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\begin{document}

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\section*{noetherian ring}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{algebra}{}\paragraph*{{Algebra}}\label{algebra}

[[!include higher algebra - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{general}{General}\dotfill \pageref*{general} \linebreak
\noindent\hyperlink{a_homological_characterization}{A homological characterization}\dotfill \pageref*{a_homological_characterization} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A Noetherian (or often, as below, noetherian) [[ring]] (or [[rng]]) is one where it is possible to do [[induction]] over its ideals, because the ordering of ideals by reverse inclusion is [[well-founded relation|well-founded]].

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

(In this section, ``ring'' means [[rng]], where the presence of a multiplicative identity is not assumed unless we say ``unital ring''.)

A (left) \textbf{noetherian ring} $R$ is a [[ring]] for which every ascending chain of its (left) [[ideals]] stabilizes. In other words, it is noetherian if its underlying $R$-module ${}_R R$ is a noetherian object in the category $R Mod$ of left $R$-modules (recall that a left ideal is simply a submodule of ${}_R R$). Similarly for right noetherian rings. Left noetherianness is independent of right noetherianness. A ring is noetherian if it is both left noetherian and right noetherian.

An equivalent condition is that all (left) ideals are finitely generated.

A dual condition is artinian: an \textbf{[[artinian ring]]} is a ring satisfying the descending chain condition on ideals. The symmetry is severely broken if one considers unital rings: for example every unital artinian ring is noetherian; artinian rings are intuitively much smaller than generic noetherian rings.

Spectra of noetherian rings are glued together to define [[noetherian scheme|locally noetherian schemes]].

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{example}
\label{FieldIsNoetherianRing}\hypertarget{FieldIsNoetherianRing}{}
Every [[field]] is a noetherian ring.

\end{example}
\begin{example}
\label{PIDIsNoetherianRing}\hypertarget{PIDIsNoetherianRing}{}
Every [[principal ideal domain]] is a noetherian ring.

\end{example}
\begin{example}
\label{PolynomialAlgebraOverNoetherianRingIsNoetherian}\hypertarget{PolynomialAlgebraOverNoetherianRingIsNoetherian}{}
For $R$ a [[Noetherian ring]] (e.g. a [[field]] by example \ref{FieldIsNoetherianRing}) then

\begin{enumerate}%
\item the [[polynomial algebra]] $R[X_1, \cdots, X_n]$


\item the [[formal power series algebra]] $R[ [ X_1, \cdots, X_n ] ]$



\end{enumerate}
over R in a [[finite number]] $n$ of [[coordinates]] are Noetherian.

\end{example}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\hypertarget{general}{}\subsubsection*{{General}}\label{general}

One of the best-known properties is the Hilbert basis theorem. Let $R$ be a (unital) ring.

\begin{theorem}
\label{}\hypertarget{}{}
\textbf{(Hilbert)} If $R$ is left Noetherian, then so is the [[polynomial|polynomial algebra]] $R[x]$. (Similarly if ``right'' is substituted for ``left''.)

\end{theorem}
\begin{proof}
(We adapt the proof from \href{https://en.wikipedia.org/wiki/Hilbert%27s_basis_theorem#First_Proof}{Wikipedia}.) Suppose $I$ is a left ideal of $R[x]$ that is not finitely generated. Using the [[axiom of dependent choice]], there is a [[sequence]] of polynomials $f_n \in I$ such that the left ideals $I_n \coloneqq (f_0, \ldots, f_{n-1})$ form a strictly increasing chain and $f_n \in I \setminus I_n$ is chosen to have degree as small as possible. Putting $d_n \coloneqq \deg(f_n)$, we have $d_0 \leq d_1 \leq \ldots$. Let $a_n$ be the leading coefficient of $f_n$. The left ideal $(a_0, a_1, \ldots)$ of $R$ is finitely generated; say $(a_0, \ldots, a_{k-1})$ generates. Thus we may write

\begin{equation}
a_k = \sum_{i=0}^{k-1} r_i a_i
\label{kill}\end{equation}
The polynomial $g = \sum_{i=0}^{k-1} r_i x^{d_k - d_i} f_i$ belongs to $I_k$, so $f_k - g$ belongs to $I \setminus I_k$. Also $g$ has degree $d_k$ or less, and therefore so does $f_k - g$. But notice that the coefficient of $x^{d_k}$ in $f_k - g$ is zero, by \eqref{kill}. So in fact $f_k - g$ has degree less than $d_k$, contradicting how $f_k$ was chosen.

\end{proof}
\hypertarget{a_homological_characterization}{}\subsubsection*{{A homological characterization}}\label{a_homological_characterization}

\begin{theorem}
\label{}\hypertarget{}{}
For a unital ring $R$ the following are equivalent:

\begin{enumerate}%
\item $R$ is left Noetherian
\item Any small direct sum of injective left $R$-modules is injective.
\item $\operatorname{Ext}^k_R(A, \cdot)$ commutes with small direct sums for any finitely generated $A$.

\end{enumerate}
\end{theorem}
Direct sums here can be replaced by filtered colimits.

\begin{proof}
$1 \Rightarrow 2$: assume that $R$ is Noetherian and $I_\alpha$ are injective modules. In order to verify that $I := \bigoplus_\alpha I_\alpha$ is injective it is enough to show that for any ideal $\mathfrak{j}$ any morphism of left modules $f : \mathfrak{j} \to I$ factors through $\mathfrak{j} \to R$. Since $R$ is Notherian, $\mathfrak{j}$ is finitely generated, so the image of $f$ lies in a finite sum $I_{\alpha_1} \oplus \dots \oplus I_{\alpha_n}$. Thus an extension to $R$ exists by the injectivity of each $I_{\alpha_k}$.

$2 \Rightarrow 1$: if $R$ is not left Noetherian then there is a sequence of left ideals $\mathfrak{j}_1 \subsetneq \mathfrak{j}_2 \subsetneq \dots$. Take $\mathfrak{j} := \bigcup_k \mathfrak{j}_k$. The obvious map $j \to \prod_k (\mathfrak{j} / \mathfrak{j}_k)$ factors through $\bigoplus_k (\mathfrak{j} / \mathfrak{j}_k)$, since any element lies in all but finitely many $\mathfrak{j}_k$. Now take any injective $I_k$ with $0 \to \mathfrak{j} / \mathfrak{j}_k \to I_k$. The map $\mathfrak{j} \to \bigoplus_k I_k$ cannot extend to the whole $R$, since otherwise its image would be contained in a sum of finitely many $I_k$. Therefore, $\bigoplus_k I_k$ is not injective.

$2 \Rightarrow 3$: $\operatorname{Ext}^k_R(A, \bigoplus_\alpha X_\alpha)$ can be computed by taking an injective resolution of $\bigoplus_\alpha X_\alpha$. Since direct sums of injective modules are assumed to be injective, we can take a direct sum of injective resolutions of each $X_\alpha$. It remains to note that Hom out of a finitely generated module commutes with arbitrary direct sums.

$3 \Rightarrow 2$: Follows from the fact that $I$ is injective iff $\operatorname{Ext}^1_R(R / \mathfrak{i}, I) = 0$ for any ideal $\mathfrak{i}$.

\end{proof}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[Noetherian module]]


\item [[Noetherian poset]]


\item [[Noetherian E-∞ ring]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item \href{http://en.wikipedia.org/wiki/Noetherian_ring}{wikipedia}, \emph{[[noetherian object]]}

\end{itemize}
[[!redirects noetherian ring]] [[!redirects noetherian rings]] [[!redirects Noetherian ring]] [[!redirects Noetherian rings]]



\end{document}