symmetric monoidal (∞,1)-category of spectra
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.
Every ring $R$ has a canonical $R$-$R$-bimodule structure, with left action $\alpha_L:R \times R \to R$ and right action $\alpha_R:R \times R \to R$ defined as the multiplicative binary operation on $R$ and biaction $\alpha:R \times R \times R \to R$ defined as the ternary product on $R$:
Let $\mathrm{TwoSidedIdeals}(R)$ be the category of two-sided ideals in $R$, whose objects are two-sided ideals $I$ in $R$, sub-$R$-$R$-bimodules of $R$ with respect to the canonical bimodule structure on $R$, and whose morphisms are $R$-$R$-bimodule monomorphisms.
An ascending chain of two-sided ideals in $R$ is a direct sequence of two-sided ideals in $R$, a sequence of two-sided ideals $A:\mathbb{N} \to \mathrm{TwoSidedIdeals}(R)$ with the following dependent sequence of $R$-$R$-bimodule monomorphisms: for natural number $n \in \mathbb{N}$, a dependent $R$-$R$-bimodule monomorphism $i_n:A_{n} \hookrightarrow A_{n+1}$.
A ring $R$ is Noetherian if it satisfies the ascending chain condition on its two-sided ideals: for every ascending chain of two-sided ideals $(A, i_n)$ in $R$, there exists a natural number $m \in \mathbb{N}$ such that for all natural numbers $n \geq m$, the $R$-$R$-bimodule monomorphism $i_n:A_{n} \hookrightarrow A_{n+1}$ is an $R$-$R$-bimodule isomorphism.
Let $\mathrm{LeftIdeals}(R)$ be the category of left ideals in $R$, whose objects are left ideals $I$ in $R$, sub-left-$R$-modules of $R$ with respect to the canonical left module structure $(-)\cdot(-):R \times R \to R$ on $R$, and whose morphisms are left $R$-module monomorphisms.
An ascending chain of left ideals in $R$ is a direct sequence of left ideals in $R$, a sequence of left ideals $A:\mathbb{N} \to \mathrm{LeftIdeals}(R)$ with the following dependent sequence of left $R$-module monomorphisms: for natural number $n \in \mathbb{N}$, a dependent left $R$-module monomorphism $i_n:A_{n} \hookrightarrow A_{n+1}$.
A ring $R$ is left Noetherian if it satisfies the ascending chain condition on its left ideals: for every ascending chain of left ideals $(A, i_n)$ in $R$, there exists a natural number $m \in \mathbb{N}$ such that for all natural numbers $n \geq m$, the left $R$-module monomorphism $i_n:A_{n} \hookrightarrow A_{n+1}$ is an left $R$-module isomorphism.
Let $\mathrm{RightIdeals}(R)$ be the category of right ideals in $R$, whose objects are right ideals $I$ in $R$, sub-right-$R$-modules of $R$ with respect to the canonical right module structure $(-)\cdot(-):R \times R \to R$ on $R$, and whose morphisms are right $R$-module monomorphisms.
An ascending chain of right ideals in $R$ is a direct sequence of right ideals in $R$, a sequence of right ideals $A:\mathbb{N} \to \mathrm{RightIdeals}(R)$ with the following dependent sequence of right $R$-module monomorphisms: for natural number $n \in \mathbb{N}$, a dependent right $R$-module monomorphism $i_n:A_{n} \hookrightarrow A_{n+1}$.
A ring $R$ is right Noetherian if it satisfies the ascending chain condition on its right ideals: for every ascending chain of right ideals $(A, i_n)$ in $R$, there exists a natural number $m \in \mathbb{N}$ such that for all natural numbers $n \geq m$, the right $R$-module monomorphism $i_n:A_{n} \hookrightarrow A_{n+1}$ is an right $R$-module isomorphism.
Every field is a noetherian ring.
Every principal ideal domain is a noetherian ring.
For $R$ a Noetherian ring (e.g. a field by example ) then
the polynomial algebra $R[X_1, \cdots, X_n]$
the formal power series algebra $R[ [ X_1, \cdots, X_n ] ]$
over R in a finite number $n$ of coordinates are Noetherian.
Spectra of noetherian rings are glued together to define locally noetherian schemes.
One of the best-known properties is the Hilbert basis theorem. Let $R$ be a (unital) ring.
(Hilbert) If $R$ is left Noetherian, then so is the polynomial algebra $R[x]$. (Similarly if “right” is substituted for “left”.)
(We adapt the proof from 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
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 (1). So in fact $f_k - g$ has degree less than $d_k$, contradicting how $f_k$ was chosen.
For a unital ring $R$ the following are equivalent:
Direct sums here can be replaced by filtered colimits.
$1 \Rightarrow 2$: 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 $\mathfrak{i}$.
A dual condition is artinian: an 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.
Noetherian E-∞ ring?
Last revised on August 6, 2022 at 01:29:25. See the history of this page for a list of all contributions to it.