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.

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$p$-torsion of quotient rings

Let $A$ be an $p$-adically complete (isomorphic to the inverse limit system) integral domain and $I$ be a prime ideal such that both $A$ and $A/I$ has no $p$-torsion. Is it possible that $A/I^2$ (or higher power $A/I^n$) has $p$-torsion? Basically,…
CO2
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Counterexample to going down Theorem

Let $k$ be a field. Let $R=k[X,Y],\ A=k[X,Y,\frac{X}{Y}]$. Show that going down doesn't hold for $A/R$. My approach: I took the map $\phi: \operatorname{Spec} A\to \operatorname{Spec} R,\ q\mapsto q\cap R$ to verify the preimages of $q$ in $R$.…
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Is every local domain dominated by some Noetherian local domain?

I have known that every Noetherian local domain can be dominated by a DVR in its own fraction field. And there exist some fields without any discrete valuation. So I wonder if every local domain can be dominated by a DVR in a larger field? I've…
pop1
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Confusion with notion $\operatorname{Spec}((k(p))\otimes_A B)$

I read this notion from Atiyah-Macdonald exercise 3.21: $f:A\to B$ is a ring homomorphism. $f^*:\operatorname{spec}(B)\to \operatorname{spec}(A)$ is a mapping induced by $f$ and defined by $f^*(q):=f^{-1}(q)$ for $q \in \operatorname{Spec}(B)$. If…
Y.Wayne
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How do I compute the length of this module?

Let $k$ be an algebraically closed field. Let $A = k[x,y,z,w]$, $\mathfrak p = (x, w)$, $\mathfrak a = (x^d, w)$, for some $d \in \mathbb N$. How do I compute $(A/\mathfrak a)_\mathfrak p$? How do I compute the length of $(A/\mathfrak a)_\mathfrak…
isekaijin
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Show that $R[x]/(x^2-c)$ is a Dedekind domain

I am trying to solve the following question. Let $R$ be a principal ideal domain (PID). Suppose that $2$ is a unit in $R$. Let $c_1, ..., c_t$ be irreducible elements of $R$ and let $c=c_1 ... c_t$. Show that the ring $R[x]/(x^2-c)$ is a Dedekind…
Orlly
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Proposition 9.2 in Atiyah's commutative algebra book

Proposition 9.2. Let $A$ be a Noetherian local domain of dimension one, $m$ its maximal ideal,$k=A/m$ its residue field. Then the following are equivalent : ii) $A$ is integrally closed; iii) $m$ is a principal ideal; Proof. Let $a\in m$ and $a\neq…
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a problem discrete valuation ring in Atiyah's commutative algebra

An integral domain $A$ is a discrete valuation ring if there is a discrete valuation $v$ of its field of fractions $K$ such that $A$ is the valuation of $v$. By (5.18), $A$ is a local ring, and its maximal ideal of $m$ is the set of all $x \in K$…
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integral closure in its fraction field

Let $R$ be a Noetherian domain with fraction field $K$. Let $\overline{R}$ be the integral closure of $R$ in $K$. Is it possible to prove that $\overline{R}$ is a finitely generated $R$-module? The condition of the original problem is that $R$ is a…
MatrixBi
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Question on a proof of dimension property

In this note proposition 1.3 claimed that : If $k$ is a field, $R$ is a finitely generated $k$-algebra which is a domain then $\dim R$ is finite and any saturated chain of prime ideals has length equal to $\dim R$. I understand mostly the proof,…
Arsenaler
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On a theorem of dimension of affine algebra

In the book of commutative algebra of G.Kemper (can be found here), the author has proved the following theorem : Let $A$ be an affine algebra and let $P_{0}\subsetneq P_{1}\subsetneq \cdots\subsetneq P_{n}$ be a maximal chain of prime ideals $P_i…
Arsenaler
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What is wrong with my computation of $H^2(G,Z/4Z)$

Let $G=Z_2= \{1,2\}$, cyclic of order 2. Let $A = Z/4Z$ and suppose $G$ acts on $A$ by inversion. Show $|H^2(G,A)|=2$ by direct computation. So consider $0\rightarrow C^0(G,A)\rightarrow C^1(G,A)\xrightarrow{d_1} C^2(G,A)\xrightarrow{d_2} C^3(G,A)$.…
Jun Xu
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Exterior Algebra's Property

Suppose for the $R$ module $M$, we consider its exterior algebra $\bigwedge(M)$, which is the $$\bigwedge(M)=\frac{T(M)}{A(M)}$$ where the two sided ideal $A(M)$ is generated by all the elements of the form $m \otimes m $ for $m \in M$. Further…
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If $I\pi ^{-k}\subseteq D$ for all $k\leq m$ but $I\pi ^{-m-1}\nsubseteq D$, then $I\pi ^{-m}\nsubseteq (\pi )$

I'm working through a proof and I've got stuck on one detail, it seems like it is supposed to be totally obvious, but need help to figure out why. Let $D$ be a Noetherian integrally closed domain with unique prime ideal $(\pi)$. For an ideal $I$…
harajm
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finite length is stronger than finiteness of a module

Let $A$ be a commutative ring and $M$ an $A$-module. I realized recently that the property of $M$ having finite length is stronger than $M$ being finitely generated. Here is my reasoning: Suppose $M$ has finite length but it is not…
Manos
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