Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [commutative-algebra]

Questions about commutative rings, their ideals, and their modules.

Commutative algebra is the area of mathematics that deals with commutative rings and their ideals, as well as modules over commutative rings.

Many results and tools of commutative algebra are cornerstones of algebraic geometry. Important tools of commutative algebra include localization and completion of rings and modules.

16857 questions
1
vote
0 answers

refining a presentation of a quotient ring

Suppose that we have a commutative ring $R$ which i) is local ii) is the quotient of a regular ring and iii) it is a $k$-algebra, where $k$ is a field. I am trying to prove that in that case we can choose a presentation $R = S/I$, where $S$ is a…
Manos
  • 25,833
1
vote
1 answer

existence of a finite free resolution

Problem 2.1.26 in Bruns and Herzog, CMR, reads as follows: "Let $R$ be a Cohen-Macaulay local ring of dimension $d$ and $M$ a finite $R$-module. Deduce that the $d$-th syzygy of $M$ in an arbitrary finite free resolution is either $0$ or a maximal…
Manos
  • 25,833
1
vote
1 answer

localizations of a direct sum module

Consider the $\mathbb{Z}$-module $M=\bigoplus{\mathbb{Z}/p\mathbb{Z}}$, where the direct sum is taken over the set of all prime numbers. How do I show that the localizations $M_\mathfrak{p}$ are finitely generated $\mathbb{Z}_\mathfrak{p}$-modules…
Dave
  • 615
1
vote
4 answers

Localization with maximal ideal

Let R be local ring with maximal ideal P. Show that every element of R\P is invertible. Now let e be an element of R satisfying $e^2=e$. How can we prove $e \in {0,1}$.
S786
  • 1,001
1
vote
1 answer

Tensor of quotients

I would like to prove the following: let $A$ be a ring and $I,J\subset A$ two ideals. Then: $$A/I\otimes_AA/J\cong A/(I+J)$$ I have seen a proof using the Yoneda lemma (even though I haven't understood it fully yet), but I've been told there's also…
Daniel Robert-Nicoud
  • 29,795
1
vote
2 answers

isomorphic ideals and projective dimensions of quotients

Let $R$ be a Noetherian ring and $I,J$ proper ideals that are isomorphic as $R$-modules. Can we conclude that the projective dimensions of $R/I,R/J$ are equal?
Manos
  • 25,833
1
vote
1 answer

Maximal regular sequences of different length

This question is Exercise 1.2.20 in the book: Winfried Bruns, H. Jürgen Herzog, Cohen-Macaulay Rings, Cambridge University Press, 1998. Let $k$ be a field and $R=k[[X]][Y]$. Deduce that $X, Y$ and $1-XY$ are maximal $R$-regular sequences. I can…
Strongart
  • 4,767
1
vote
1 answer

regarding a certain homomorphism of finite free modules

Consider the following situation: let $(R,m)$ be a local Noetherian ring and $f: F_1 \rightarrow F_0$ a morphism of finite free $R$-modules with $\operatorname{rank}(F_i) = r_i, \, i=0,1$. Suppose that the image of $f$ contains a free direct summand…
Manos
  • 25,833
1
vote
1 answer

Bruns and Herzog Proposition 1.4.11

Proposition 1.4.11 in Bruns and Herzog Cohen-Macaulay Rings reads: Proposition: Let $R$ be a Noetherian ring and $\phi: F \rightarrow G$ a homomorphism of finite free $R$-modules. Then $rank (\phi) = r$ if and only if $grade(I_r(\phi)) \ge 1$ and…
Manos
  • 25,833
1
vote
1 answer

Localization of two rings which is an integral extension, then integral extension still holds?

Question seems simple, but I just can't find the solution. Let $A/B$ be an integral ring extension and let $P$ be a prime ideal of $B$. By lying-over theorem, there is $Q$, a prime ideal of $A$, lying over $P$. Then the ring of fractions of $A$…
user83008
  • 11
1
vote
1 answer

$\mathbb Z_p[T]/(T^a,p^b)\cong\mathbb Z_p[[T]]/(T^a,p^b)$

Let $a,b\in \mathbb N,$ then $$\mathbb Z_p[T]/(T^a,p^b)\cong\mathbb Z_p[[T]]/(T^a,p^b)$$ 1.What is this isomorphism ? 2.How to prove that $|\mathbb Z_p[[T]]/(T,p)^t|=p^{t(t+1)/2}$ Now Let $X$ be a $\mathbb Z_p[[T]]-$module. 3.we can view $X$ as…
Med
  • 1,598
1
vote
0 answers

Use the degree of Hilbert polynomial to define the dimension of a ring

Let $R$ be a local ring , $\mathfrak{m}$ the maximal ideal, $q$ is $\mathfrak m$-primary . Then we can prove that there exists a polynomial $F_q(t)\in \mathbb{Q}(t)$ such that $F_q(n)=\mathcal{l}(R/\mathfrak q^n)$ for $n>>0$ . Let $d_q(R)$ be the…
molan
  • 699
1
vote
2 answers

For fractional ideal, why $AB=R$ implies $B=A^{-1}$?

Let $A,B$ be two fractional ideals of $R$ (an integral domain). Could anyone tell me why $AB=R$ implies $B=A^{-1}$?
hxhxhx88
  • 5,257
1
vote
1 answer

definition of the grade of an ideal

Definition: Let $A$ be a Noetherian ring and $M$ a non-zero finite $A$-module. Then the grade of $M$ is defined as $grade(M) = \inf_i \left\{Ext^i(M,A) \neq 0\right\}$. Matsumura (CRT) p.132, says that if $I$ is an ideal of $A$, then by grade of…
Manos
  • 25,833
1
vote
0 answers

Lying over "from above"

Let $B/A$ be a integral extension of (commutative unital) rings. The "Lying over" theorem states that for any prime ideal $P$ in $A$ there is a prime ideal $Q$ in $B$ such that $Q\cap A=P$. The usual proof runs as follows (forgive me for the excess…
Matemáticos Chibchas
  • 6,061
1 2 3
…
99 100