8 Artinian Rings

Definition

We say $A$ is an Artinian ring if it satisfies d.c.c. for ideals.

Examples.

• field
• $k[x]/(x^2)$ (it has only $3$ ideals)
• $k[x,y]/(xy)$ is not Artinian, because $(xy,y^2)\supseteq (xy,y^3)\supseteq \cdots$ is not d.c.c.

Proposition

Let $A$ be an Artinian ring. Then each prime ideal is maximal.

\begin{proof} Let $p\subseteq A$ be an prime ideal. Then $B=A/p$ is an Artinian integral domain. We claim that $B$ is a field.

Let $x\in B$ with $x\ne 0$. Then $(x)\supseteq (x^2)\supseteq \cdots$ terminates. Then $x^n=x^{n+1}y$ for some $y\in B$ and so $xy=1$ by $B$ integral domain.

Therefore, $p$ is maximal, by $A/p$ being a field. \end{proof}

Corollary

Let $A$ be an Artinian ring. Then $\mathrm{Nil}(A)=\mathrm{Jac}(A)$.

Proposition

If $A$ is an Artinian ring, then it has finitely many maximal ideals.

\begin{proof} Let $S=\{m_1\cap \cdots \cap m_r:m_i\text{ is maximal}\}$ be the set of finite intersections of maximal ideals. By d.c.c. $S$ has a minimal element, denoted by $m_1\cap \cdots \cap m_n$. Now let $m\subseteq A$ be a maximal ideal. Then $m\cap ({\cap }_{i=1}^n{m}_i)={\cap }_{i=1}^n{m}_i$ and so $m\supseteq {\cap }_{i=1}^n{m}_i$. By ^tepoo1, $m\supseteq m_i$ for some $i$ and so $m=m_i$. So $\mathrm{Spec}(A)=\{m_1,\cdots ,m_n\}$. \end{proof}

Proposition

If $A$ is an Artinian ring, then $N=\mathrm{Nil}(A)$ is nilpotent.

\begin{proof} By d.c.c, the chain $N\supseteq N^2\supseteq \cdots$ terminates and so $\alpha =N^k=N^{k+1}=\cdots$. We aim to show $\alpha =(0)$.

If $\alpha \ne 0$, define $\Sigma =\{\beta \subseteq A:\beta \text{ is an ideal, }\alpha \beta \ne 0\}$. Since $A\in \Sigma$, the ideal $\Sigma \ne 0$ and so it has a minimal element $c$. Since $c\alpha \ne 0$, there exists $x\in c$ such that $x\alpha \ne 0$. It deduces that $(x)\in \Sigma$ and $c=(x)$ by minimality.

Note that $x\alpha \ne 0$ yields $(x\alpha )\alpha =x\alpha \ne 0$ and $x\alpha \in \sigma$. By minimality, $x\alpha =(x)$ and $x=xy$ for some $y\in \alpha =N^k$. Now

$$x=xy=x{y}^2=\cdots =x{y}^n=0$$

for some $n$, which contradicts with $(x)\in \Sigma$. \end{proof}

Definition

A chain of prime ideals is $p_0\subset p_1\subset \cdots \subset p_n\subset \cdots$. Its length is the number of jumps. Define the (Krull-)dimension of $A$ to be the maximal length of chaim of prime ideal.

Examples. 我觉得我完全没学会怎么算这玩意

• For a field $k$, $\dim k=0$.
• $(0)\subseteq (p)$ is the longest prime chain of $\mathbb{Z}$, so $\dim \mathbb{Z}=1$.
• In $\mathbb{C}[x]$, one have $(0)\subseteq (x-a)$ and so $\dim \mathbb{C}[x]=1$. Remark that Every non-zero prime ideal in a Principal Ideal Domain (PID) is a maximal ideal.
• In $\mathbb{C}[x]/(x^2)$, $(\overline{x})$ is the unique prime ideal and so $\dim \mathbb{C}[x]/(x^2)=0$.
• Chain of prime ideal of $\mathbb{C}[x,y]/(xy)$ is $\overline{(x)}=\overline{(x,xy)}\subseteq \overline{(x,y-a)}$ and $\dim \mathbb{C}[x,y]/(xy)=1$.
• $\dim (A\times B)=\max \{\dim A,\dim B\}$. Recall that $\mathrm{Spec}(A\times B)=\mathrm{Spec}(A)\bigsqcup \mathrm{Spec}(B)$.
• $\dim (\mathbb{C}[x,y])=2$ (not easy) (he said it use ^d13b03)

Theorem

$A$ is Artinian iff $A$ is Noetherian and $\dim A=0$.

\begin{proof} Recall that if $(0)=m_1\cdots m_n$ is a product of some maximal ideals, then $A$ Noetherian iff $A$ Artinian by ^3gn38j.

"→" By ^fsei3o, one have $\dim A=0$. By ^eii4v9, $N^k=0$ for some $k$. By ^steden, $N$ is the intersection of all prime ideals, which is a finite intersection. It deduces that $(m_1\cdots m_k{)}^k\subseteq {\cap }_{i=1}^n{m}_i^k=(0)$, verifying assumption of ^3gn38j.

"←" By ^k8wbyb, $(0)={\cap }_{i=1}^n{q}_i$ with distinct $r(q_i)=p_i$. Since $\dim A=0$, we have $p_i$ are maximal ideals. It deduces that $\mathrm{Nil}(A)=r((0))={\cap }_{i=1}^n{p}_i$. Since $A$ is Noetherian, $N$ is nilpotent. Then $(0)=N^k\supseteq (p_1\cdots p_n{)}^k$ verifying assumption of ^3gn38j. \end{proof}

Example. Note that ${\prod }_{i=1}^{\infty }k$ is dimension $0$ but not Artinian.

Proposition

If $A$ is a Noetherian ring and $m$ is a maximal ideal, then exactly one of following is true.

• $m^n\ne m^{n+1}$ for all $n⩾0$;
• $m^n=0$ for some $n⩾0$ in which case $A$ is Artinian.

\begin{proof} If i) does not hold, then $m^n=m^{n+1}$. By Nakayama lemma, $m^n=0$. By ^3gn38j, $A$ is Artinian. \end{proof}

Theorem

$A$ is Artinian iff $A={\prod }_{i=1}^n{A}_i$ is a finite product of Artinian local rings.

\begin{proof} "←" Each $A_i$ is Noetherian and dimension $0$. Then by ^myy86x, $A$ is also Noetherian and dimension $0$. It deduces that $A$ is Artinian.

"→" By ^steden, $A$ has finitely many maximal ideals $m_1,\cdots ,m_n$. By proof of ^myy86x, we have $(m_1\cdots m_n{)}^k=0$.

Note that $\{m_i{\}}_{i=1}^n$ are pairwise coprime and so for $\{m_i^k\}$ are also coprime, because

$$r(m_i^k+m_j^k)=r(r(m_i^k)+r(m_j^k))=A$$

by ^xs5hv8. Then by ^ss3fn4, we have ${\prod }_{i=1}^n{m}_i^k={\cap }_{i=1}^n{m}_i^k$. It deduces that

$$A\simeq A/(0)=A/\prod _{i=1}^n{m}_i^k=A/{\cap }_{i=1}^n{m}_i^k\simeq \prod _{i=1}^{n}A/m_i^k.$$

Claim that $A/m_i^k$ is Artinian local. It is easy to check $A/m_i^k$ is Artinian. Additionally, note that $\overline{m_j}=A/m_i^k$ for all $j\ne i$. Hence ${\overline{m}}_i$ is the unique maximal ideal of $A/m_i^k$. Now we finish the proof. \end{proof}

Remark. Given a Noetherian local ring $(A,m)$, $m/m^2$ is a $k=A/m$-vector space and so ${\dim }_{k}m/m^2<\infty$.

Proposition

Suppose $(A,m)$ is Artinian local. Then TFAE:

• all ideals of $A$ are principal;
• $m$ is principal;
• ${\dim }_{k}m/m^2⩽1$.

\begin{proof} i)→ii) Obvious.

ii)→iii) $m=(x)$ yields $m/m^2=k\cdot \overline{x}$. Then ${\dim }_{k}m/m^2⩽1$.

iii)→i) ${\dim }_{k}m/m^2⩽1$ yields that there exists $x\in m$ such that $\overline{x}\in m/m^2$ generate $m/m^2$ as a $k$-vector space.

Note that ^w0idub yields that $m=(x)$. Let $\alpha \subseteq A$ be a proper ideal. Since $\mathrm{Nil}(A)=m$, we have $m^k=0$ for a big enough $k$ by ^eii4v9. It deduces that $m^k\subseteq \alpha \subseteq m$.

So there exists $r$ such that $\alpha \subseteq m^r=(x^r)$ but $\alpha ⊈m^{r+1}=(x^{r+1})$. Then there is $y\in \alpha$ such that $y\notin (x^{r+1})$. But $y\in (x^r)$ yields $y=x^{r}a$. It deduces $a\notin (x)=m$ and $a$ is a unit. Thus $\alpha \supseteq (y)=(x^r)\supseteq \alpha$ and so $\alpha =(x^r)$ is principal. \end{proof}