Questions tagged [commutative-algebra]
Questions about commutative rings, their ideals, and their modules.
2,515
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Prove that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw) \subset \mathbb{A}^4$, is not a unique factorization domain
I want to show that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw)\subset\mathbb{A}^4$, is not a unique factorization domain.
Morally, all we need to do is find some nonzero element that can ...
5
votes
1
answer
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Why $\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$ implies that $\gcd(f(t)-a,g(t)-b)=t-c$, for some $a,b,c \in \mathbb{C}$?
Assume that $f(t),g(t) \in \mathbb{C}[t]$ satisfy the following two conditions:
(1) $\deg(f) \geq 2$ and $\deg(g) \geq 2$.
(2) $\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$.
In this question it was ...
3
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1
answer
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Problem on idempotent finitely generated ideal
I have a question.
Could you please help me to solve this?
Thanks in advance
Let $\mathfrak a$ be a finitely generated ideal of $A$, commutative ring with identity, such that $\mathfrak a^2 = \...
71
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answers
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Using Gröbner bases for solving polynomial equations
In my attempts to understand just how computer algebra systems "do things", I tried to dig around a bit on Gröbner bases, which are described almost everywhere as "a generalization of the Euclidean ...
47
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5
answers
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Example of modules that are projective but not free; torsion-free but not free
Free modules are projective, and projective modules are direct summands of free modules.
Are there examples of projective modules that are not free?
(I know this is not possible for modules of ...
42
votes
6
answers
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Tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$
I want to understand the tensor product $\mathbb C$-algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$. Of course it must be isomorphic to $\mathbb{C}\times\mathbb{C}.$ How can one construct an ...
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1
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Does localisation commute with Hom for finitely-generated modules?
Question. Let $R$ be a ring, $\mathfrak{p}$ a prime, $M$ a finitely-generated $R$-module, and $N$ any $R$-module. Is the natural map
$$\textrm{Hom}_R(M, N)_\mathfrak{p} \to \textrm{Hom}_{R_\mathfrak{p}...
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6
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A non-noetherian ring with noetherian spectrum
Question 1: Does such a ring exist?
Note: The definition of a noetherian topological space is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite number of ...
27
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3
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Inverse limit of modules and tensor product
Let $(M_n)_n$ be an inverse system of finitely generated modules over a commutative ring $A$ and $I\subset A$ an ideal.
When is the canonical homomorphism
$$\left(\varprojlim\nolimits_n M_n\right)\...
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Compactness of $\operatorname{Spec}(A)$
In an exercise in Atiyah-Macdonald it asks to prove that the prime spectrum $\operatorname{Spec}(A)$ of a commutative ring $A$ as a topological space $X$ (with the Zariski Topology) is compact.
Now ...
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answer
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Finitely generated modules over a Noetherian ring are Noetherian
I'm trying to prove that if the ring $R$ is Noetherian then every finitely generated $R$-module is Noetherian.
First of all, it is known that every module is a homomorphic image of a free module, so ...
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2
answers
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Projective module over a PID is free? [duplicate]
A common result is that finitely generated modules over a PID $R$ are projective iff they are free.
Is the same true that an arbitrary projective module over a PID is free? I can't find this fact ...
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3
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Is Orzech's generalization of the surjective-endomorphism-is-injective theorem correct?
In math.stackexchange answer #239445, Makoto Kato quoted a statement from the paper
Morris Orzech, Onto Endomorphisms are Isomorphisms, Amer. Math. Monthly 78 (1971), 357--362.
The statement (...
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2
answers
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Is any prime element irreducible?
I have seen many proofs about a prime element is irreducible, but up to now I am thinking whether this result is true for any ring.
Recently, I got this proof:
Suppose that $a$ is prime, and that $...
17
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2
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Hom and tensor with a flat module
Let $A$ be a commutative noetherian ring. Let $M, N$ be $A$-modules, and assume that $M$ is finite over $A$. Let $P$ be a flat $A$-module.
Is it true that there is an isomorphism
$\operatorname{Hom}...
14
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3
answers
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The radical of a monomial ideal is also monomial
I have problems with this:
I need to prove that in the polynomial ring the radical of an ideal generated by monomials is also generated by monomials.
I found a proof on internet that uses the ...
11
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2
answers
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Integral closure $\tilde{A}$ is flat over $A$, then $A$ is integrally closed
Question. Let $A$ be an integral domain and $\tilde{A}$ be its integral closure in the field of fractions $K$. Assume that $\tilde{A}$ is a finitely generated $A$-module. I want to prove that if $\...
11
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2
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In a finite ring extension there are only finitely many prime ideals lying over a given prime ideal [duplicate]
I'm trying to solve the exercise 6.7 of Miles Reid's Undergraduate Commutative Algebra (pag 93).
How can I prove that if $B$ is a finite ring extension of $A$, there are only finitely many prime ...
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4
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Explicit examples of infinitely many irreducible polynomials in k[x]
My question is the following.
Is it possible to give examples of infinitely many irreducible polynomials in a polynomial ring $k[x]$ with $k$ a field?
I'm interested in this because I'm ...
9
votes
2
answers
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If $p\in R[X_1,\dots,X_n]$ is irreducible, is it still irreducible in $R[X_1,\dots,X_n,\dots,X_N]$?
It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.
The standard approach essentially goes something along the ...
9
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1
answer
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Equivalent characterizations of discrete valuation rings
Let $R$ be a commutative ring with identiy, then the following are equivalent:
$R$ is a DVR
$R$ is a local Euclidean domain that is not a field.
$R$ is a local PID that is not a field.
$R$ is a local ...
8
votes
1
answer
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When $\operatorname{Hom}_{R}(M,N)$ is finitely generated as $\mathbb Z$-module or $R$-module?
Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is ...
7
votes
2
answers
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Converse to Chinese Remainder Theorem
So as seen on this question Converse of the Chinese Remainder Theorem, we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, ...
5
votes
1
answer
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Why is $\mathbb{C}[x,y]/(y^2 - x^3 + 1)$ normal?
A problem on an algebra qual reads
Show that the ring $R = \mathbb{C}[x,y]/(y^2 - x^3 +1)$ is a Dedekind
domain. (Hint: compare $R$ with the subring $\mathbb{C}[x]$.)
$R$ is clearly Noetherian. ...
5
votes
1
answer
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Primary ideals in Noetherian rings
For an $R$-module $M$ I have the following definition for a submodule $N\subset M$ to be $\mathfrak{p}$-primary: this is the case when $\text{Ass}(M/N) = \{\mathfrak{p}\}$, that is, $M/N$ is coprimary ...
4
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1
answer
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Normality of localizations in polynomial rings?
Normality of a ring here refers to being equal to it's integral closure in it's field of fractions.
The problem is:
Let $A=\mathbb{C}[x,y]/(y^2-x^3-x^2)$. Show that $A_m$ is normal for every ...
2
votes
1
answer
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$\mathfrak{a}$-adic completion of an $A$-module is a topological $\hat{A}$-module with which topology? Why does completion induce a continuous map?
Let $A$ be a commutative ring with identity and $\mathfrak{a}$ and ideal of $A$. Then $A$ has a topological structure which is defined by the following chain of ideals
\begin{equation*}
A \supseteq \...
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votes
6
answers
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Reference request: introduction to commutative algebra
My goal is to pick up some commutative algebra, ultimately in order to be able to understand algebraic geometry texts like Hartshorne's. Three popular texts are Atiyah-Macdonald, Matsumura (...
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3
answers
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Commutative property of ring addition
I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring:
Let $R\neq \...
28
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2
answers
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In a Dedekind domain every ideal is either principal or generated by two elements.
Prove that in a Dedekind domain every ideal is either principal or generated by two elements.
Help me some hints.
Thanks a lot!
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3
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Why is the ring of holomorphic functions not a UFD?
Am I correct or not? I think that a ring of holomorphic functions in one variable is not a UFD, because there are holomorphic functions with an infinite number of $0$'s, and hence it will have an ...
27
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2
answers
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When is a tensor product of two commutative rings noetherian?
In particular, I'm told if $k$ is commutative (ring), $R$ and $S$ are commutative $k$-algebras such that $R$ is noetherian, and $S$ is a finitely generated $k$-algebra, then the tensor product $R\...
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6
answers
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Irreducibility of Polynomials in $k[x,y]$
I'm working through some Hartshorne problems and have noticed that in order to do certain problems properly one must prove a given polynomial $f\in k[x,y]$ is irreducible.
For example, in problem I....
22
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3
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What are the integers $n$ such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed?
I was recently reading about integral ring extensions. One of the first examples given is that $\mathbb{Z}$ is integrally closed in its quotient field $\mathbb{Q}$. Another is that $\mathbb{Z}[\sqrt{5}...
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2
answers
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Spectrum of $\mathbb{Z}^\mathbb{N}$
Is anything known about the spectrum of $\mathbb{Z}^{\mathbb{N}}$? Notice that the fiber of $\mathrm{Spec}(\mathbb{Z}^{\mathbb{N}}) \to \mathrm{Spec}(\mathbb{Z})$ at a non-zero prime ideal $(p)$ is ...
17
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3
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Finitely generated ideals in a Boolean ring are principal, why?
The classical book on commutative algebra Introduction to Commutative Algebra, by Atiyah and Macdonald, has the following as exercise I.11.
A ring is Boolean if $x^2=x$ for any $x$ of $A$. In a ...
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3
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Show $\mathbb{Z}[\sqrt{6}]$ is a Euclidean domain
I'm attempting to modify the proof the $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain to prove a similar result for $\mathbb{Z}[\sqrt{6}]$. The idea is to prove that $\mathbb{Q}[\sqrt{6}]$ is Euclidean ...
16
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5
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Integral domain that is not a factorization domain
I am looking for rings that are integral domains but not factorization domains, that is, rings in which it is not possible to express a nonzero nonunit element as a product of irreducible elements.
...
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Injectivity of Homomorphism in Localization
Let $\alpha:A\to B$ be a ring homomorphism, $Q\subset B$ a prime ideal, $P=\alpha^{-1}(Q)\subset A$ a prime ideal. Consider the natural map $\alpha_Q:A_P\to B_Q$ defined by $\alpha_Q(a/b)=\alpha(a)/\...
14
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2
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Primary ideals confusion with definition
In a commutative ring, can I say
An ideal $\mathfrak q$ in a ring $A$ is primary if $\mathfrak q \neq A $ and if $ xy \in \mathfrak q \Rightarrow $ either $ x \in \mathfrak q$ or $y^n \in \mathfrak q ...
13
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1
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Finding a space $X$ such that $\dim C(X)=n$.
Let $n\in \mathbb{N}$ . Is there some topological space $X$ such that $C(X)$ is a finite dimensional ring with $\dim C(X) = n$?
Here, $C(X):=\{ f:X \to \mathbb{R} \mid f$ is continuous$\}$ and $\dim ...
12
votes
3
answers
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Is every Artinian module over an Artinian ring finitely generated?
I know that if $R$ is Artinian, then a f.g. $R$-module is Artinian. Is f.g. a necessary condition?
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Ideal of the twisted cubic
The twisted cubic is the image of the morphism $\phi : \mathbb{P}^1 \to \mathbb{P}^3 , (x:y) \mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(...
11
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2
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Associated Prime Ideals in a Noetherian Ring; Exercise 6.4 in Matsumura
Let $I$ and $J$ be ideals of a Noetherian ring $A$. If $JA_P\subseteq IA_P$ for every $P\in \operatorname{Ass}_A(A/I)$, then $J\subseteq I$.
I'm reading Matsumura's Commutative Ring Theory book on my ...
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1
answer
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Does this "extension property" for polynomial rings satisfy a universal property?
On page 151 of Paolo Aluffi's Algebra: Chapter 0, an important property of the polynomial ring $\mathbb{Z}[x_1, \cdots, x_n]$ is introduced, namely that it's initial in the category of set functions ...
10
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1
answer
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Is $R/N(R)$ a faithfully flat $R$-module?
I'm studying recently faithfully flat modules and I'd like to know the following:
Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of ...
9
votes
3
answers
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The vanishing ideal $I_{K[x,y]}(A\!\times\!B)$ is generated by $I_{K[x]}(A) \cup I_{K[y]}(B)$?
Let $K$ be a field, $x=(x_1,\ldots,x_m)$, $y=(y_1,\ldots,y_n)$, $A\!\subseteq\!\mathbb{A}^m_K$, $B\!\subseteq\!\mathbb{A}^n_K$. Does there hold $$I_{K[x,y]}(A\!\times\!B)=\langle\langle I_{K[x]}(A) \...
9
votes
2
answers
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Finite number of elements generating the unit ideal of a commutative ring
Let $A$ be a commutative ring with $1$.
Let $f_1,\dots,f_r$ be elements of $A$.
Suppose $A = (f_1,\dots,f_r)$.
Let $n > 1$ be an integer.
Can we prove that $A = (f_1^n,\dots,f_r^n)$ without using ...
8
votes
1
answer
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Existence of valuation rings in an algebraic function field of one variable
The following theorem is a slightly modified version of Theorem 1, p.6 of Chevalley's Introduction to the theory of algebraic functions of one variable.
He proved it using Zorn's lemma.
However, Weil ...
8
votes
1
answer
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"Instructive" proof of "If I is maximal among ideals not ..., then I is prime"
In this question all rings are commutative with identity.
Consider the following well-known statement:
(*) Let $R$ be a ring and $S$ a multiplicatively closed subset of $R$. Suppose $I$ is an ...