Can an Artinian Ring Fail to Be Noetherian?

A commutative Artinian ring cannot fail to be Noetherian. The answer changes for noncommutative rings, where “Artinian” and “Noetherian” may refer to left ideals or right ideals. In fact, there are rings that are right Artinian but not left Noetherian.

Here is a concrete construction, followed by checks of both properties. It also illustrates why the sidedness convention matters: the example is not Artinian on both sides.

The commutative case and the sidedness issue

A ring is left Artinian if descending chains of left ideals eventually stop; right Artinian is defined using right ideals. Left Noetherian and right Noetherian use ascending chains instead. For commutative rings, left and right ideals are the same, and the Artinian condition implies the Noetherian condition.

For associative rings with identity, the Hopkins–Levitzki theorem says that a left Artinian ring is left Noetherian, and a right Artinian ring is right Noetherian. Thus, if “Artinian” means Artinian on both sides, the ring is also Noetherian on both sides. A one-sided example must be Artinian on one side and fail Noetherianity on the other.

A right Artinian, not left Noetherian example

Let $A\subset B$ be an infinite field extension, so $B$ is infinite-dimensional as a vector space over $A$. Define

$$R=\begin{pmatrix} A & B \\ 0 & B \end{pmatrix} =\left\{\begin{pmatrix}a&m\\0&b\end{pmatrix}:a\in A,\ m,b\in B\right\}.$$

Use the usual matrix addition and multiplication, with the inclusion $A\subset B$ used in the upper-right entry. The ring has an identity, the diagonal matrix with entries $1$ and $1$. Its multiplication is

$$\begin{pmatrix}a&m\\0&b\end{pmatrix} \begin{pmatrix}a'&m'\\0&b'\end{pmatrix} = \begin{pmatrix}aa'&am'+mb'\\0&bb'\end{pmatrix}.$$

Let

$$J=\left\{\begin{pmatrix}0&m\\0&0\end{pmatrix}:m\in B\right\}.$$

This is a two-sided ideal and $J^2=0$. The quotient is $R/J\cong A\times B$, a product of fields.

Why it is right Artinian

As a right $R$-module, $J$ is just the one-dimensional right vector space $B_B$: multiplication on the right acts on its upper-right entry by the lower-right entry of the matrix. In particular, $J$ has finite length as a right $R$-module.

The quotient $R/J\cong A\times B$ also has finite length as a right $R$-module: it is the direct sum of the two simple modules associated with the two diagonal entries. From the exact sequence of right modules

$$0\longrightarrow J\longrightarrow R_R\longrightarrow (R/J)_R\longrightarrow 0,$$

$R_R$ has finite length. A module of finite length satisfies both chain conditions on submodules, so $R$ is right Artinian (and right Noetherian).

Why it is not left Noetherian

Choose elements $b_1,b_2,\ldots$ of $B$ that are linearly independent over $A$. For each positive integer $n$, let

$$I_n=\left\{\begin{pmatrix}0&m\\0&0\end{pmatrix}:m\in\operatorname{span}_A(b_1,\ldots,b_n)\right\}.$$

Each $I_n$ is a left ideal: multiplying an element of $I_n$ on the left by a matrix in $R$ multiplies its upper-right entry by an element of $A$. The linear independence makes the inclusions strict, giving

$$I_1\subsetneq I_2\subsetneq I_3\subsetneq\cdots.$$

Thus $R$ is not left Noetherian.

It is also not left Artinian. For each positive integer $n$, define

$$K_n=\left\{\begin{pmatrix}0&m\\0&0\end{pmatrix}:m\in\operatorname{span}_A(b_n,b_{n+1},\ldots)\right\}.$$

Each $K_n$ is a left ideal for the same reason as $I_n$, and linear independence gives the strict descending chain

$$K_1\supsetneq K_2\supsetneq K_3\supsetneq\cdots.$$

The right Artinian property does not imply the corresponding left-sided properties.

Common mistakes

• Ignoring which side is meant. In a noncommutative ring, a chain of left ideals and a chain of right ideals are different tests. State the side whenever it matters.
• Applying the commutative theorem too broadly. Artinian implies Noetherian for commutative rings, and on the same side for rings with identity. It does not turn right Artinian into left Noetherian.
• Calling the example Artinian on both sides. The construction is right Artinian but not left Noetherian. It answers the one-sided interpretation, not a claim of being Artinian on both sides.
• Confusing two-sided ideals with one-sided ideals. The ascending chain above consists of left ideals. In noncommutative algebra, “Noetherian” is often used for left Noetherian or right Noetherian rings; if a text means the ascending chain condition on two-sided ideals, check its definition separately.