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

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6
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1answer
854 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." ...
1
vote
1answer
241 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
606 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 ...
8
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? ...
11
votes
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 ...
11
votes
1answer
3k 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
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?
14
votes
2answers
533 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
420 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
1answer
218 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 $a=ub$.
5
votes
2answers
586 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 $R[[x_1,\ldots,x_n]]/(x_1-...
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 ...
6
votes
1answer
694 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 ...
9
votes
3answers
375 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, ...
8
votes
1answer
186 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 ...
5
votes
1answer
298 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
2answers
659 views

Finitely many prime ideals lying over $\mathfrak{p}$

Let $A$ be a commutative ring with identity and $B$ a finitely generated $A$-algebra that is integral over $A$. If $\mathfrak{p}$ is a prime ideal of $A$, there are only finitely many prime ideals $P$...
8
votes
2answers
878 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 $$(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$$...
3
votes
2answers
857 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 ...
3
votes
1answer
466 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, ...
3
votes
3answers
255 views

If $a,b$ is an $R$-sequence, then $ax-b$ is prime (Eisenbud, Exercise 10.4)

This is the exercise mentioned above: Let $a,b$ be regular sequence over a domain $R$. Prove that $ax-b$ is a prime of $R[x]$. Thank you for your answer!
2
votes
1answer
138 views

Lemma 1.3.4(b) in Bruns and Herzog

My question refers to the proof of the second of the following lemma given in Cohen-Macaulay rings by Bruns and Herzog. Lemma 1.3.4 (Bruns and Herzog): Let $(R,\mathfrak m,k)$ be a local ring, and ...
10
votes
2answers
763 views

Intersection of finitely generated ideals

Let $I$, $J$ be finitely generated ideals in a ring $A$ (commutative with identity). I know that the intersection need not be finitely generated: can somebody give me an example? Thanks.
4
votes
2answers
178 views

Noetherian ring and the same prime divisors

Let $A$ be a Noetherian ring and let $x \in A$ be an element which is neither a unit nor a zerodivisor. Why the ideals $xA$ and $x^{n}A$ (where $n \in \mathbb{N}$) have the same prime divisors? i.e. $$...
0
votes
2answers
691 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 ring ...
41
votes
5answers
3k 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 ...
91
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 $...
32
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 ...
17
votes
2answers
1k views

$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$

Let $A$ be an integral domain of finite Krull dimension. Let $\mathfrak{p}$ be a prime ideal. Is it true that $$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$$ where $\dim$ ...
9
votes
4answers
7k views

Can you construct a field with 4 elements?

Can you construct a field with 4 elements? can you help me think of any examples?
13
votes
1answer
388 views

Geometrical interpretation of $I(X_1\cap X_2)\neq I(X_1)+I(X_2)$, $X_i$ algebraic sets in $\mathbb{A}^n$

Edit: I should point out that I'm working over an algebraically closed field $k$. Let $X_1,X_2\subset\mathbb{A}^n$ be affine algebraic sets. Show that $I(X_1\cap X_2)=\sqrt{I(X_1)+I(X_2)}$. Show ...
12
votes
3answers
1k 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
1answer
166 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 $A$. ...
5
votes
2answers
214 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 ...
10
votes
2answers
1k 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\...
10
votes
2answers
1k views

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 ...
9
votes
2answers
2k views

Is fibre product of varieties irreducible (integral)?

Let $k$ be an algebraically closed field and $X,Y$ varieties (i.e. integral, separated schemes of finite type over $k$). Is the fibre product $X \times_k Y$ necessary irreducible or integral? I ...
7
votes
2answers
160 views

$M_a =\{ f\in C[0,1] |\ f(a)=0 \}$ for $a$ $\in$ $[0,1]$. Is $M_a$ finitely generated in $C[0,1]$?

Let $C[0,1]$ denote the ring of continuous functions on $[0,1]$. Consider the maximal ideal $M_a =\{ f\in C[0,1] | f(a)=0 \}$ for $a$ $\in$ $[0,1]$. Is $M_a$ finitely generated? Note: It may seem ...
6
votes
1answer
224 views

In an extension of finitely generated $k$-algebras the contraction of a maximal ideal is also maximal

Let $k$ be a field and let $A \subset B$ be two finitely generated $k$-algebras. Prove that the contraction of any maximal ideal of $B$ is a maximal ideal of $A$. thank you very much again!
6
votes
3answers
1k views

Tensor product of 2 coordinate rings

For the term variety, I mean the irreducible algebraic set. My question is, if $V$ and $W$ are 2 varieties over a field $\Bbbk$, then does $\Bbb{k}[V]\otimes \Bbb{k}[W]$ has special structure? I try ...
6
votes
3answers
2k 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(...
4
votes
1answer
299 views

Prove that the normalisation of $A=k[X,Y]/(Y^2-X^2-X^3)$ is $k[t]$ where $t=Y/X$ (Reid, Exercise 4.5)

This is a problem about finding the normalisation of a quotient polynomial ring. So I have to find the integral closure of the ring in its field of fractions. The problem statement is as follows: ...
12
votes
3answers
883 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
1k views

An ideal that is maximal among non-finitely generated ideals is prime.

I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over. Let $R$ be a commutative ring with ...
6
votes
1answer
246 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 tried ...
3
votes
2answers
257 views

$X$ compact Hausdorff space, characterize the maximal ideals of $C(X)$

I know this question has been asked before, but I think I'm very close to a new solution and wanted to know if it is a viable approach. Let $C(X)$ be the ring of continuous functions $X \rightarrow \...
8
votes
2answers
1k 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 ...
7
votes
2answers
774 views

Does every Noetherian domain have finitely many height 1 prime ideals?

Let $A$ be a Noetherian domain. Is the set $\{P\subset A \mid P \mbox{ prime ideal, } \dim A_P=1\}$ always finite? I can prove for $f \neq 0, f\in A$, the set $\{P\subset A \mid \dim A_P=1, f\in P\}$...
7
votes
1answer
369 views

If $B$ is finitely generated as a $k$-algebra, and $\phi:A\to B$ is a $k$-algebra map, is $\phi^{-1}(M)$ maximal for any maximal $M\subset B$?

Suppose that $A$ and $B$ are commutative rings containing a field $k$, and $B$ is finitely generated $k$-algebra. Let $\phi: A\rightarrow B$ be a ring homomorphism with $\phi|_k =\mathrm{Id}$. I am ...