathfrak{p}} }= \bigcap_{\mathfrak{p}' \supset \mathfrak{a}R_\mathfrak{p} \text{ prime}} \mathfrak{p}'$ it would be enough to show that $\mathfrak{p}R_{\mathfrak{p}}$ is the only [[prime ideal]] of $R_{\mathfrak{p}}$ containing $\mathfrak{a}R_{\mathfrak{p}}$. > > > > So $\mathfrak{a}R_{\mathfrak{p}} \subset \mathfrak{n} \in \text{Spec }R_{\mathfrak{p}}$. Then $\mathfrak{a} \subset(\mathfrak{a}R_{\mathfrak{p}})^{c} \subset \mathfrak{n}^{c} \subset \mathfrak{p}$. The first two inclusions are just by definition of inverse image, the last can be seen e.g. by [[extension and contraction under localization|here]]. But $\mathfrak{p}$ was a *minimal* prime over $\mathfrak{a}$, so $\mathfrak{n}^{c}=\mathfrak{p}$, hence $\mathfrak{n}^{ce}=\mathfrak{p}R_{\mathfrak{p}}$. But $\mathfrak{n}^{ce}=\mathfrak{n}$, [[extension and contraction under localization|since every ideal in a ring of fractions is extended]]. > > > > This finishes the proof. > > > Summary: 1. Let $\mathfrak{p}$ be minimal prime ideal over $\mathfrak{a}$. Explain why the result reduces to showing $\mathfrak{a}^{e}=\mathfrak{a}R_{\mathfrak{p}}$ is $\mathfrak{m}=\mathfrak{p}R_{\mathfrak{p}}$-primary 2. (small) successive wants: suffices to want $\sqrt{ \mathfrak{a}R_{\mathfrak{p}} }=\mathfrak{p}R_{\mathfrak{p}}$, and then to want that $\mathfrak{p}R_{\mathfrak{p}}$ is the only prime ideal containing $\mathfrak{a}R_{\mathfrak{p}}$ 3. Let $\mathfrak{n}$ be some prime containing $\mathfrak{a}R_{\mathfrak{p}}$... show (at some point using the minimality of $\mathfrak{p}$ assumption) that $\mathfrak{n}^{ce}=\mathfrak{p}_{R\mathfrak{p}}$, and thus $\mathfrak{n}=\mathfrak{p}R_{\mathfrak{p}}$ > [!proof]- Proof. ([[dimension theorem for Noetherian local rings]]) > > - [ ] bring over from notes > > **$\delta(A) \geq d\big( G_{\mathfrak{m}}(A) \big)$.** > > Let $\mathfrak{q}$ be an $\mathfrak{m}$-primary ideal of $A$. By [[primary sandwiches for Noetherian local rings]], there exists $t \geq 1$ such that $\mathfrak{m}^{t} \subset \mathfrak{q} \subset \mathfrak{m}$ and thus $\ell\left( \frac{A}{\mathfrak{m}^{}} \right) \leq \ell\left( \frac{A}{\mathfrak{q}} \right) \leq \ell (\frac{A}{\mathfrak{m}^{t}})$. So $\ell\left( \frac{A}{\mathfrak{m}^{n}} \right)\leq \ell\left( \frac{A}{\mathfrak{q}^{n}} \right) \leq \ell\left( \frac{A}{\mathfrak{m}^{tn}} \right)$ for all $n \geq 0$. So $\text{deg }\ell(A /\mathfrak{q}^{n})$ ---- #### Notation: - $G_{\mathfrak{m}}(A)$ is the [[associated graded ring to an ideal|associated graded ring]] to the [[maximal ideal]] $\mathfrak{m}$ [^1]: Cf. [[Poincare Series and Hilbert-Serre]]. ----- #### 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 > ```