Dependencies

Let $E$ be an additive subgroup of $R$ which is multiplicatively closed. Let $I_1,\dots ,I_n$ be ideals such that $I_3,\dots ,I_n$ are prime. Then if $E$ is not contained in any one of the $I_j$, then $E$ is not contained in their union.

Let $R$ be a ring and ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ be prime ideals. Let $x$ be some element of $R$ and let $J$ be an ideal. Define $I:=(x)+J$. If $I\not\subset {\mathfrak{p}}_j$ for all $j$, then there exists some $y\in J$ such that $x+y\notin {\mathfrak{p}}_j$ for all $j$

Let ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n$ be prime ideals. Let $I$ be an ideal contained in in their union. Then $I\subset {\mathfrak{p}}_i$ for some $i$.

Let $I_1,\dots ,I_n$ be ideals, let $\mathfrak{p}$ be a prime ideal containing the intersection of them all. Then $I_i\subset \mathfrak{p}$ for some $i$. If $\mathfrak{p}$ is exactly the intersection, then $I_i=\mathfrak{p}$ for some $i$.

Let $(R,\mathfrak{m},k)$ be a local ring. Let $M$ be an $R$-module. If the elements $x_1,\dots ,x_n$ are elements in $M$ that form a basis in the projection $M/\mathfrak{m}M$, then $x_1,ldots,x_n$ generate $M$.

Let $R$ be a ring, and ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ be prime ideals. Let $I$ be a finitely generated ideal of $R$. Then if $I\not\subset {\mathfrak{p}}_j$ for all $j\in \{1,\dots ,m\}$, then there exists an $x\in I$ such that $x\notin {\mathfrak{p}}_j$ for all $j$.

Let $x\in R$ be an element which is not in any minimal prime ideal. Then $\dim R/(x)\le \dim R-1$

The annihilator of a module M is the ideal of elements $r\in R$ such that $rm=0$ for all $m\in M$.

The set ${\mathrm{Assoc}}_R(M)$ is the set of primes satisfying the preivious proposition

An ideal $I$ is decomposable there exists a primary decomposition with intersection $I$.

The depth of a module $M$ is . . .

A free resolution of $M$ over $R$ is a projective resolution whose components $P^i$ are free.

The global dimension of a ring $R$ is the supremum over all $R$-modules of $\text{proj dim}M$.

The height of an ideal $I\subset R$ is the infimum of the heights of prime ideals containing $I$.

Let $R$ be a ring, and $\mathfrak{p}$ be a prime ideal in $R$. Then the height of $\mathfrak{p}$ is the sup of the lengths of chains

The Krull dimension of the $R$ is the supremum of heights of prime ideals $\mathfrak{p}\subset R$.

The length of a projective resolution $P_{\cdot }$ is the highest $i$ such that $P^i$ is nonzero. If there exists no such $i$, the length is infinity

Let $(R,\mathfrak{m},k)$ be a local (noetherian?) ring. Let $\nu (R)$ be the minimum of the lengths of generating sets of $\mathfrak{m}$-primary ideals.

A minimal primary decomposition is a primary decomposition such that

1. The ideals $\sqrt{{\mathfrak{q}}_i}$ are distinct

2. $$\bigcap _{i\ne j}^{}{\mathfrak{q}}_j\not\subset {\mathfrak{q}}_i$$

   for all $1\le i\le n$.

Given a prime ideal $\mathfrak{p}$ of the ring $R$, an ideal $I$ is $\mathfrak{p}$-primary if for all $x,y\in I$, either $x\in I$ or $y^n\in I$ for some $n$. (This is already in mathlib)

A primary decomposition of an ideal $I\subset R$ is an expression of $I$ as an intersection of primary ideals, that is a collection of ideals ${\mathfrak{q}}_i$, $1\le i\le n$, such that

An ideal $I$ is $\mathfrak{p}$-primary if it’s radical is equal to $\mathfrak{p}$.

Let $M$ be an $R$-module. Then $\text{proj dim}M$ is the minimum length of a projective resolution of $M$. It lives in the set $\mathbb{N}\cap \mathrm{\infty }$.

Let $R$ be a ring. Let $M$ be an $R$-module. A projective resolution of $M$ over $R$ Is an exact sequence

. That is to say, a projective resolution is a quasi-isomorphism of complexes between the inclusion of $M$ into chain complexes over $R$ (i.e. the complex which has $M$ in degree zero, trivial everywhere else, with trivial maps) and a bounded—below complex whose components are projective $R$-modules.

Let $M$ be an $R$-module. An element $x\in R$ is $M$-regular if it does not annihilate any element in $M$. In colon notation, it is non a member of the ideal $(0:M)$. If $x$ is an $R$-regular element, we simply say it is a regular element.

A local ring $R$ is regular if $\mathfrak{m}$ is generated by a system of parameters.

A ring $R$ is regular if $R_{\mathfrak{p}}$ is regular for every $\mathfrak{p}\in R$.

Let $M$ be an $R$-module. An $M$-regular sequence is a weak $M$-regular sequence such that $M/(x_1,\dots ,x_n)M\ne 0$. If $M=R$, we simply call this a regular sequence.

The support of a module is the set of prime ideals $\mathfrak{p}\subset R$ such that $M_{\mathfrak{p}}\ne 0$.

Let $M$ be an $R$-module. Then a weak $M$-regular sequence is a sequence $x_1,\dots ,x_n$ with $x_i\in R$ for all $i$, such that $x_i$ is a $M/(x_1,\dots ,x_{i-1})M$-regular element for all $i$. If $M=R$, we say that $x_1,\dots ,x_n$ is a weak regular sequence.

The zero locus of the annihilator of a module is the same as the support of that module. I.e. the set of prime ideals containing the ${\mathrm{Ann}}_R(M)$ is the support of $M$.

The assocated primes of $M$ are precisely the primes which are minimal in the support of $M$.

An element $x\in R$ is a zero-divisor iff it is in an associated prime. (geometrically, it vanishes at an associated point)

Let $R$ a ring and $\mathfrak{p}$ a prime ideal, then

A finitely generated projective module over a regular local ring is free

If there exists a projective resolution of $M$ with finite length, then $\text{proj dim}M<\mathrm{\infty }$.

Any decomposable ideal has a minimal primary decomposition

Let $I_1,\dots ,I_n$ be $\mathfrak{p}$-primary ideals. Then their intersection is $\mathfrak{p}$-primary.

Localization by a multiplicative set is an exact functor.

Let $M$ be an $R$-module. Let $S$ be a multiplicative set. Let $\dots \to A_2\to A_1\to A_0\to M\to 0$ be a free resoltion of $M$. Then $\dots S^{-1}(A_2)\to S^{-1}(A_1)\to S^{-1}(A_0)\to S^{-1}M\to 0$ is a free resolution of $S^{-1}M$.

Any resolution of $M$ over $R$ has length at least that of the minimal resolution.

BH 1.3.4

State Nakayama’s Lemma here.

(See BH 1.2.2) Let $R$ be a ring, and ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ be prime ideals. Let $M$ be an $R$-module, and let $x_1,\dots ,x_n\in M$. Let $N$ be the submodule generated by $x_1,\dots ,x_n$.

If $N_{{\mathfrak{p}}_j}\not\subset {\mathfrak{p}}_j{M}_{{\mathfrak{p}}_j}$ for all $j\in \{1,\dots ,m\}$, then there exist $a_2,\dots ,a_n\in R$ such that $x_1+\sum _{i=2}^n{a}_i{x}_i\notin {\mathfrak{p}}_j{M}_{{\mathfrak{p}}_j}$ for all $j\in \{1,\dots ,m\}$.

BH 1.3.5

If $\mathfrak{q}$ is a primary ideal, then $\sqrt{q}$ is a prime ideal.

Let $R$ be a regular local ring. Then $R$ has finite global dimension. That is, any finitely generated module $R$ has finite projective dimension.

Every regular ring is an integral domain.

$R$ is regular if and only if the minimal number of generators of its maximal ideal is equal to the (Krull) dimension of $R$

Let $f:M\to M$ be a surjection of modules (over a local ring?). Then $f$ is an isomorphism.

Let $R$ a ring, and $M$ an $R$-module. Let $\mathfrak{p}$ be a prime ideal, then the following are equivalent:

1. $\mathfrak{p}={\mathrm{Ann}}_R(m)$ for some element $m\in M$.

2. $R/\mathfrak{p}$ embeds into $M$.

The associated primes of $M$ are in the support of $M$.

There exists a free resolution $F_{\cdot }$ of $M$ such that length of $F_{\cdot }$ is equal to the projective dimension of $M$.

The following are equivalent Firstly, $R$ is regular. Secondly, the zariski cotangent space is a vector space of dimension $\dim R$.

A local ring $R$ is regular if and only if its maximal ideal is generated by a regular sequence.

Let $R$ be a regular local ring. Then $R/I$ is regular local if and only if $I$ is generated by a (regular) system of parameters (I.e. a generating set for $\mathfrak{m}$).

depth + projdim = depth

Let $(R,\mathfrak{m},k$ be a noetherian local ring. The following are equivalent:

1. $R$ is regular.

2. $R$ has finite global dimension.

3. $\text{proj dim}k<\mathrm{\infty }$

Let $(R,\mathfrak{m},k)$ be a local noetherian ring. Let $n\in \mathbb{N}$. Then the following are equivalent

1. $\dim R=n$

2. $\mathrm{height}\mathfrak{m}=n$

3. $\nu (R)=n$.

Let $(R,\mathfrak{m},k)$ be a local noetherian ring. Let $I$ be a nonzero ideal with finite projective dimension. If $I/I^2$ is a free $R$-module, then $I$ is generated by a regular sequence.

Let $R$ a noetherian ring, and $I$ a proper ideal of height $n$. Then for all $1\le i\le n$, there exist elements $x_1,\dots ,x_i$ such that $\mathrm{height}(x_1,\dots ,x_i)=i$.

Let $I=(x_1,\dots ,x_n)$ be an ideal. Then each minimal prime over $I$ has height at most $n$.

Let $R$ be a noetherian ring, and $I$ a prinicpal ideal. Then each minimal prime ideal over $I$ (i.e. minimal among primes containing I) has height at most 1.

Let $R$ be a ring, and $M$ an $R$-module. Let $x$ be an $M$-regular element. Then $\dim M/xM\le \dim M-1$.

Any regular sequence is part of a system of parameters

Let $R$ be a regular local ring, and let $\mathfrak{p}$ be a prime ideal in $R$. Then $R_{\mathfrak{p}}$ is a regular local ring.