Questions tagged [commutative-algebra]

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

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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 ...
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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 ...
user237522's user avatar
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3 votes
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 = \...
M.Subramani's user avatar
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71 votes
8 answers
10k views

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 ...
J. M. ain't a mathematician's user avatar
47 votes
5 answers
23k views

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 ...
ShinyaSakai's user avatar
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42 votes
6 answers
11k views

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 ...
Mikhail Borovoi's user avatar
36 votes
1 answer
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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}...
Zhen Lin's user avatar
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33 votes
6 answers
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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 ...
fosco's user avatar
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27 votes
3 answers
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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)\...
Cyril's user avatar
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26 votes
2 answers
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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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25 votes
1 answer
15k views

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 ...
AlexCon's user avatar
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20 votes
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 ...
Hana Bailey's user avatar
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20 votes
3 answers
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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 (...
darij grinberg's user avatar
18 votes
2 answers
12k views

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 $...
Hassan Muhammad's user avatar
17 votes
2 answers
4k views

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}...
the L's user avatar
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14 votes
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 ...
Arkj's user avatar
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11 votes
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 $\...
Karatuğ Ozan Bircan's user avatar
11 votes
2 answers
3k views

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 ...
Corra's user avatar
  • 225
10 votes
4 answers
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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 ...
Adrián Barquero's user avatar
9 votes
2 answers
696 views

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 ...
Adelaide Dokras's user avatar
9 votes
1 answer
3k views

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 ...
Lukas Heger's user avatar
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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 ...
user avatar
7 votes
2 answers
2k views

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 $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, ...
MadMonty's user avatar
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5 votes
1 answer
851 views

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. ...
Eric Auld's user avatar
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5 votes
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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 ...
Maanroof's user avatar
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4 votes
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 ...
Koto's user avatar
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2 votes
1 answer
655 views

$\mathfrak{a}$-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 \...
Peter Hu's user avatar
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77 votes
6 answers
21k views

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 (...
30 votes
3 answers
5k views

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 \...
Jeyekomon's user avatar
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28 votes
2 answers
9k views

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!
Truong's user avatar
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28 votes
3 answers
4k views

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 ...
Myshkin's user avatar
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27 votes
2 answers
5k views

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\...
Heidi's user avatar
  • 933
23 votes
6 answers
5k views

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....
Brian Fitzpatrick's user avatar
22 votes
3 answers
7k views

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}...
yunone's user avatar
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18 votes
2 answers
2k views

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 ...
Martin Brandenburg's user avatar
17 votes
3 answers
7k views

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 ...
awllower's user avatar
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16 votes
3 answers
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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 ...
chris's user avatar
  • 2,659
16 votes
5 answers
9k views

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. ...
user avatar
15 votes
3 answers
3k views

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)/\...
Edward Hughes's user avatar
14 votes
2 answers
3k views

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 ...
Ram's user avatar
  • 1,658
13 votes
1 answer
1k views

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 ...
number's user avatar
  • 333
12 votes
3 answers
4k views

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?
Aliakbar's user avatar
  • 3,157
11 votes
3 answers
6k views

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(...
Martin Brandenburg's user avatar
11 votes
2 answers
990 views

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 ...
Vladimir's user avatar
  • 2,879
10 votes
1 answer
2k views

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 ...
Alf's user avatar
  • 2,597
10 votes
1 answer
1k views

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 ...
user93721's user avatar
  • 101
9 votes
3 answers
2k views

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) \...
Leo's user avatar
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9 votes
2 answers
745 views

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 ...
Makoto Kato's user avatar
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8 votes
1 answer
2k views

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
839 views

"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 ...
Tom Bachmann's user avatar
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