## Tag `00J4`

## 10.52. Artinian rings

Artinian rings, and especially local Artinian rings, play an important role in algebraic geometry, for example in deformation theory.

Definition 10.52.1. A ring $R$ is

Artinianif it satisfies the descending chain condition for ideals.Lemma 10.52.2. Suppose $R$ is a finite dimensional algebra over a field. Then $R$ is Artinian.

Proof.The descending chain condition for ideals obviously holds. $\square$Lemma 10.52.3. If $R$ is Artinian then $R$ has only finitely many maximal ideals.

Proof.Suppose that $\mathfrak m_i$, $i = 1, 2, 3, \ldots$ are maximal ideals. Then $\mathfrak m_1 \supset \mathfrak m_1\cap \mathfrak m_2 \supset \mathfrak m_1 \cap \mathfrak m_2 \cap \mathfrak m_3 \supset \ldots$ is an infinite descending sequence (because by the Chinese remainder theorem all the maps $R \to \oplus_{i = 1}^n R/\mathfrak m_i$ are surjective). $\square$Lemma 10.52.4. Let $R$ be Artinian. The radical $\text{rad}(R)$ of $R$ is a nilpotent ideal.

Proof.Denote the radical $I$. Note that $I \supset I^2 \supset I^3 \supset \ldots$ is a descending sequence. Thus $I^n = I^{n + 1}$ for some $n$. Set $J = \{ x\in R \mid xI^n = 0\}$. We have to show $J = R$. If not, choose an ideal $J' \not = J$, $J \subset J'$ minimal (possible by the Artinian property). Then $J' = J + Rx$ for some $x \in R$. By NAK, Lemma 10.19.1, we have $IJ' \subset J$. Hence $xI^{n + 1} \subset xI \cdot I^n \subset J \cdot I^n = 0$. Since $I^{n + 1} = I^n$ we conclude $x\in J$. Contradiction. $\square$Lemma 10.52.5. Any ring with finitely many maximal ideals and locally nilpotent radical is the product of its localizations at its maximal ideals. Also, all primes are maximal.

Proof.Let $R$ be a ring with finitely many maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_n$. Let $I = \bigcap_{i = 1}^n \mathfrak m_i$ be the radical of $R$. Assume $I$ is locally nilpotent. Let $\mathfrak p$ be a prime ideal of $R$. Since every prime contains every nilpotent element of $R$ we see $ \mathfrak p \supset \mathfrak m_1 \cap \ldots \cap \mathfrak m_n$. Since $\mathfrak m_1 \cap \ldots \cap \mathfrak m_n \supset \mathfrak m_1 \ldots \mathfrak m_n$ we conclude $\mathfrak p \supset \mathfrak m_1 \ldots \mathfrak m_n$. Hence $\mathfrak p \supset \mathfrak m_i$ for some $i$, and so $\mathfrak p = \mathfrak m_i$. By the Chinese remainder theorem (Lemma 10.14.3) we have $R/I \cong \bigoplus R/\mathfrak m_i$ which is a product of fields. Hence by Lemma 10.31.5 there are idempotents $e_i$, $i = 1, \ldots, n$ with $e_i \bmod \mathfrak m_j = \delta_{ij}$. Hence $R = \prod Re_i$, and each $Re_i$ is a ring with exactly one maximal ideal. $\square$Lemma 10.52.6. A ring $R$ is Artinian if and only if it has finite length as a module over itself. Any such ring $R$ is both Artinian and Noetherian, any prime ideal of $R$ is a maximal ideal, and $R$ is equal to the (finite) product of its localizations at its maximal ideals.

Proof.If $R$ has finite length over itself then it satisfies both the ascending chain condition and the descending chain condition for ideals. Hence it is both Noetherian and Artinian. Any Artinian ring is equal to product of its localizations at maximal ideals by Lemmas 10.52.3, 10.52.4, and 10.52.5.Suppose that $R$ is Artinian. We will show $R$ has finite length over itself. It suffices to exhibit a chain of submodules whose successive quotients have finite length. By what we said above we may assume that $R$ is local, with maximal ideal $\mathfrak m$. By Lemma 10.52.4 we have $\mathfrak m^n =0$ for some $n$. Consider the sequence $0 = \mathfrak m^n \subset \mathfrak m^{n-1} \subset \ldots \subset \mathfrak m \subset R$. By Lemma 10.51.6 the length of each subquotient $\mathfrak m^j/\mathfrak m^{j + 1}$ is the dimension of this as a vector space over $\kappa(\mathfrak m)$. This has to be finite since otherwise we would have an infinite descending chain of sub vector spaces which would correspond to an infinite descending chain of ideals in $R$. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 12065–12194 (see updates for more information).

```
\section{Artinian rings}
\label{section-artinian}
\noindent
Artinian rings, and especially local Artinian rings,
play an important role in algebraic geometry, for example
in deformation theory.
\begin{definition}
\label{definition-artinian}
A ring $R$ is {\it Artinian} if it satisfies the
descending chain condition for ideals.
\end{definition}
\begin{lemma}
\label{lemma-finite-dimensional-algebra}
Suppose $R$ is a finite dimensional algebra over a field.
Then $R$ is Artinian.
\end{lemma}
\begin{proof}
The descending chain condition for ideals obviously holds.
\end{proof}
\begin{lemma}
\label{lemma-artinian-finite-nr-max}
If $R$ is Artinian then $R$ has only finitely many maximal ideals.
\end{lemma}
\begin{proof}
Suppose that $\mathfrak m_i$, $i = 1, 2, 3, \ldots$ are maximal ideals.
Then $\mathfrak m_1 \supset \mathfrak m_1\cap \mathfrak m_2
\supset \mathfrak m_1 \cap \mathfrak m_2 \cap \mathfrak m_3 \supset \ldots$
is an infinite descending sequence (because by the Chinese
remainder theorem all the maps $R \to \oplus_{i = 1}^n R/\mathfrak m_i$
are surjective).
\end{proof}
\begin{lemma}
\label{lemma-artinian-radical-nilpotent}
Let $R$ be Artinian. The radical $\text{rad}(R)$ of $R$ is
a nilpotent ideal.
\end{lemma}
\begin{proof}
Denote the radical $I$.
Note that $I \supset I^2 \supset I^3 \supset \ldots$ is a descending
sequence. Thus $I^n = I^{n + 1}$ for some $n$.
Set $J = \{ x\in R \mid xI^n = 0\}$. We have to show $J = R$.
If not, choose an ideal $J' \not = J$, $J \subset J'$ minimal (possible
by the Artinian property). Then $J' = J + Rx$ for some $x \in R$.
By NAK, Lemma \ref{lemma-NAK}, we have $IJ' \subset J$.
Hence $xI^{n + 1} \subset xI \cdot I^n \subset J \cdot I^n = 0$.
Since $I^{n + 1} = I^n$ we conclude $x\in J$. Contradiction.
\end{proof}
\begin{lemma}
\label{lemma-product-local}
Any ring with finitely many maximal ideals and
locally nilpotent radical is the product of its localizations
at its maximal ideals. Also, all primes are maximal.
\end{lemma}
\begin{proof}
Let $R$ be a ring with finitely many maximal ideals
$\mathfrak m_1, \ldots, \mathfrak m_n$.
Let $I = \bigcap_{i = 1}^n \mathfrak m_i$
be the radical of $R$. Assume $I$ is locally nilpotent.
Let $\mathfrak p$ be a prime ideal of $R$.
Since every prime contains every nilpotent
element of $R$ we see
$ \mathfrak p \supset \mathfrak m_1 \cap \ldots \cap \mathfrak m_n$.
Since $\mathfrak m_1 \cap \ldots \cap \mathfrak m_n \supset
\mathfrak m_1 \ldots \mathfrak m_n$
we conclude $\mathfrak p \supset \mathfrak m_1 \ldots \mathfrak m_n$.
Hence $\mathfrak p \supset \mathfrak m_i$ for some $i$, and so
$\mathfrak p = \mathfrak m_i$. By the Chinese remainder theorem
(Lemma \ref{lemma-chinese-remainder})
we have $R/I \cong \bigoplus R/\mathfrak m_i$
which is a product of fields.
Hence by Lemma \ref{lemma-lift-idempotents}
there are idempotents $e_i$, $i = 1, \ldots, n$
with $e_i \bmod \mathfrak m_j = \delta_{ij}$.
Hence $R = \prod Re_i$, and each $Re_i$ is a
ring with exactly one maximal ideal.
\end{proof}
\begin{lemma}
\label{lemma-artinian-finite-length}
A ring $R$ is Artinian if and only if it has finite length
as a module over itself. Any such ring $R$ is both Artinian and
Noetherian, any prime ideal of $R$ is a maximal ideal, and $R$ is equal
to the (finite) product of its localizations at its maximal ideals.
\end{lemma}
\begin{proof}
If $R$ has finite length over itself then it satisfies both
the ascending chain condition and the descending chain
condition for ideals. Hence it is both Noetherian and Artinian.
Any Artinian ring is equal to product of its localizations
at maximal ideals by Lemmas \ref{lemma-artinian-finite-nr-max},
\ref{lemma-artinian-radical-nilpotent}, and \ref{lemma-product-local}.
\medskip\noindent
Suppose that $R$ is Artinian. We will show $R$ has finite
length over itself. It suffices to exhibit a chain of
submodules whose successive quotients have finite length.
By what we said above
we may assume that $R$ is local, with maximal ideal $\mathfrak m$.
By Lemma \ref{lemma-artinian-radical-nilpotent} we have
$\mathfrak m^n =0$ for some $n$.
Consider the sequence
$0 = \mathfrak m^n \subset \mathfrak m^{n-1} \subset
\ldots \subset \mathfrak m \subset R$. By Lemma
\ref{lemma-dimension-is-length} the length of each subquotient
$\mathfrak m^j/\mathfrak m^{j + 1}$ is the dimension of this
as a vector space over $\kappa(\mathfrak m)$. This has to be
finite since otherwise we would have an infinite descending
chain of sub vector spaces which would correspond to an
infinite descending chain of ideals in $R$.
\end{proof}
```

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