HW6

Suppose $A$ is Noetherian ring. Prove that $A[[X]]$, the formal power series ring, is also Noetherian ring.

\begin{proof} For any $f\in A[[x]]$, define the leading coefficient of $f$ is the coefficient of the lowest degree term and define the lowest degree of $f$ as $\mathrm{deg}f$. For an ideal $I\subseteq A[[x]]$, we aim to show $I$ is finitely generated. Let $f_1$ be the element with the minimal degree $d_i$ and the leading coefficient $a_i$. We then choose $f_{i+1}\in I$ to be an element of minimal degree such that its leading coefficient $a_{i+1}\notin (a_1,\cdots ,a_i)\subseteq A$. Since $A$ is Noetherian, the chain

$$(a_1)\subseteq (a_1,a_2)\subseteq \cdots \subseteq (a_1,\cdots ,a_n)\subseteq \cdots$$

terminate of some $k\in \mathbb{N}$. We aim to show $I$ is generated by $f_1,\cdots ,f_k$.

Let $g\in I$ be an element of degree $d$ and leading coefficient $a$. Then by the construction of $(a_1,\cdots ,a_k)$, $a\in (a_1,\cdots ,a_k)$ and we can assume that $a={\sum }_{i=1}^k{c}_{i,0}{a}_i$ with $c_{i,0}\in R$.

Case 1. If $d⩾d_k$, then $d⩾d_i$ for all $i\in \{1,\cdots ,k\}$. It deduces that $g_0={\sum }_{i=1}^k{c}_{i0}{X}^{d-d_i}{f}_i$ has degree $d$ and leading coefficient $a$ and so $\mathrm{deg}(g-g_0)>d$. Similarly define $g_1,\cdots ,g_r\in (f_1\cdots ,f_k)$ such that $\mathrm{deg}(g-g_0-g_1)>d+1$, $\cdots$, $\mathrm{deg}(g-{\sum }_{i=0}^r{g}_i)>d+r$. Take $r\to \infty$, and we have

$$g=\sum _{r=0}^{\infty }{g}_r=\sum _{r=0}^{\infty }\sum _{i=1}^k{c}_{ir}{X}^{d+r-d_i}{f}_i.$$

Therefore, $g\in (f_1,\cdots ,f_k)$.

Case 2. If $d<d_k$, then there is a smallest $m\in \{1,\cdots ,k\}$ such that $a\in (a_1,\cdots ,a_m)$ and $d⩾d_m$. So we can assume $c_{i,0}=0$ for $i>m$. Now define $h={\sum }_{i=1}^m{c}_{i,0}{X}^{d-d_i}{f}_i$, and $\mathrm{deg}g-h>d$. We replace $g$ by $g-h$ and repeat this procedure. After finitely many steps, we produce $\tilde{g}$ such that $\mathrm{deg}g⩾d_k$. Then we get back to case 1 and so $g\in (f_1,\cdots ,f_k)$. \end{proof}

Text-Ex. 7.1. Let $A$ be a non-Noetherian ring and let $\Sigma$ be the set of ideals in $A$ which are not finitely generated. Show that $\Sigma$ has maximal elements and that the maximal elements of $\Sigma$ are prime ideals. (Let $a$ be a maximal element of $\Sigma$, and suppose that there exist $x,y\in A$ such that $x\notin a$ and $y\notin a$ and $xy\in a$. Show that there exists a finitely generated ideal $a_0\subseteq \mathfrak{a}$ such that $a_0+(x)=a+(x)$, and that $a=a_0+x\cdot (a:x)$. Since $(a:x)$ strictly contains $a$, it is finitely generated and therefore so is $a$.) Hence a ring in which every prime ideal is finitely generated is Noetherian (I. S. Cohen).

\begin{proof} Since $A$ is non-Noetherian, $\Sigma$ is non-empty and $\Sigma$ is a partially ordered set containing upper bounds for every chain. By Zorn’s lemma, $\Sigma$ has at least one maximal element. Let $a$ be a maximal element of $\Sigma$. Now we aim to show $a$ is prime. Otherwise there exist $x,y\in A$ such that $xy\in a$ and $x,y\notin a$. Since $a+(x)⊋a$, it is finitely generated.

Assume that $a+(x)=⟨a_i+r_{i}x:1⩽i⩽n⟩$, and define $a_0=⟨a_i:1⩽i⩽n⟩$. Then $a_0\subseteq a$ is finitely generated with $a_0+(x)=a+(x)$. Now we aim to show $a=a_0+x\cdot (a:x)$. Since $a_0\subseteq a$ and $x\cdot (a:x)\subseteq a$, we have $a\supseteq a_0+x\cdot (a:x)$. On the other hand, for any $\alpha \in a$, $\alpha +x={\sum }_{i=1}^n{s}_i(a_i+r_{i}x)={\sum }_{i=1}^n{s}_i{a}_i+s_i{r}_{i}x$, it deduces that ${\sum }_{i=1}^n{s}_i{r}_i-1\in (a:x)$ and so $\alpha ={\sum }_{i=1}^n{s}_i{a}_i+({\sum }_{i=1}^n{s}_i{r}_i-1)x\in a_0+x\cdot (a:x)$. Thus $a\subseteq a_0+x\cdot (a:x)$. Now we have proved that $a=a_0+x\cdot (a:x)$. Since $y\in (a:x)$ and $(a:x)\supseteq a$, we have $(a:x)$ strictly contains $a$, and so it is finitely generated. Therefore $a$ is finitely generated. \end{proof}

2. Let $A$ be a Noetherian ring and let $f={\sum }_{n=0}^{\infty }{a}_n{x}^n\in A[[x]]$. Prove that $f$ is nilpotent if and only if each $a_n$ is nilpotent.

\begin{proof} If $f$ is nilpotent, then $f^m=0$ for some $m$. Let $k$ be the smallest number such that $a_k$ is non-nilpotent. Recall that in Text-Ex. 1.2 we have proved $g\in A[x]$ is nilpotent iff all coefficients of $g$ are nilpotent. So $f$ can be written as $f=g_{k-1}+a_k{x}^k+\cdots$, where $g_{k-1}$ is nilpotent. Assume that $g_{k-1}^{m^{\prime }}=0$, then

$$0=f^{m+m^{\prime }}=a_k^{m+m^{\prime }}{x}^{(m+m^{\prime })k}+\text{higher degree terms}$$

and so $a_k$ is nilpotent, which is a contradiction. Therefore, all $a_n$ are nilpotent.

Conversely, assume that $f={\sum }_{n=0}^{\infty }{a}_n{x}^n$ with $a_n$ nilpotent. There is a chain

$$(a_0)\subseteq (a_0,a_1)\subseteq \cdots \subseteq (a_0,\cdots ,a_n)\subseteq \cdots$$

and there exists $N$ such that $(a_0,\cdots ,a_N)=(a_n:n\in \mathbb{N})$ as $A$ is Noetherian. Then $f$ can be written as

$$f=\sum _{n=0}^{\infty }{a}_n{x}^n=\sum _{n=0}^{\infty }\left(\sum _{i=0}^N{r}_{i,n}{a}_i{x}^n\right)=\sum _{i=0}^N{a}_i{f}_i(x)$$

for some $f_i\in A[[x]]$. Let $k\in \mathbb{N}$ such that $a_i^k=0$ for all $i\in \{1,\cdots ,N\}$. Then $f^{k(N+1)}=({\sum }_{i=0}^N{a}_i{f}_i(x){)}^{k(N+1)}=0$ and so $f$ is nilpotent. \end{proof}

8. If $A[x]$ is Noetherian, is $A$ necessarily Noetherian?

\begin{proof} Assume that $A$ is not Noetherian, then there exists an ideal $I\subseteq A$ which is not finitely generated. Then $I[x]\subseteq A[x]$ is also an ideal of $A[x]$ and so $I[x]$ is finitely generated. Suppose $I[x]=⟨f_1,\cdots ,f_t⟩$ for some $f_i\in A[x]$. Define $S:=\{a_{ij}:1⩽i⩽t,0⩽j⩽\mathrm{deg}{f}_i\}$, then $S$ is a finite set and $S\subseteq I$. For any $i\in I\subseteq I[x]$, $i={\sum }_{i=1}^t{r}_i{f}_i$ for some $r_i\in A[x]$ and so $i\in ⟨S⟩$. It deduces $I={⟨S⟩}_A$, which is impossible. Therefore, $A$ is Noetherian. \end{proof}

10. Let $M$ be a Noetherian $A$-module. Show that $M[x]$ (Chapter 2, Exercise 6) is a Noetherian $A[x]$-module.

\begin{proof} By Exercise 2.6, $M[x]\simeq A[x]{\otimes }_{A}M$. Define $f:A[x]\times M\to A[x]{\otimes }_{A}M,(f,m)\mapsto f\otimes m$. By Hilbert basis theorem, $A[x]$ is Noetherian and so $A[x]\times M$ is a Noetherian $A$-module. Since $A[x]{\otimes }_{A}M$ is a quotient module of $A[x]\times M$, $A[x]{\otimes }_{A}M$ is also Noetherian $A$-module. For each submodule $L\subseteq A[x]{\otimes }_{A}M$, $L$ is finitely generated over $A$. So $L$ is also finitely generated over $A[x]$. It deduces that $A[x]{\otimes }_{A}M\simeq M[x]$ is a Noetherian $A[x]$-module. \end{proof}

Remark. 这道题我写错了，因为 $(f,m)\mapsto f\otimes m$ is not a morphism between $A$-modules.

(Nullstellensatz, strong form) Text-Ex.7.14. Let $k$ be an algebraically closed field, let $A$ denote the polynomial ring $k\left[t_1,\dots ,t_n\right]$ and let $a$ be an ideal in $A$. Let $V$ be the variety in $k^n$ defined by the ideal $\mathfrak{a}$, so that $V$ is the set of all $x=\left(x_1,\dots ,x_n\right)\in k^n$ such that $f(x)=0$ for all $f\in \mathfrak{a}$. Let $I(V)$ be the ideal of $V$, i.e. the ideal of all polynomials $g\in A$ such that $g(x)=0$ for all $x\in V$. Then $I(V)=r(\mathfrak{a})$. (It is clear that $r(\mathfrak{a})\subseteq I(V)$. Conversely, let $f\notin r(\mathfrak{a})$, then there is a prime ideal $p$ containing $\mathfrak{a}$ such that $f\notin \mathfrak{p}$. Let $\overline{f}$ be the image of $f$ in $B=A/p$, let $C=B_f=$ $B[1/f]$, and let $m$ be a maximal ideal of $C$. Since $C$ is a finitely generated $k$ algebra we have $C/\mathrm{m}\cong k$, by (7.9). The images $x_i$ in $C/\mathrm{m}$ of the generators $t_i$ of $A$ thus define a point $x=\left(x_1,\dots ,x_n\right)\in k^n$, and the construction shows that $x\in V$ and $f(x)\ne 0$.)

\begin{proof} Define $V=\{x\in k^n:f(x)=0,\forall f\in \mathfrak{a}\}$ and $I(V)=\{g\in A:g(x)=0,\forall x\in V\}$. For any $f\in r(\mathfrak{a})$, there is $n\in \mathbb{N}$ such that $f^n\in \mathfrak{a}$ and $f^n(x)=0=f(x)$ for all $x\in V$. So $f\in I(V)$ and $r(\mathfrak{a})\subseteq I(V)$.

Now we aim to show $r(\mathfrak{a})\supseteq I(V)$. Take $f\notin r(\mathfrak{a})={\cap }_{P\supseteq \mathfrak{a},\text{prime }P}P$, then there exists a prime ideal $P\supseteq \mathfrak{a}$ and $f\notin P$. Let $\overline{f}$ be the image of $f$ in $B:=A/P$, and let $C=B_f=\{b/\overline{f^n}:b\in B\}$. Let $m$ be a maximal ideal of $C$. Since $C$ is a finitely generated $k$-algebra and $C/m$ is a field, one have $C/m$ is a finite extension of $k$. Since $k$ is algebraically closed, $C/m\simeq k$. Define the image of $t_i$ in $C/m$ as $x_i$, and we can identify $x:=(x_1,\cdots ,x_n)$ as an element in $k^n$.

For any $g\in \mathfrak{a}$, $g(x_1,\cdots ,x_n)$ goes to $0$ under the map $A\to A/P$, as $g\in \mathfrak{a}\subseteq P$. So $(x_1,\cdots ,x_n)\in V(\mathfrak{a})$. On the other hand, $f(x_1,\cdots ,x_n)\ne 0$ by $\overline{f}\notin m$. Thus $f\notin I(V)$ and $r(\mathfrak{a})\supseteq I(V)$. Now we finish the proof. \end{proof}