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

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11
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2answers
2k 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 ...
13
votes
2answers
2k 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 ...
4
votes
2answers
2k views

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? ...
10
votes
1answer
2k 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 ...
13
votes
2answers
493 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 ...
6
votes
1answer
357 views

Normalization of a quotient ring of polynomial rings (Reid, Exercise 4.6)

I solved all parts of Exercise 4.6 of the book Undergraduate Commutative Algebra of Miles Reid except the last one. Let $A=k[X]$ and $f\in A$ has a square factor but it is not a square polynomial ...
5
votes
2answers
176 views

One-dimensional Noetherian UFD is a PID

I am looking for a reference which has a self-contained (elementary, that is, at the "undergraduate algebra level") proof of the the fact that any one-dimensional Noetherian UFD is a PID. Does anyone ...
5
votes
1answer
190 views

Associated elements in a ring

Please help me to find elements $a,b$ in a ring $R$ such that $a\mid b$ and $b\mid a$, but there does not exist any unit $u$ in $R$ such that $a=ub$.
13
votes
2answers
1k views

A non-noetherian ring with all localizations noetherian

If for a ring $A$ every localization $A_\mathfrak{p}$ by a prime $\mathfrak{p}\subseteq A$ is noetherian, is it true that $A$ is noetherian? I believe not but I can't find a good counterexample.
10
votes
3answers
487 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, $a$ ...
5
votes
2answers
566 views

Completion of a Noetherian ring R at the ideal $ (a_1,\ldots,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 ...
16
votes
2answers
1k views

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 ...
13
votes
2answers
615 views

What is a primary decomposition of the ideal $I = \langle xy, x - yz \rangle$?

Given the ring $k[x,y,z]$, where $k$ is a field, and an ideal $I=(xy,x-yz)$, find a primary decomposition of $I$. I tried to draw the graph of the variety of $I$ and get a decomposition of ...
9
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 ...
4
votes
2answers
322 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 $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, ...
6
votes
1answer
574 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 ...
5
votes
4answers
621 views

Examples of Non-Noetherian Valuation Rings

For valuation rings I know examples which are Noetherian. I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind? I am very ...
5
votes
1answer
740 views

Constructing Idempotent Generator of Idempotent Ideal

Exercise 2.1 in Matsumura's Commutative Ring Theory reads as follows: "Let $A$ be a commutative ring and $I$ an ideal that is finitely generated and $I=I^2$. Then $I$ is generated by an idempotent." ...
5
votes
1answer
282 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 ...
5
votes
1answer
1k 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?
4
votes
3answers
901 views

Dedekind's theorem on the factorisation of rational primes

Let $K$ be an algebraic number field, and suppose its ring of integers is $\mathcal{O}_K = \mathbb{Z}[\theta]$ for some $\theta \in \mathcal{O}_K$. Let $f \in \mathbb{Z}[X]$ be the minimal polynomial ...
3
votes
2answers
748 views

Principal prime ideals are minimal among prime ideals in a UFD

Fulton, "Algebraic Curves," Exercise 1.39(a): Let $R$ be a UFD, and $P = (t)$ a principal, proper, prime ideal. Show there is no prime ideal $Q$ with $0 \subset Q \subset P$. After being ...
1
vote
1answer
891 views

A maximal ideal among those avoiding a multiplicative set is prime

Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime. If $R$ is an integral domain, ...
8
votes
2answers
726 views

How to prove that this subring is not noetherian?

Consider the subring $R=k[x,xy,xy^2,\ldots]$ of $k[x,y]$. I want to prove that $R$ is not noetherian. An ascending chain of ideals is the following ...
8
votes
3answers
318 views

sufficient conditions for being a PID

Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, ...
7
votes
2answers
719 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
1k 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 ...
5
votes
1answer
362 views

Commutative integral domain with d.c.c. is a field [duplicate]

If $R$ is a commutative integral domain and it also satisfies descending chain condition on its ideals then how will we show that such ring $R$ will be a field?
3
votes
1answer
442 views

Use Nakayama's Lemma to show that $I$ is principal, generated by an idempotent. [duplicate]

Let $I$ be a finitely generated ideal in $R$, such that $I^2 = I$. Using the fact there exists $x\in R$ such that $e = 1 - x\in I$ and $xI = 0$, use Nakayama's Lemma to show that $I$ is principal, ...
0
votes
2answers
561 views

Every finitely generated algebra over a field is a Jacobson ring

Knowing that the polynomial ring in $n$ variables over a field $k$ is a Jacobson ring, how can we prove from it that every finitely generated $k$-algebra is a Jacobson ring? EDIT: We define a ...
11
votes
1answer
242 views

What is the intuition behind the name “Flat modules”?

I am studying Atiyah and MacDonald's book "Introduction to Commutative Algebra" and I have just read the definition of a flat module. It seems to me that if they have called that kind of modules ...
3
votes
2answers
276 views

If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).

If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat). The converse, although not immediately apparent, can be proven quite easily. But ...
1
vote
1answer
148 views

A relation involving an endomorphism of a finitely generated module

Suppose $R$ is a commutative ring, $I$ is an ideal of $R$, and $M$ a finitely generated $R$-module. (Usually in this problem $R$ includes $1_R$.) Let $\phi : M \to M$ be an $R$-homomorphism, and ...
85
votes
4answers
3k 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 ...
34
votes
5answers
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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 ...
23
votes
2answers
3k views

Tensor products commute with direct limits

This is Exercise 2.20 in Atiyah-Macdonald. How can we prove that $\varinjlim (M_i \bigotimes N) \cong (\varinjlim M_i) \bigotimes N$ ? Atiyah gives a suggestion, he says that one should obtain a map ...
30
votes
3answers
4k 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
3answers
1k 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 ...
11
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1answer
1k views

Every maximal ideal is principal. Is $R$ principal?

Let $R$ be a commutative ring with 1. If every maximal ideal of $R$ is principal, is $R$ a principal ideal ring?
13
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2answers
1k views

Fields finitely generated as $\mathbb Z$-algebras are finite?

Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra. (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite?
12
votes
3answers
982 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 ...
9
votes
3answers
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If a ring is Noetherian, then every subring is finitely generated?

Let $R$ be a commutative ring with $1$, and let $K$ be a field. We know that $R$ is Noetherian iff every ideal of $R$ is finitely generated as an ideal. Question 1: If $R$ is Noetherian, is every ...
9
votes
1answer
157 views

Infinite linear independent family in a finitely generated $A$-module

So I'm stuck with this problem. Let $A$ be a nonzero commutative ring (with unit). I have several questions that are really close to each other. 1) Let $M$ be a finitely generated module over $A$. ...
8
votes
3answers
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If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. [duplicate]

Possible Duplicate: Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field I need to prove this result, but the only starting point I think of is to ...
12
votes
3answers
719 views

Ring of holomorphic functions

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 ...
6
votes
3answers
560 views

$R^n \cong R^m$ iff $n=m$

How can i show that two $R$-modules of finite rank are isomorphic if and only if they have the same rank, i.e., $R^n \cong R^m$ iff $n=m$.
10
votes
2answers
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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 ...
8
votes
2answers
962 views

Product of two primitive polynomials

I'm having troubles with one of the problems in the book Introduction to Commutative Algebra by Atiyah and MacDonald. It's on page 11, and is the last part of the second question. Given $R$ a ...
8
votes
2answers
788 views

A counterexample to the going down theorem

I will appreciate any enlightenment on the following which must be an exercise in a certain textbook. (I don't recognize where it comes from.) I understand that the going down property does not hold ...
6
votes
1answer
234 views

Zero divisors in $A[x_1,x_2,\dots,x_r]$

I am trying to show that if $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor then there exists $a$ in $A-\{0\}$ such that $af=0$ in $A[x_1,x_2,\ldots, x_r]$. What I have ...