---- > [!definition] Definition. ([[Artinian ring]]) > A [[ring]] $R$ is said to be **Artinian** if it is [[Artinian module|Artinian]] as an $R$-[[module]] over itself. ^definition > [!equivalence] > - See those in [[Artinian module]] > - Here is one that is false for general modules (see examples/counterexamples in notes) but true for rings: $R$ is Artinian if and only if $R$ is [[Noetherian module|Noetherian]] and $\text{dim }R=0$.[^1] ^equivalence > [!basicproperties] > - Every [[prime ideal]] of $R$ is [[maximal ideal|maximal]], and there are finitely many prime ideals > - $\text{Nil }R=J(R)$ the [[Jacobson radical]] ^properties [^1]: An Artinian ring $R$ must be Noetherian because > [!proof] > ^proof **Every prime ideal of $R$ is maximal.** Let $\mathfrak{p} \in \text{Spec }R$. Then $\frac{R}{\mathfrak{p}}$ is an [[integral domain]]. It is Artinian, by [[the correspondence theorem for rings]]. Suppose it is not a [[field]]: fix some. nonzero, [[unit|nonunit]] $f \in R$ with no inverse: for all $g \in R-\{ 0 \}$, $fg \neq 0$ and $fg \neq 1$. In particular, $f^{n} \notin \{ 0,1 \}$ for all $n \geq 1$. This produces an indefinite [[ascending chain condition|descending chain]] of proper [[ideal|ideals]] $\langle f \rangle \supset \langle f ^{2} \rangle \supset \dots $ which cannot happen because $\frac{R}{\mathfrak{p}}$ is Artinian. Thus, $\frac{R}{\mathfrak{p}}$ is a [[field]]. **$\text{Nil }R=J(R)$.** The [[nilradical of a ring|nilradical]] $\text{Nil }R$ [[nilradical equals intersection of all prime ideals|equals]] the intersection of all [[prime ideal|prime ideals]]. The [[Jacobson radical]] equals the intersection of all [[maximal ideal|maximal ideals]]. But we have shown that prime and maximal ideals coincide for Artinian rings. **$R$ has finitely many prime (equivalently, maximal) ideals.** This is clear if $R=0$. Assume $R \neq 0$. Let $\Sigma$ be the set of all ideals of $R$ of the form $\mathfrak{m}_{1} \cap \dots \cap \mathfrak{m}_{r}$. By Artinianity, $\Sigma$ has a minimal (by inclusion) element $\mathfrak{m}_{1} \cap \dots \cap \mathfrak{m}_{n}=\mathfrak{a}$. Now, for any maximal ideal $\mathfrak{m}$, the intersection $\mathfrak{m} \cap \mathfrak{a} \subset \mathfrak{a}$, but must not be a proper subset by minimality, so $\mathfrak{m} \cap \mathfrak{a}=\mathfrak{a}$. This means $\mathfrak{a} \subset \mathfrak{m}$. A basic property in [[prime ideal]] then implies $\mathfrak{m}=\mathfrak{m}_{i}$ for some $i$. (Recall prime $\iff$ maximal in $R$.) **$\text{Nil }R$ is a [[nilpotent ideal]].** By Artinianity, $\big( (\text{Nil }R)^{\ell} \big)_{\ell \geq 1}$ stabilizes in some finite step $\ell'$ at an [[ideal]] $\mathfrak{a}$. We claim that $\mathfrak{a}=0$. Suppose not. Let $\Sigma$ be the collection of ideals $\mathfrak{d}$ of $R$ satisfying $\mathfrak{a} \mathfrak{d}\neq 0$. By Artinianity, $\Sigma$ has an inclusion-minimal element $\mathfrak{c}$. Since $\mathfrak{a} \mathfrak{c} \neq 0$, we may fix $x \in \mathfrak{c}$ such that $ax \neq 0$ for some $a \in \mathfrak{a}$. Then $\langle x \rangle \mathfrak{a} \neq 0$. Since $\langle x \rangle \subset \mathfrak{c}$, minimality of $\mathfrak{c}$ implies $\langle x \rangle=\mathfrak{c}$. Moreover, $x \mathfrak{a}=\langle \{ xa : a \in \mathfrak{a} \} \rangle=\langle x \rangle$, because $x \mathfrak{a} \subset \mathfrak{c}$ and $x \mathfrak{a} \neq 0$ by definition of $x$. Thus, there exists $y \in \mathfrak{a}$ such that $x=xy$. Because $\mathfrak{a}=(\text{Nil }R)^{\ell} \subset \text{Nil }R$, we have $y \in \text{Nil }R$, say, $y^{m}=0$ for some $m \geq 1$. Then $x=xy^{m}=0$. So $\mathfrak{c}=0$ — contradiction.v ---- #### ---- #### 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 > ```