representation, ∞-representation?
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.
(In this section, “ring” means rng, where the presence of a multiplicative identity is not assumed unless we say “unital ring”.)
A (left) 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 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 locally noetherian schemes.
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.
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 $\mathfrak{i}$.
Noetherian E-∞ ring?
Last revised on September 29, 2017 at 07:12:53. See the history of this page for a list of all contributions to it.