Artinian Rings and Length Examples

By For every field $F$, $F[x]$ is a principal ideal domain, every ideal of $k[x]$ is principal. The ideals of $R$ correspond by Correspondence theorem: ideals of $R\mathrm{/}I$ correspond to ideals of $R$ containing $I$ to the ideals of $k[x]$ containing $(x^n)$, hence to the principal ideals $(f)$ with $f$ dividing $x^n$. Up to multiplication by a unit, these are exactly $(x^r)$ for $0\le r\le n$. Therefore the ideals of $R$ are precisely $R=(x^0)⊋(x)⊋\cdots ⊋(x^{n-1})⊋(x^n)=0$, so $(x)$ is the unique maximal ideal.

The chain in step 1.1 shows directly that $R$ is Artinian and that $(x{)}^n=0$. For each $0\le r<n$, the quotient $(x^r)\mathrm{/}(x^{r+1})$ is generated by the class of $x^r$ and annihilated by $x$, so it is one-dimensional over the residue field $R\mathrm{/}(x)\cong k$. Hence each quotient has length $1$.

Applying Module length is additive in short exact sequences successively to $0\to (x^{r+1})\to (x^r)\to (x^r)\mathrm{/}(x^{r+1})\to 0$ for $0\le r<n$ shows that ${\mathrm{\ell }}_R(R)=n$. Thus $R$ is a local Artinian ring of length $n$.

The ideals $(3)$ and $(4)$ of $\mathbb{Z}$ are comaximal, so Chinese remainder theorem for pairwise comaximal ideals gives $\mathbb{Z}\mathrm{/}12\mathbb{Z}\cong \mathbb{Z}\mathrm{/}3\mathbb{Z}\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$. The first factor is a field, hence local, and the second has unique proper nonzero ideal $(2)$, so it is also local.

Let $P$ be a prime ideal of $\mathbb{Z}\mathrm{/}3\mathbb{Z}\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$. Since $(1,0)(0,1)=(0,0)\in P$, primality forces $(1,0)\in P$ or $(0,1)\in P$. In the first case every $(a,0)=(a,0)(1,0)$ lies in $P$, so $P=\mathbb{Z}\mathrm{/}3\mathbb{Z}\times Q$ for an ideal $Q$ of $\mathbb{Z}\mathrm{/}4\mathbb{Z}$; primality then forces $Q=(2)$. In the second case $P=0\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$. Under the inverse of the isomorphism in step 1.1, these two primes pull back to $(2)\mathrm{/}(12)$ and $(3)\mathrm{/}(12)$ in $\mathbb{Z}\mathrm{/}12\mathbb{Z}$.

So $\mathbb{Z}\mathrm{/}12\mathbb{Z}$ is exhibited concretely as the product of its two local Artinian factors $\mathbb{Z}\mathrm{/}3\mathbb{Z}$ and $\mathbb{Z}\mathrm{/}4\mathbb{Z}$, and its two prime ideals are exactly the two coordinate pullbacks found in step 2.1.

The only ideals of a field are $0$ and the whole field, so $0<K$ is a composition series of the regular module ${}_{K}K$.

By Composition series and length of a module, that composition series has exactly one factor, so ${\mathrm{\ell }}_K(K)=1$.

Every ideal of $\mathbb{Z}$ is an additive subgroup, so Every subgroup of $(\mathbb{Z},+)$ is $⟨n⟩=n\mathbb{Z}$ for exactly one natural number $n$ makes it principal; then A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member shows that $\mathbb{Z}$ is Noetherian. Also If $R$ is Noetherian then $R[x_1,\dots ,x_n]$ is Noetherian for every $n\in \mathbb{N}$ applied to the field $k$ shows that $k[x]$ is Noetherian.

In $\mathbb{Z}$ the principal ideals $(2)⊋(2^2)⊋(2^3)⊋\cdots$ form a strict descending chain, because $2^n\in (2^n)\setminus (2^{n+1})$ for every $n\ge 1$. Likewise $(x)⊋(x^2)⊋(x^3)⊋\cdots$ is a strict descending chain of ideals in $k[x]$. Therefore neither ring is Artinian.

These two standard examples show that Noetherianity alone does not imply Artinianness.

By The truncated polynomial ring $k[x]\mathrm{/}(x^n)$ is local Artinian of length $n$, every quotient $(x^m)\mathrm{/}(x^{m+1})$ with $0\le m<n$ is one-dimensional over $k$. In particular each such quotient has length $1$, and the module $R\mathrm{/}(x)\cong k$ also has length $1$.

For every $1\le m<i$, the natural projection $R\mathrm{/}(x^{m+1})\to R\mathrm{/}(x^m)$ has kernel $(x^m)\mathrm{/}(x^{m+1})$, so there is a short exact sequence $0\to (x^m)\mathrm{/}(x^{m+1})\to R\mathrm{/}(x^{m+1})\to R\mathrm{/}(x^m)\to 0.$ Starting from the base case ${\mathrm{\ell }}_R(R\mathrm{/}(x))=1$ from step 1.1 and applying Module length is additive in short exact sequences inductively, one gets ${\mathrm{\ell }}_R(R\mathrm{/}(x^{m+1}))={\mathrm{\ell }}_R(R\mathrm{/}(x^m))+1=m+1$ for every $m<i$.

Taking $m=i-1$ in step 2.1 gives ${\mathrm{\ell }}_R(R\mathrm{/}(x^i))=i$.

Every element $x\in R$ is idempotent, because $x^2=x$ coordinatewise in $\mathbb{Z}\mathrm{/}2$. Let $\mathfrak{p}$ be a prime ideal of $R$. Then $R\mathrm{/}P$ is an integral domain if and only if $P$ is a prime ideal makes $R\mathrm{/}\mathfrak{p}$ an integral domain, and every class $\bar{x}\in R\mathrm{/}\mathfrak{p}$ still satisfies ${\bar{x}}^2=\bar{x}$. So $\bar{x}(\bar{x}-1)=0$ forces $\bar{x}=0$ or $\bar{x}=1$. Thus $R\mathrm{/}\mathfrak{p}$ has exactly two elements and is a field, so $\mathfrak{p}$ is maximal.

For each $n\ge 0$, let $e_n$ be the sequence with $1$ in coordinate $n$ and $0$ elsewhere, and let $I_n:=R{e}_0+\cdots +R{e}_n$. Then $I_0⊊I_1⊊I_2⊊\cdots$ is a strict ascending chain of ideals, because $e_{n+1}\in I_{n+1}\setminus I_n$ for every $n$. Therefore $R$ is not Noetherian.

Let ${\mathfrak{p}}_0=\{x\in R:x_0=0\}\mathrm{.}$ The first-coordinate projection $R\to \mathbb{Z}\mathrm{/}2$ is a surjective ring homomorphism with kernel ${\mathfrak{p}}_0$, so $R\mathrm{/}{\mathfrak{p}}_0\cong \mathbb{Z}\mathrm{/}2$ is a field and therefore an integral domain. Thus $R\mathrm{/}P$ is an integral domain if and only if $P$ is a prime ideal makes ${\mathfrak{p}}_0$ a prime ideal. By step 1.1 every prime ideal of $R$ is maximal, so no strict chain of prime ideals can have length greater than $0$. Since ${\mathfrak{p}}_0$ provides a prime ideal, chains of length $0$ do occur. Therefore Krull dimension of a nonzero ring gives $\dim R=0$. Together with step 2.1, this ring is zero-dimensional but not Noetherian, so it is a concrete witness that the Noetherian hypothesis in the prime-maximal Artinian criterion cannot be dropped.