7 Noetherian Rings

Highlight: Hilbert basis theorem, Hilbert Nullstellensatz

Proposition

If $A$ is Noetherian, then $A/\alpha$ is also Noetherian.

\begin{proof} By ^7j4294. \end{proof}

Proposition

Let $A\subseteq B$ be a subring, where $A$ is Noetherian and $B$ is a finitely generated $A$-module. Then $B$ is a Noetherian ring.

\begin{proof} By ^8ezl57, $B$ is a Noetherian $A$-module. For any $B$-submodule $N$ of $B$, it is also a $A$-submodule and so is finitely generated over $A$. Then $N$ is finitely generated over $B$. Thus $B$ is a Noetherian $B$-module. \end{proof}

Proposition

For a Noetherian ring $A$ and a multiplicatively closed set $S$, $S^{-1}A$ is Noetherian.

\begin{proof} By ^961e0a, ideals in $S^{-1}A$ are of form $S^{-1}I$ with $I\subseteq A$ ideal. But $I$ is finitely generated $A$-module. Then $S^{-1}I=S^{-1}A{\otimes }_{A}I$ is a finitely generated $S^{-1}A$-module. \end{proof}

Corollary

If $A$ is Noetherian, then $A_p$ is Noetherian.

Remark. There exists a ring $A$ such that $A_p$ is Noetherian for all prime ideals $p$, but $A$ is not Noetherian. Indeed, let $A=({\prod }_{n=1}^{\infty }\mathbb{F})$ with $\mathbb{F}=\mathbb{Z}/2\mathbb{Z}$, which is non-Noetherian, because $\mathbb{F}\times 0\times 0\cdots \subseteq \mathbb{F}\times \mathbb{F}\times 0\times \cdots \subseteq \cdots$ does not terminate. All prime ideals are in the form of $p=\mathbb{F}\times \mathbb{F}\times \cdots \times \{0\}\times \mathbb{F}\times \cdots$ and the corresponding $A_p=\mathbb{F}$ is Noetherian. See here.

Hilbert basis theorem

Let $A$ be a Noetherian ring. Then $A[x]$ is also a Noetherian ring.

\begin{proof} Let $\alpha \subseteq A[x]$ be an ideal. Let $I=\{\text{leading coefficients of }f(x)\in \alpha \}$. Note that $I$ is an ideal of $A$, and then $I$ is finitely generated, that is, $I=(a_1,\cdots ,a_n)A$. For each $i$, let $f_i=a_i{x}^{r_i}+\text{lower terms}\in \alpha$. Let $r=\max \{r_i{\}}_{i=1}^n$. Then we get a sub-ideal ${\alpha }^{\prime }=(f_1,\cdots ,f_n)A[x]\subseteq \alpha$.

Let $f\in \alpha$ with $f=a{x}^m+\text{lower terms}$ and $a\in I$. If $m⩾r$, then one can do division using $f_1,\cdots ,f_n$, namely, if $a={\sum }_{i=1}^n{u}_i{a}_i$ with $u_i\in A$, then one can construct

$$f-\sum _{i=1}^n{u}_i{f}_i{x}^{m-r_i}\in \alpha$$

and its degree $<\mathrm{deg}f$. Repeat this procedure, and we get $f=g+h$ where $g\in {\alpha }^{\prime }$ and $\mathrm{deg}h<r$.

Let $M={\oplus }_{i=1}^{r-1}A{x}^i\subseteq A[x]$ be a $A$-submodule. By the decomposition $f=g+h$, we have $\alpha ={\alpha }^{\prime }+\alpha \cap M$ as $A$-modules.

But now, $M$ is a Noetherian $A$-module and $\alpha \cap M$ is also Noetherian $A$-module. So $\alpha \cap M=(g_1,\cdots ,g_m{)}_A$. It is easy to see $\alpha =(f_1,\cdots ,f_n,g_1,\cdots ,g_m{)}_{A[x]}$ and so $\alpha$ is finitely generated. \end{proof}

Remark. We have proved it here. Similarly, we can prove that if $A$ is Noetherian, then $A[[x]]$ is also Noetherian. The proof is similar, using “lowest term coefficient”. See here.

Corollary

If $A$ is Noetherian, then $A[x_1,\cdots ,x_n]$ is also Noetherian.

Corollary

Let $B$ be a finitely generated $A$-algebra. If $A$ is Noetherian, then so is $B$. In particular, every finitely generated ring, and every finitely generated algebra over a field, is Noetherian.

\begin{proof} There exists surjective map $A[x_1,\cdots ,x_n]\twoheadrightarrow B$.

Or see ^gmjk45. \end{proof}

Proposition

Let $A\subseteq B\subseteq C$ be rings. Suppose that $A$ is Noetherian, that $C$ is finitely generated as an A-algebra and that $C$ is finitely generated as a B-module. Then $B$ is finitely generated as an A-algebra.

Example & Counterexample.

• $k\subseteq k[x^n]\subseteq k[x]$.
• $k\subseteq B\subseteq k[x]$, where $B=k[x^2+x^3,x^4+x^6,\cdots ]$. Since $B$ is not finitely generated $A$-algebra, $C$ is not finitely generated as a $B$-module.

\begin{proof} Write $C=A[x_1,\cdots ,x_m]$ with $x_i\in C$, and write $C={⟨y_1,\cdots ,y_n⟩}_B$ as $B$-module with $y_i\in C$. Then

$$\begin{array}{rl}& x_i=\sum _{j=1}^n{b}_{ij}{y}_j,b_{ij}\in B(1)\\ & y_i{y}_j=\sum _{k=1}^n{b}_{ijk}{y}_k,b_{ijk}\in B(2)\end{array}$$

Let $B_0=A[b_{ij},b_{ijk}]\subseteq B$ be a subring, which is a finitely generated $A$-algebra. So now, it suffices to show $B$ is a finitely generated $B_0$-algebra. But in fact, one can show that $B$ is a finitely generated $B_0$-module.

Indeed, for any $b\in B\subseteq C=A[x_1,\cdots ,x_m]$, $B$ is a polynomial in $x_1,\cdots ,x_m$ over $A$ and so $b$ is a polynomial in $y_1,\cdots ,y_n$ over $B$ by (1). Furthermore, $b$ is a linear combination of $y_1,\cdots ,y_n$ over $B_0$ by (2). Remark that $y_j\in C$ NOT in $B$, so they are not a set of generator of $B$ over $B_0$. But the above argument shows $B\subseteq {⟨y_1,\cdots ,y_n⟩}_{B_0}\subseteq C$.

Since $A$ is a Noetherian ring, by ^b4d5c1, $B_0=A[b_{ij},b_{ijk}]$ is also Noetherian ring. Then it yields that ${⟨y_1,\cdots ,y_n⟩}_{B_0}\supseteq B$ is a Noetherian $B_0$-module. So $B$ is a Noetherian $B_0$-module, proving the claim. \end{proof}

Proposition

Let $k$ be a field, and let $E$ be a finitely generated $k$-algebra. If $E$ is also a field, then $E/k$ is a finite extension.

\begin{proof} Write $E=k[x_1,\cdots ,x_n]$ where $x_i\in E$. Since $E$ is also a field, $E=k(x_1,\cdots ,x_n)$. If $E/k$ is not a finite extension, then not all $x_i$ are algebraic over $k$. We can reorder $x_i$ such that $x_1,\cdots ,x_r$ are algebraic independent over $k$, and $x_{r+1},\cdots ,x_n$ are algebraic over $k(x_1,\cdots ,x_r):=F$. So we get $k\subseteq F\subseteq E$ where $E$ is a finitely generated $k$-algebra and $E$ is finitely generated $F$-module. By ^983c2e, $F$ is a finitely generated $k$-algebra.

We claim that it is impossible. That is, it is impossible to write $k(x_1,\cdots ,x_r)=k[y_1,\cdots ,y_m]$. Indeed, write $y_j=f_j(x_1,\cdots ,x_r)/g_j(x_1,\cdots ,x_r)$ where $f_j,g_j\in k[x_1,\cdots ,x_r]$, then $1/(g_1\cdots g_m+1)\notin \text{RHS}$. \end{proof}

Corollary

Let $k$ be a field, and let $A$ be a finitely generated $k$-algebra. If $m\subseteq A$ is a maximal ideal, then $A/m$ is a finitely extension of $k$.

\begin{proof} Let $E=A/m$ in ^f58c35. \end{proof}

Hilbert Nullstellensatz, weak form

Let $k$ be a algebraically closed field. If $m\subseteq k[x_1,\cdots ,x_n]$ is a maximal ideal, then $m=(x_1-a_1,\cdots ,x_n-a_n)$ with $a_i\in k$.

Remark. “null”=zero, “stellen”=point, “satz”=theorem

\begin{proof} It has been proved in ^6rh0cw.

By ^0346fb, $k[x_1,\cdots ,x_n]/m$ is a finite extension over $k$. Since $\overline{k}=k$, one have $k[x_1,\cdots ,x_n]/m\stackrel{\theta }{\simeq }k$. Write $\theta (x_i)=a_i\in k$. Then $(x_1-a_1,\cdots ,x_n-a_n)\subseteq m$. But LHS is maximal, which yields that $(x_1-a_1,\cdots ,x_n-a_n)=m$. \end{proof}

Equivalent form of ^8sohrm

Let $k$ be a algebraically closed field. Let $I=(f_1,\cdots ,f_m)⊊k[x_1,\cdots ,x_n]$ be a proper ideal. Then these $f_i$ have a common solution in $k^n$.

\begin{proof} Since $I$ is proper, $I\subseteq m$ for some maximal ideal $m$. By ^8sohrm, $I\subseteq (x_1-a_1,\cdots ,x_n-a_n)$ and so $(a_1,\cdots ,a_n)$ is a common solution for any $f\in I$. \end{proof}

Remark. For an ideal $I\subseteq k[x_1,\cdots ,x_n]$, solution of $I$ is empty iff $I=k[x_1,\cdots ,x_n]$.

Text-Ex. 7.14.

It is a strong form of Hilbert Nullstellensatz.

Let $\alpha ⊊k[x_1,\cdots ,x_n]$ be an ideal. Define $V=\{(a_1,\cdots ,a_n)\in k^n:f(a_1,\cdots ,a_n)=0,\forall f\in \alpha \}$ as the solution set of $\alpha$. Define $I(V)=\{g\in k[x_1,\cdots ,x_n]:g(a_1,\cdots ,a_n)=0,\forall (a_1,\cdots ,a_n)\in \alpha \}$. Then $I(V)=r(\alpha )$.

\begin{proof} See here. \end{proof}

Primary Decomposition

Definition

Say an ideal $\alpha$ is irreducible, if for any $\alpha =\beta \cap \gamma$, then $\alpha =\beta$ or $\alpha =\gamma$. Say $\alpha$ is reducible, if $\alpha$ can be written as $\alpha =\beta \cap \gamma$ with $\beta ,\gamma ⊋\alpha$.

Lemma

Let $A$ be a Noetherian ring. Then any ideal is a finite intersection of irreducible ideals, that is, for any ideal $\alpha$, $\alpha$ can be written as $\alpha ={\cap }_{i=1}^n{\beta }_i$.

\begin{proof} Let $S=\{\alpha \subseteq A:\alpha \text{ is not a finite intersection of irreducible ideals}\}$. If $S\ne \varnothing$, then by $A$ Noetherian $S$ has a maximal element $\alpha$. Since $\alpha$ is reducible, $\alpha$ can be written as $\alpha =\beta \cap \gamma$ and both $\beta ,\gamma$ are finite intersection of irreducible ideals, leading to contradiction. \end{proof}

Lemma

For a Noetherian ring $A$, an irreducible ideal is primary.

\begin{proof} Note that $\alpha \subseteq A$ is irreducible iff $\overline{0}\subseteq A/\alpha$ is irreducible, and $\alpha \subseteq A$ is primary iff $\overline{0}\subseteq A/\alpha$ is primary. So WLOG just consider the case where $0\subseteq A$ is an irreducible ideal. We aim to show $0$ irreducible yields $0$ primary. Suppose $xy=0$ and $y\ne 0$, it is enough to show $x^n=0$ for some $n$. Consider the chain

$$\mathrm{Ann}(x)\subseteq \mathrm{Ann}(x^2)\subseteq \cdots .$$

As $A$ Noetherian, there exists $n$ such that $\mathrm{Ann}(x^n)=\mathrm{Ann}(x^{n+1})=\cdots$.

We claim that $(x^n)\cap (y)=(0)$. For any $a\in (x^n)\cap (y)$, we have $ax=0$ by $xy=0$ and $a=x^{n}b$ for some $b\in A$. It deduces that $b{x}^{n+1}=0$ and $b\in \mathrm{Ann}(x^{n+1})=\mathrm{Ann}(x^n)$ and so $a=x^{n}b=0$. Now we prove the claim.

Since $(0)$ is irreducible and $(y)\ne 0$, we have $(x^n)=(0)$ and so we finish the proof. \end{proof}

Theorem

A Noetherian proper ideals have primary decomposition.

\begin{proof} By ^98d425 and ^3e8c56. \end{proof}

Proposition

Let $A$ be a Noetherian ring, and let $\alpha$ be an ideal of $A$. Then $\alpha \supseteq (r(\alpha ){)}^n$ for some $n$.

\begin{proof} Suppose $r(\alpha )$ is finitely generated by $x_1,\cdots ,x_k$ with $x_i^n\in \alpha$ for big enough $n$. Then $\alpha \supseteq (r(\alpha ){)}^{nk+1}$ and we have done. \end{proof}

Corollary

For a Noetherian ring $A$, then there exists $n$ such that $(\mathrm{Nil}(A){)}^n=0$.

\begin{proof} Take $\alpha =(0)$ in ^384015. \end{proof}

Corollary

Let $A$ be a Noetherian ring, and let $m\subseteq A$ be a maximal ideal. For an ideal $q\subseteq A$, TFAE:

• $q$ is $m$-primary
• $r(q)=m$
• $m^n\subseteq q\subseteq m$ for some $n$

\begin{proof} i)→ii) by definition.

ii)→i) By ^vsnd02.

ii)→iii) by ^384015.

iii)→ii) obvious. \end{proof}