Every Noetherian ring can be embedded into an Artinian ring

Posted: September 9, 2022 in Basic Algebra, Rings
Tags: Artinian rings, embedding, irreducible ideal, irreducible ring, localization of commutative ring, nilpotent, Noetherian ring, zero-divisor

All rings in this post are commutative with $1.$

Noetherian rings are very important in ring theory but Artinian rings, which are Noetherian by Akizuki theorem, are nicer because, as we proved in Fact 4 in this post, they have a nicer structure. Oh, and don’t forget that not every Noetherian ring is Artinian; an example is the ring of integers. In this post, we’re going to prove that every Noetherian ring can be embedded into an Artinian ring, i.e. there is an Artinian ring with a subring isomorphic to the Noetherian ring.

Note. The reference for this post is section 3.2, volume I, of my favorite Jew’s book Ring Theory, and in case you’re wondering who my favorite Jew is, his name is Louis Rowen. I have slightly modified the proof given in his book, which by the way is a great one.

Definition. Let $R$ be a ring. An ideal $I$ of $R$ is said to be irreducible if $I=J \cap K,$ with $J,K$ ideals of $R,$ implies that $I=J$ or $I=K.$ We say that $R$ is an irreducible ring if $(0)$ is an irreducible ideal of $R.$

Lemma 1. Let $R$ be a ring.

i) If $I$ is an irreducible ideal of $R,$ then $R/I$ is an irreducible ring.

ii) If $R$ is Noetherian, then every ideal of $R$ is a finite intersection of some irreducible ideals of $R.$

Proof. i) Let $J,K$ be some ideals of $R$ containing $I.$ It is clear that $(J \cap K)/I \subseteq J/I \cap K/I$ and thus if $J/I \cap K/I=(0),$ then $(J \cap K)/I=(0)$ giving $J \cap K=I.$ Hence, since $I$ is irreducible, either $I=J$ or $I=K$ and so either $J/I=(0)$ or $K/I=(0),$ proving that $R/I$ is an irreducible ring.

ii) Suppose the claim is not true, i.e. there exists an ideal $I$ of $R$ that is not a finite intersection of some irreducible ideals of $R.$ Let $T$ be the set of all ideals of $R$ that are not finite intersection of some irreducible ideals of $R.$ So $I \in T$ and hence $T \ne \emptyset.$ Thus, since $R$ is Noetherian, $T$ has a maximal element, say $L.$ Clearly $L$ can’t be irreducible itself because then it wouldn’t be in $T.$ Thus there exist ideals $J,K$ of $R$ such that $J,K \supset L$ and $L=J \cap K.$ So $J,K \notin T,$ by maximality of $L$ in $T,$ Thus each $J$ and $K$ is a finite intersection of some irreducible ideals of $R.$ But then $L=J \cap K$ would also be a finite intersection of some irreducible ideals of $R,$ contradiction. $\ \Box$

Lemma 2. Every zero-divisor of an irreducible Noetherian ring is nilpotent.

Proof. For an element $z \in R,$ let $\text{ann}(z):=\{r \in R: \ rz=0\}.$ Clearly $\text{ann}(z)$ is an ideal of $R.$ Now, let $x \in R$ be a zero-divisor of $R,$ and consider the ascending chain of ideals $\text{ann}(x) \subseteq \text{ann}(x^2) \subseteq \cdots.$ Since $R$ is Noetherian, the chain will eventually stop, i.e. there exists a positive integer $n$ such that

$\text{ann}(x^n)=\text{ann}(x^{n+1}) = \cdots .$

Let $y:=x^n.$ Then $\text{ann}(y)=\text{ann}(y^2)$ and $y$ is a zero-divisor of $R.$ Thus there exists $0 \ne a \in R$ such that $ay=0.$ Now, if $r \in Ra \cap Ry,$ then $r=r_1a=r_2y,$ for some $r_i \in R,$ and hence $r_2y^2=r_1ay=0.$ Thus $r_2 \in \text{ann}(y^2)=\text{ann}(y)$ and so $r=r_2y=0.$ So we have shown that $Ra \cap Ry=(0)$ and thus, since $R$ is irreducible, either $Ra=(0)$ or $Ry=(0).$ But $Ra \ne (0),$ because $a \ne 0,$ and so $Ry=(0),$ which gives $x^n=y=0, \ \Box$

Notation. Let $R$ be a ring, and let $Z$ be the set of zero-divisors of $R.$ Let $S:=R \setminus Z.$ We denote by $Q(R)$ the localization of $R$ by $S,$ i.e. $Q(R):=S^{-1}R.$ Note that since no element of $S$ is a zero-divisor in $R,$ the function $g: R \to Q(R)$ defined by $g(x)=x/1,$ for all $x \in R,$ is an injective ring homomorphism. So $R$ is isomorphic to a subring of $Q(R).$ In other words, $R$ is embedded into $Q(R).$

Theorem. Every Noetherian ring can be embedded into an Artinian ring.

Proof. By Lemma 1, ii), there exist irreducible ideals $I_1, \cdots , I_n$ of $R$ such that $\displaystyle (0)=\bigcap_{k=1}^nI_k.$ Thus the map

$\displaystyle f: R \to \bigoplus_{k=1}^nR/I_k, \ \ \ \ f(x)=(x+I_1, \cdots , x+ I_n), \ \ \ \forall x \in R,$

is an injective ring homomorphism. In other words, $R$ is embedded into $\displaystyle \bigoplus_{k=1}^nR/I_k.$ On the other hand, since $R$ is Noetherian and each $I_k$ is irreducible, $R/I_k$ is a Noetherian irreducible ring, by Lemma 1, i). Hence, by Lemma 2, every zero-divisor of $R/I_k$ is nilpotent. Therefore, by Corollary 2 in this post, $Q(R/I_k)$ is an Artinian ring and $R/I_k$ is embedded into $Q(R/I_k).$ Thus $R,$ which, as we’ve already shown, is embedded into $\displaystyle \bigoplus_{k=1}^nR/I_k,$ is embedded into $\displaystyle \bigoplus_{k=1}^nQ(R/I_k),$ which is an Artinian ring because each $Q(R/I_k)$ is an Artinian ring. $\ \Box$