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

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21
votes
1answer
949 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) ...
11
votes
3answers
1k 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
2answers
799 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 ...
8
votes
2answers
1k 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 ...
5
votes
4answers
3k 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 ...
13
votes
2answers
924 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 ...
19
votes
4answers
2k 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: ...
3
votes
2answers
598 views

In a finite ring extension there are only finitely many prime ideals lying over a given prime ideal

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 ...
16
votes
3answers
550 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 $e \in A$ ...
7
votes
1answer
656 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 ...
24
votes
9answers
3k views

Motivation for Tensor Product

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 ...
4
votes
2answers
711 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 ...
8
votes
5answers
725 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 ...
13
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 ...
5
votes
3answers
372 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) ...
5
votes
3answers
364 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 ...
14
votes
1answer
448 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 ...
9
votes
2answers
891 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 ...
14
votes
4answers
1k views

Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$ Can you tell ...
6
votes
1answer
471 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 ...
6
votes
1answer
489 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?
3
votes
0answers
520 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 ...
4
votes
2answers
241 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 ...
5
votes
3answers
315 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
1answer
794 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 ...
5
votes
1answer
830 views

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 ...
6
votes
2answers
570 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 ...
3
votes
1answer
802 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 = ...
7
votes
2answers
677 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 ...
6
votes
2answers
509 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 ...
5
votes
2answers
656 views

$x$ not nilpotent implies that there is a prime ideal not containing $x$.

Let $\mathscr{N}(R)$ denote the set of all nilpotent elements in a ring $R$. I have actually done an exercise which states that if $x \in \mathscr{N}(R)$, then $x$ is contained in every prime ideal ...
15
votes
3answers
810 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 + ...
25
votes
3answers
2k 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 ...
26
votes
3answers
705 views

Ideals of $\mathbb{Z}[X]$

Is it possible to classify all ideals of $\mathbb{Z}[X]$? By this I mean a preferably short enumerable list which contains every ideal exactly once, preferably specified by generators. The prime ...
4
votes
3answers
697 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
2answers
559 views

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 ...
8
votes
1answer
687 views

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 ...
3
votes
1answer
380 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 ...
11
votes
3answers
535 views

Examples of rings with ideal lattice isomorphic to $M_3$, $N_5$

In thinking about this recent question, I was reading about distributive lattices, and the Wikipedia article includes a very interesting characterization: A lattice is distributive if and only if ...
3
votes
2answers
438 views

Completion of a Noetherian ring R at $(a_1,\ldots,a_n)$ is isomorphic to $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$

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 ...
2
votes
1answer
1k views

Submodule of free module over a p.i.d. free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
2
votes
1answer
253 views

A proposition on a Dedekind domain

I need a proof of the following proposition(?). Actually I think I came up with a proof. But it's nice to confirm it and/or to know other proofs. Thanks. Proposition Let $A$ be a Dedekind domain. Let ...
3
votes
1answer
218 views

Multidimensional Hensel lifting

I have a question about a practical application of (some) generalised form of Hensel's Lemma. I cannot find it stated in an appropriate form in Bourbaki or anywhere else, so here goes ... Let $p$ be ...
3
votes
2answers
467 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
3answers
319 views

Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice

Let $A$ be an integral domain which is finitely generated over a field $k$. Let $f \neq 0$ be a non-invertible element of $A$. Can one prove that there exists a prime ideal of $A$ containing $f$ ...
0
votes
1answer
125 views

If $R$ is a local ring with $\operatorname{emb dim} (R) = \text{depth}(R) + 1$ then $R$ is Gorenstein

$(R,m)$ is a Noetherian local ring such that $\operatorname{emb dim}(R) = \text{depth}(R) + 1$. Show that $R$ is a Gorenstein ring. (Here $\operatorname{emb dim}(R)$ means "number of minimal system of ...
3
votes
1answer
179 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 ...
0
votes
1answer
803 views

The set of all nilpotent element is an ideal of R

Given that R is commutative ring with unity. I want show that set of all nilpotent element is an ideal of R.i know how to show ideal if set is given but here set is not given to me. Can anyone help ...
67
votes
4answers
2k views

Does $R[x] \cong S[x]$ imply $R \cong S$?

This is a very simple question but I believe it's nontrivial. I would like to know if the following is true: If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ ...
25
votes
5answers
2k views

Why should I care about adjoint functors

I am comfortable with the definition of adjoint functors. I have done a few exercises proving that certain pairs of functors are adjoint (tensor and hom, sheafification and forgetful, direct image and ...