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

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Comaximal ideals in a commutative ring

Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r ...
31
votes
1answer
2k views

Classification of prime ideals of $\mathbb{Z}[X]$

Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable. My question: Is every prime ideal of $\mathbb{Z}[X]$ one of following types? If yes, how would you prove this? (1) $(0)$ (2) ...
13
votes
4answers
2k views

Surjective endomorphisms of finitely generated modules are isomorphisms

My Problem: Let $M$ be a finitely generated $A$-module and $T$ an endomorphism. I want to show that if $T$ is surjective then it is invertible. My attempt: Let $m_1,...,m_n$ be the generators of ...
10
votes
4answers
6k views

A ring is a field iff the only ideals are $(0)$ and $(1)$

Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$. I can't figure out where to start other that I ...
29
votes
4answers
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Why does a minimal prime ideal consist of zerodivisors?

Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors. This can be proved using localization, when $A$ is noetherian: ...
15
votes
2answers
1k views

A subring of the field of fractions of a PID is a PID as well.

Let $A$ be a PID and $R$ a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the field of fractions of $A$. How to show $R$ is also a PID? Any ...
12
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2answers
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$A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$

I'm trying to prove that if $A\neq 0$ is a commutative ring and there is an injective $A$-module homomorphism $A^m\hookrightarrow A^n$ then $m\leq n$ must necessarily hold. This is exercise 2.11 ...
11
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4answers
4k views

Why are maximal ideals prime?

Could anyone explain to me why maximal ideals are prime? I'm approaching it like this, let $R$ be a commutative ring with $1$ and $A$ be a maximal ideal. Let $a,b\in R:ab\in A$ I'm trying to ...
18
votes
3answers
1k views

If $\mathop{\mathrm{Spec}}A$ is not connected then there is a nontrivial idempotent

I'm solving a problem from Atiyah-Macdonald. I have to show that if $X=\mathop{\mathrm{Spec}}A$ is not connected then $A$ contains idempotents $e \neq 0,1$. The converse is easy. If $e \in A$ ...
10
votes
2answers
872 views

Ideal class group of a one-dimensional Noetherian domain

Let $A$ be a one-dimensional Noetherian domain. Let $K$ be its field of fractions. Let $B$ be the integral closure of $A$ in $K$. Suppose $B$ is finitely generated $A$-module. It is well-known that B ...
11
votes
5answers
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Proving that surjective endomorphisms of Noetherian modules are isomorphisms and a semi-simple and noetherian module is artinian.

I am revising for my Rings and Modules exam and am stuck on the following two questions: $1.$ Let $M$ be a noetherian module and $ \ f : M \rightarrow M \ $ a surjective homomorphism. Show that $f ...
4
votes
2answers
794 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 ...
17
votes
5answers
2k views

Showing the set of zero-divisors is a union of prime ideals

I'm working on an exercise from Atiyah and MacDonald's Commutative Algebra, and have hit a bump on Exercise 14 of Chapter 1. In a ring $A$, let $\Sigma$ be the set of all ideals in which every ...
7
votes
1answer
795 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 ...
21
votes
4answers
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Commutative non Noetherian rings in which all maximal ideals are finitely generated

In commutative rings we have the following Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated. From this Theorem I am looking for commutative rings $R$ in which ...
13
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4answers
1k views

An integral domain whose every prime ideal is principal is a PID

Does anyone has a simple proof of the following fact: An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
15
votes
4answers
847 views

Why is ideal more important than subring?

I have read that subgroups, subrings, submodules, etc. are substructures. But if you look at the definition of the Noetherian rings and Noetherian modules, Noetherian rings are defined with ideals ...
5
votes
2answers
895 views

Classification of prime ideals of $\mathbb{Z}[X]/(f(X))$

Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable. Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial. Let $A = \mathbb{Z}[X]/(f(X))$. Let $\theta$ = $X$ (mod $f(X)$). My ...
6
votes
1answer
644 views

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 ...
5
votes
3answers
452 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) ...
4
votes
2answers
284 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 ...
28
votes
9answers
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Why is the tensor product important when we already have direct and semidirect products?

Can anyone explain me as to why Tensor Products are important, and what makes Mathematician's to define them in such a manner. We already have Direct Product, Semi-direct products, so after all why do ...
14
votes
3answers
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Complement of maximal multiplicative set is a prime ideal

Let $R$ be a commutative ring with identity. I've been trying to prove the following: If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$. I have spent ...
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votes
3answers
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Ring of trigonometric functions with real coefficients

Let $R$ be the ring of functions that are polynomials in $\cos t$ and $\sin t$ with real coefficients. Prove that $R$ is isomorphic to $\mathbb R[x,y]/(x^2+y^2-1)$. Prove that $R$ is not a unique ...
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1answer
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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 = ...
6
votes
3answers
428 views

On the ring generated by an algebraic integer over the ring of rational integers

Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial. Let $\theta$ be a root of $f(X)$. Let $A = \mathbb{Z}[\theta]$. Let $p$ be a prime number. Suppose $p$ does not divide the discriminant ...
17
votes
3answers
928 views

Motivation for Eisenstein Criterion

I have been thinking about this for quite sometime. Eisentein Criterion for Irreducibility: Let $f$ be a primitive polynomial over a commutative unique factorization domain $R$, say $$f(x)=a_0 + ...
16
votes
1answer
639 views

When does the modular law apply to ideals in a commutative ring

Let $R$ be a commutative ring with identity and $I,J,K$ be ideals of $R$. If $I\supseteq J$ or $I\supseteq K$, we have the following modular law $$ I\cap (J+K)=I\cap J + I\cap K$$ I was wondering ...
10
votes
2answers
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Methods to check if an ideal of a polynomial ring is prime or at least radical

I am looking for methods to check whether a given ideal in $K[x_0,\dots,x_n]$ is prime. I mean something you can effectively use in some concrete non-trivial example. To be more explicit, I am working ...
11
votes
3answers
1k views

About the localization of a UFD

I was wondering, is the localization of a UFD also a UFD? How would one go about proving this? It seems like it would be kind of messy to prove if it is true. If it is not true, what about ...
5
votes
1answer
597 views

Irreducible Components of the Prime Spectrum of a Quotient Ring and Primary Decomposition

Recently I encountered a problem (the first exercise from chapter four of Atiyah & McDonald's Introduction to Commutative Algebra) stating that if $\mathfrak{a}$ is a decomposable ideal of $A$ (a ...
9
votes
1answer
950 views

Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?

If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a commutative ring, does it follow that $m \geq n$? This is obviously true for vector spaces over a field, but how would one show ...
6
votes
2answers
852 views

$\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain

How can I prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain? Also, I need to prove that its field of fractions is isomorphic to the field of rational functions ...
7
votes
2answers
804 views

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 ...
7
votes
1answer
679 views

Is each power of a prime ideal a primary ideal?

I want to show that each power of a prime ideal is a primary ideal or I have to think about a counterexample?
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vote
1answer
199 views

Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements from $k[X_1,X_2,X_3,X_4]$?

Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements in the ring $R=k[X_1,X_2,X_3,X_4]$? Can it be generated with three elements? (Here $k$ is a field.) Thanks for any help.
2
votes
0answers
571 views

The group of invertible fractional ideals of a Noetherian domain of dimension 1

Let $A$ be a Noetherian domain of dimension 1. Let $I(A)$ be the group of invertible fractional ideals of $A$. Let $P$ be a maximal ideal of $A$. Let $I(A_P)$ be the group of invertible fractional ...
13
votes
4answers
5k views

Example of modules that are projective but not free; torsion-free but not free

Free modules are projective, and projective modules are direct summand of free modules. Is there any example of projective modules that are not free? (I know this is not possible for modules of ...
4
votes
2answers
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Why is the localization of a commutative Noetherian ring still Noetherian?

This is an unproven proposition I've come across in multiple places. Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian. Why is this? ...
9
votes
2answers
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Tensor product of domains is a domain

I'm reading Milne's Algebraic Geometry course notes, version 5.22, as a companion to an algebraic geometry course I'm taking now. Proposition 4.15 states: Let $A$ and $B$ be $k$-algebras, which are ...
9
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1answer
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Definition of a finitely generated $k$ - algebra

In Miles Reid's Undergraduate Commutative Algebra he defines a ring $B$ to be finite as an $A$ - algebra if it is finite as an $A$ - module. Now what I don't understand is suppose we look at the ...
13
votes
2answers
441 views

The bijection between homogeneous prime ideals of $S_f$ and prime ideals of $(S_f)_0$

It is well-known that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the ...
5
votes
3answers
508 views

Is noetherianity a local property?

Let $R$ be a ring with finitely many maximal ideals such that $R_{\mathfrak m}$ ($\mathfrak m$ maximal ideal) is noetherian ring for all $\mathfrak m$. Is $R$ noetherian? I think $R$ has to be ...
8
votes
1answer
1k views

Ring of Polynomials is a Principal Ideal Ring implies Coefficient Ring is a Field?

I read this proof that if $D$ is an integral domain and $D[X]$ is a principal ideal domain, then $D$ is a field. My question is if the requirements can be relaxed a bit, namely: Is it true that ...
5
votes
2answers
492 views

Completion of a Noetherian ring

How can we prove that if $R$ is a commutative Noetherian ring, $\mathfrak{m} = (a_1,\ldots,a_n)$ is an ideal, then the completion of $R$ at $\mathfrak{m}$ is isomorphic to ...
9
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3answers
455 views

If $R$ is a commutative ring with identity, and $a, b\in R$ are divisible by each other, is it true that they must be associates?

Thank you very much! My problem is: If $R$ is a commutative ring with identity, and $a, b$ are its elements that are divisible by each other, is it true that they must be associates? Here, $a$ ...
6
votes
1answer
336 views

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 ...
4
votes
1answer
237 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 ...
4
votes
1answer
857 views

The total ring of fractions of a reduced Noetherian ring is a direct product of fields

This is question 6.5 in Matsumura's "Commutative ring theory": How can I prove that the total ring of fractions of a reduced Noetherian ring is a direct product of fields?
7
votes
2answers
705 views

Lying-over theorem without Axiom of Choice

This question is motivated by this and this. Can the following proposition be proved without Axiom of Choice? Proposition: Let $k$ be a field. Let $A$ and $B$ be commutative algebras without ...