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

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8
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7answers
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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 ...
38
votes
1answer
4k 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? $(0)$ $(f(X))$, where ...
18
votes
4answers
3k 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 ...
15
votes
4answers
10k 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 ...
14
votes
2answers
2k views

$A^m\hookrightarrow A^n$ 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 ...
31
votes
3answers
3k views

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: ...
17
votes
5answers
6k 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 ...
17
votes
2answers
2k 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 ...
26
votes
4answers
3k views

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 ...
28
votes
6answers
2k views

Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$

This seems to be one of those tricky examples. I only know one proof which is quite complicated and follows by localizing $\mathbb{Z}[\sqrt[3]{2}]$ at different primes and then showing it's a DVR. ...
20
votes
4answers
1k views

Why are ideals more important than subrings?

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 ...
12
votes
5answers
2k views

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 ...
21
votes
3answers
2k 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
907 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 ...
17
votes
5answers
3k 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 ...
5
votes
3answers
1k views

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 ...
8
votes
1answer
796 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 ...
32
votes
9answers
4k views

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
3k 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).
5
votes
2answers
964 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 ...
6
votes
1answer
672 views

Tensor product of two finitely-generated modules over a local ring is zero

If $R$ is a local ring and $M$ and $N$ are finitely generated $R$-modules such that $M\otimes N=0$ then how does it follow from Nakayama's lemma that either $M=0$ or $N=0$? This is an exercise ...
5
votes
3answers
510 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) ...
8
votes
1answer
876 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 ...
19
votes
1answer
781 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 ...
5
votes
2answers
967 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 ...
4
votes
2answers
311 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 ...
13
votes
3answers
2k 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 ...
16
votes
3answers
2k views

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 ...
7
votes
1answer
761 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 ...
13
votes
2answers
822 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 ...
10
votes
3answers
493 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, $a$ ...
2
votes
1answer
1k views

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
447 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 ...
19
votes
4answers
7k 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 ...
17
votes
3answers
997 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 + ...
10
votes
2answers
2k views

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 ...
10
votes
3answers
1k views

Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$

Krull dimension of a ring $R$ is the supremum of the number of strict inclusions in a chain of prime ideals. Question 1. Considering $R = \mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 ...
12
votes
1answer
1k 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 ...
8
votes
4answers
1k views

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 ...
7
votes
3answers
648 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 ...
7
votes
3answers
996 views

Verifying that the ideal $(x^3-y^2)$ is prime

How to prove that the ideal $I=(x^3-y^2)$ in $k[x,y]$ is prime? I have constructed a map from $k[x,y]$ to $k[t]$, which maps $x$ to $t^2$, and $y$ to $t^3$. Then, I want to show that the kernel ...
11
votes
2answers
925 views

Must $k$-subalgebra of $k[x]$ be finitely generated?

Suppose $k$ is a field, $A$ is a $k$-subalgebra of the polynomial ring $k[x]$. Must $A$ be a finitely generated $k$-algebra? Thanks.
6
votes
2answers
951 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 ...
4
votes
4answers
318 views

$\mathbb{C}[x,y]/(f,g)$ is an artinian ring, if $\gcd(f,g)=1$. [closed]

This problem extends the fact that $\mathbb{C}[x,y]/(x^n,y^m)$ is artinian ring. Let $f,g \in \mathbb{C}[x,y]$ such that $\gcd(f,g)=1$. Show that $\mathbb{C}[x,y]/(f,g)$ is an artinian ring.
4
votes
2answers
1k views

A characterization of invertible fractional ideals of an integral domain

Let $A$ be an integral domain, $K$ its field of fractions. Let $M$ be a fractional ideal of $A$. I'd like to prove that $M$ is invertible if and only if $MA_P$ is a principal fractional ideal of $A_P$ ...
1
vote
1answer
236 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.
8
votes
2answers
973 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
2answers
567 views

Is quotient of a ring by a power of a maximal ideal local?

Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring? This is a well ...
7
votes
1answer
793 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?
2
votes
0answers
598 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 ...