Krull's Principal Ideal Theorem

Presentation 定理二号, Krull PIT 是维数理论中十分重要的一个定理. 用它可以刻画出Noetherian scheme 上的tangent space 的维数大于等于局部维数的性质, 从而定义那些nonsingular 的点. Hartshorne 第一章就有这个定理, 关于tangent space的讨论则被藏在了习题5.10 里面. 值得注意的是Hartshorne 定理1.8A是一个非常不平凡的结果, 这个定理对于finitely generated $k$ algebra成立, 然而对于一般的catenary ring, 即任何prime ideal chain 长度相等的ring 却不一定成立, 也挺有意思的.

Krull Principal Ideal Theorem

In general idea of Zariski tangent space, we have $\dim \mathbb{T}_{\mathfrak{p},X}\geq \dim R_{\mathfrak{p}}$ , where $X=\mathrm{Spec}(R)$ and $R$ Noetherian. We will prove this in algebraic sense by the Krull’s principal ideal theorem.

Recall the Nakayama lemma first:

Lemma 1 (Nakayama). For $R$ be a ring and $\mathfrak{R}$ be its Jacobson radical, $\mathfrak{a}$ is an ideal contained in $\mathfrak{R}$ . Then

For $M$ be a finite generated $R$ -module, $M /\mathfrak{a}M=0$ implies $M=0$ .
For $M$ be a $R$ -module, $N'$ is a finite generated submodule of $M$ and $M=\mathfrak{a}N'+N$ . Then $M=N$ .
For $\{x_{1},\dots,x_{n}\}\subset M$ satisfies $\{\bar{x_{1}},\dots,\bar{x_{n}}\}$ generates $M /\mathfrak{a}M$ , then $\{x_{1},\dots,x_{n}\}$ generates $M$ .

Proof

Proof.

Omitted. See the textbook.
$M=\mathfrak{a}N'+N$ implies $M /N=\mathfrak{a}(N' /N)$ , so $M /N$ is finitely generated. Then $M /N=\mathfrak{a}(N' /N)\subset \mathfrak{a}(M /N)\subset M /N$ , $M /N=\mathfrak{a}(M /N)$ . By (1), $M /N=0$ , i.e. $M=N$ .
Consider a map $f:R^{n}\to M$ with $f:(r_{1},\dots,r_{n})\mapsto r_{1}x_{1}+\cdots+r_{n}x_{n}$ , we want to show $f$ is surjective, i.e. $M /\mathrm{im}f=0$ . Let $\pi:M\to M /\mathfrak{a}M$ , then $\mathfrak{a}R^{n}\subset \mathrm{ker}(\pi\circ f)$ so $\pi\circ f$ factor through $R^{n}/ \mathfrak{a}R^{n}$ . Let $\bar{f}:R^{n}/\mathfrak{a}R^{n}\to M /\mathfrak{a}M$ . $\bar{f}$ maps $(\bar{r_{1}},\dots,\bar{r_{n}})$ to $\bar{r_{1}}\bar{x_{1}}+\cdots+\bar{r_{n}}\bar{x_{n}}$ is surjective. Notice that we can write $M=\mathrm{im}f+\mathfrak{a}M$ , by $(2)$ , we have $M =\mathrm{im}f$ . ◻

We give the definition of height of an prime ideal.

Definition 1. Let $\mathfrak{p}$ be an prime ideal of $R$ , we call the dimension of local ring $R_{\mathfrak{p}}$ the height of $\mathfrak{p}$ , i.e. $\mathrm{ht}(\mathfrak{p})=\dim R_{\mathfrak{p}}$ .

Theorem 1 (Principal Ideal Theorem). Let $R$ be a Noetherian ring, for $x\in R$ , then the minimal prime ideal $\mathfrak{p}$ containing $(x)$ has $\mathrm{ht}(\mathfrak{p})\leq 1$ .

Proof

Proof. Localize at $\mathfrak{p}$ we can deduce to the case of Noetherian local ring $R'=R_{\mathfrak{p}}$ and regard $\mathfrak{p}$ to be the maximal ideal of $R'$ and $(x)$ be the image in $R'$ of the principal ideal. Take $\mathfrak{q}$ be a prime ideal of $R'$ and $\mathfrak{q}\subsetneq \mathfrak{p}$ , then we want to show $\mathfrak{q}$ to be the minimal prime ideal of $R'$ , i.e. $\dim R'_{\mathfrak{q}}=0$ , $R'_{\mathfrak{q}}$ is Artinian.

Let $\mathfrak{q}'$ be the maximal ideal of $R'_{\mathfrak{q}}$ , so $\mathfrak{q}'=\mathfrak{q}R'_{\mathfrak{q}}$ . If we can show $(\mathfrak{q}')^{n}=0$ for some $n$ , then for any prime ideal $\mathfrak{a}\subset R'_{\mathfrak{q}}$ , $(\mathfrak{q}')^{n}\subset \mathfrak{a}$ , which implies $\mathfrak{q}'\subset \mathfrak{a}$ . Then we have $\mathfrak{a}=\mathfrak{q}'$ and $R'_{\mathfrak{q}}$ is Artinian.

Now we show $(\mathfrak{q}')^{n}=0$ for some $n$ . Let $\mathfrak{q}_{n}=(\mathfrak{q}^{n})^{c}=\{r\in R'|rs\in (\mathfrak{q})^{n} \text{ for some }s\notin \mathfrak{q}\}$ . Notice that $\mathfrak{q}_{n}$ is saturated and therefore $\mathfrak{q}_{n}R'_{\mathfrak{q}}=(\mathfrak{q}')^{n}$ . It’s easy to see we have $\mathfrak{q}_{1}\supset \mathfrak{q}_{2}\supset\cdots$ . $\mathfrak{p}$ is minimal over $(x)$ , we have $\dim R' /(x)=0$ , $R' /(x)$ is Artinian. The descending chain $\mathfrak{q}_{1} /(x)\supset \mathfrak{q}_{2} /(x)\supset\cdots$ is stable. Assume for $n\geq N$ , $\mathfrak{q}_{n+1}/(x)=\mathfrak{q}_{n}/(x)$ , then $\forall a\in \mathfrak{q}_{n}$ , $a+rx=b\in \mathfrak{q}_{n+1}$ for some $r\in R$ and $b\in\mathfrak{q}_{n+1}$ . Then $b-a=rx\in \mathfrak{q}_{n}$ . Since $\mathfrak{p}$ is minimal over $(x)$ , $x\notin \mathfrak{q}$ , we have $r\in \mathfrak{q}$ . Hence $\mathfrak{q}_{n}=\mathfrak{q}_{n+1}+\mathfrak{q}_{n}(x)$ . By Nakayama lemma, $\mathfrak{q}_{n}=\mathfrak{q}_{n+1}$ , therefore $\mathfrak{q}_{n}R'_{\mathfrak{q}}=\mathfrak{q}_{n+1}R'_{\mathfrak{q}}$ . $(\mathfrak{q}')^{n}=(\mathfrak{q}')^{n+1}$ , apply Nakayama lemma again, $(\mathfrak{q}')^{n}=0$ . ◻

For general case, we have:

Theorem 2. Let $R$ be a Noetherian ring, $\mathfrak{p}$ is a minimal prime ideal containing $(x_{1},\dots,x_{n})$ , then $\mathrm{ht}(\mathfrak{p})\leq n$ .

Proof

Proof. W.l.o.g., assume $R$ is a Noetherian local ring. Let $\mathfrak{q}\subset \mathfrak{p}$ with no prime ideal between $\mathfrak{p}$ and $\mathfrak{q}$ . We want to show $\mathfrak{q}$ is minimal over some $(y_{1},\dots,y_{n-1})$ . Since $\mathfrak{p}$ is minimal over $(x_{1},\dots,x_{n})$ , $(x_{1},\dots,x_{n})\not\subset\mathfrak{q}$ . Assume $x_{n}\notin \mathfrak{q}$ , then $\mathfrak{p}$ is minimal over $(\mathfrak{q},x_{n})$ . Therefore $\dim R /(\mathfrak{q},x_{n})=0$ and $R /(\mathfrak{q},x_{n})$ is Artinian, the chain $(\mathfrak{p}/(\mathfrak{q},x_{n}))\supset (\mathfrak{p} /(\mathfrak{q},x_{n}))^{2}\supset\cdots$ is stable. By Nakayama lemma $(\mathfrak{p} /(\mathfrak{q},x_{n}))^{m}=0$ for some $m$ , i.e. $\mathfrak{p}^{m}\subset (\mathfrak{q},x_{n})$ . Then $x_{i}^{m}=a_{i}x_{n}+y_{i}$ for $i=1,\dots,n-1$ , for $y_{i}\in \mathfrak{q}$ .

Now we show $\mathfrak{q}$ is minimal over $(y_{1},\dots,y_{n-1})$ . First, $\mathfrak{p}$ is minimal over $(y_{1},\dots,y_{n-1},x_{n})$ , otherwise $(y_{1},\dots,y_{n-1},x_{n})\subset\mathfrak{p}'\subsetneq \mathfrak{p}$ . Since $R /(x_{1},\dots,x_{n})$ is Artinian, $\mathfrak{p}^{k}\subset (x_{1},\dots,x_{n})$ for some $k$ , then $\mathfrak{p}^{km}\subset(y_{1},\dots,y_{n-1},x_{n})\subset \mathfrak{p}'\Rightarrow \mathfrak{p}\subset \mathfrak{p}'$ , this shows $\mathfrak{p}'=\mathfrak{p}$ . Pass to $R /(y_{1},\dots,y_{n-1})$ , $\mathfrak{p} /(y_{1},\dots,y_{n-1})$ is minimal over $(\bar{x_{n}})$ , the residue of $(x_{n})$ in $R /(y_{1},\dots,y_{n-1})$ . Then $\mathrm{ht}(\mathfrak{p} /(y_{1},\dots,y_{n-1}))\leq 1$ and $\mathrm{ht}(\mathfrak{q} /(y_{1},\dots,y_{n-1}))=0$ . So $\mathfrak{q}$ is minimal over $(y_{1},\dots,y_{n-1})$ . ◻

We also have the converse of Krull’s principal ideal theorem.

Theorem 3. Let $R$ be a Noetherian ring, $\mathfrak{p}$ is prime ideal and $\mathrm{ht}(\mathfrak{p})=n$ , then $\mathfrak{p}$ is minimal over and ideal $\mathfrak{a}$ which is generated by $n$ elements.

Proof

Proof. We use induction to prove the result.

First $n=0$ , obvious.

Assume for $n-1$ . For $\mathfrak{q}\subset \mathfrak{p}$ with no prime ideal between them, then $\mathrm{ht}(\mathfrak{q})=n-1$ . By induction hypothesis, there are $x_{1},\dots,x_{n-1}$ s.t. $\mathfrak{q}$ is minimal over $(x_{1},\dots,x_{n-1})$ . Since $R$ is Noetherian, there are finitely many minimal primes over $(x_{1},\dots,x_{n-1})$ , denote them as $\mathfrak{q}_{1},\dots,\mathfrak{q}_{m}$ . By prime avoidence, $\mathfrak{p}\neq\bigcup\limits_{i=1}^{m}\mathfrak{q}_{i}$ otherwise $\mathfrak{p}\subset \mathfrak{p}_{i}$ for some $i$ . Then there is $x_{n}\in \mathfrak{p}$ but $x_{n}\notin \mathfrak{q}_{i}$ for any $i$ . So the minimal prime ideal over $(x_{1},\dots,x_{n})$ has height at least $n$ , hence it’s $\mathfrak{p}$ . ◻

Application of Krull’s PIT

We will have the following geometric implication of Krull’s principal ideal theorem:

Theorem 4. For any $\mathfrak{p}\in X$ , $X$ is a affine variety, let $\mathfrak{m}$ be the maximal ideal of the local ring $R_{\mathfrak{p}}$ , then $$\dim \mathbb{T}_{\mathfrak{p},X}\geq \dim X$$ The equality holds if and only if $R_{\mathfrak{p}}$ is a regular local ring, i.e. $X$ is nonsingular at $\mathfrak{p}$ .

Proof

Proof. Notice that $\dim \mathbb{T}_{\mathfrak{p},X}=\dim_{R_{\mathfrak{p}}/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^{2}$ . For $\{x_{1},\dots,x_{n}\}\subset \mathfrak{m}$ s.t. $\{\bar{x_{1}},\dots,\bar{x_{n}}\}$ generates $\mathfrak{m}/\mathfrak{m}^{2}$ as $R_{\mathfrak{p}}/\mathfrak{m}$ -vector space, $\{\bar{x_{1}},\dots,\bar{x_{n}}\}$ also generates $\mathfrak{m}/\mathfrak{m}^{2}$ as $R_{\mathfrak{p}}$ -module. By Nakayama lemma, $\{x_{1},\dots,x_{n}\}$ generates $\mathfrak{m}$ . From Krull’s PIT, $\mathrm{ht}(\mathfrak{p})=\dim R_{\mathfrak{p}}\leq n$ . ◻

For example, $k[x_{1},\dots,x_{n}]_{(x_{1},\dots,x_{n})}$ is a regular local ring.

Theorem 5. Prime ideals in Noetherian ring $R$ also satisfies the descending chain condition.

Proof

Proof. For any prime $\mathfrak{p}$ , $\mathfrak{p}$ is finitely generated by $n$ elements, then $\mathrm{ht}(\mathfrak{p})\leq n$ . ◻

Theorem 6. A Noetherian integral domain $R$ is UFD if and only if every prime ideal of height 1 is principal.

Proof

Proof. ( $\Leftarrow$ ): Let $p$ be an irreducible element, $\mathfrak{q}$ is minimal over $(p)$ , then $\mathrm{ht}(\mathfrak{q})\leq 1$ . Since $R$ is domain, $\mathrm{ht}(\mathfrak{q})=1$ . Therefore $\mathfrak{q}$ is principal, $p$ is irreducible so $(p)=\mathfrak{q}$ .

( $\Rightarrow$ ): Let $\mathfrak{p}$ be an prime ideal s.t. $ht(\mathfrak{p})=1$ . By the converse of Krull’s PIT, $\mathfrak{p}$ is minimal over some $(x)$ . Let $x=cp_{1}\cdots p_{n}$ with $p_{1},\dots,p_{n}$ prime and $c$ unit. $(p_{1})\cdots(p_{n})\subset \mathfrak{p}\Rightarrow (p_{i})\subset \mathfrak{p}$ for some $i$ . $\mathrm{ht}(\mathfrak{p})=1$ implies $(p_{i})=\mathfrak{p}$ . ◻

Lemma 2 (Generalized Noether Normalization Lemma). Let $R=k[x_{1},\dots,x_{n}]$ be a polynomial ring, $\mathfrak{a}$ be an ideal of $R$ . Let $r=\inf\{\mathrm{ht}(\mathfrak{p})|\mathfrak{a}\subset \mathfrak{p}\}$ . Then we can choose $y_{1},\dots,y_{n}\in R$ such that $R$ is integral over $k[y_{1},\dots,y_{n}]$ and $\mathfrak{a}\cap k[y_{1},\dots,y_{n}]=(y_{1},\dots,y_{r})$ .

We can derive the Noether normalization lemma from above lemma. From the lemma above we can show the following result:

Lemma 3. Let $R=k[x_{1},\dots,x_{n}]$ be a polynomial ring, $\mathfrak{p}$ be a prime ideal of $R$ . Then $\dim R /\mathfrak{p}=\dim R-\mathrm{ht}(\mathfrak{p})$ .

Proof

Proof. Let $\mathrm{ht}(\mathfrak{p})=m$ , then we can find $y_{1},\dots,y_{n}\in R$ such that $R$ integral over $k[y_{1},\dots,y_{n}]$ and $\mathfrak{p}\cap k[y_{1},\dots,y_{n}]=(y_{1},\dots,y_{m})$ . Then $R /\mathfrak{p}$ is integral over $k[y_{1},\dots,y_{n}]/(\mathfrak{p}\cap k[y_1,\dots,y_{n}])$ and therefore $$\dim R /\mathfrak{p}=\dim k[y_{1},\dots,y_{n}]/(\mathfrak{p}\cap k[y_1,\dots,y_{n}])=\dim k[y_{m+1},\dots,y_{n}]=n-m$$
◻

Theorem 7. Let $k$ be a field. An affine variety $X\subset\mathbb{A}^{n}$ has dimension $n-1$ if and only if it is the zero locus of a single nonconstant irreducible polynomial in $k[x_{1},\dots,x_{n}]$ .

Proof

Proof. ( $\Leftarrow$ ): For $f\in k[x_{1},\dots,x_{n}]$ irreducible, $(f)$ is prime since $k[x_{1},\dots,x_{n}]$ is UFD, then $\mathcal{Z}(f)$ is a variety. By Krull’s PIT, we have $\mathrm{ht}(f)=1$ . Therefore $\dim\mathcal{Z}(f)=\dim k[x_{1},\dots,x_{n}] /(f)=\dim k[x_{1},\dots,x_{n}]-\mathrm{ht}(f)=n-1$ .

( $\Rightarrow$ ): Let $X=\mathcal{Z}(\mathfrak{p})$ for some $\mathfrak{p}$ prime. We have $\dim k[x_{1},\dots,x_{n}] /\mathfrak{p}=n-1$ . Notice that $k[x_{1},\dots,x_{n}]$ is UFD and $\mathfrak{ht}(\mathfrak{p})=1$ , by lemma above, we have $\mathfrak{p}$ is principal ideal. Thus $X=\mathcal{Z}(f)$ for some $f$ . ◻

We have a counterexample: A prime ideal $\mathfrak{p}$ of height 2 in $k[x_{1},\dots,x_{n}]$ may not be principal. Let $X$ be the affine algbraic set in $\mathbb{A}^{3}$ ( $\mathrm{char} k=0$ ) defined by $\{(t^{3},t^{4},t^{5})|t\in k\}$ . Then use Grobner basis we can calculate $\mathcal{I}(X)=(xz-y^{2},x^{3}-yz,z^{2}-x^{2}y)$ and $\mathrm{ht}(\mathcal{I}(X))=2$ .