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

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Why does the structure theorem for finitely generated modules over PIDs fail for arbitrary modules over a PID?

The proof that I know of the theorem goes like this: Any module $M$ is a quotient of a free module $F$ (over any ring). Any submodule $K$ of a free module $F$ over a PID $R$ is a free module, so in ...
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3answers
247 views

In $\mathbb{Z}/(n)$, does $(a) = (b)$ imply that $a$ and $b$ are associates?

[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.] Below, $R$ is a commutative ring with $1$. In John J. Watkins' Topics in ...
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1answer
234 views

If $I$ is a finitely generated ideal of $A[X]$, is $I\cap A$ necessarily finitely generated for a commutative unital ring $A$?

Let $A$ be a commutative ring with $1$ and $A[X]$ the ring of polynomials in one variable over $A$. Assume $I$ is a finitely generated ideal of $A[X]$. My question is Is $I\cap A$ necessarily ...
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191 views

Isomorphic factor rings of polynomial rings does imply isomorphic ideals?

Let $k$ be a field, $I$ and $J$ are ideals of $R=k[x_1,\dots,x_n]$. If $R/I\simeq R/J$ as rings, then $I \simeq J$ as $R$-modules holds? Thanks in advance!
9
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1answer
117 views

Curious about Hilbert-Zariski theorem involving homogeneous variety and set of zeroes.

I got myself in a confusing situation the other week while trying to read a bit of algebraic geometry. I'm hoping someone can pull me out. Suppose $k$ is a field, and $V$ a homogeneous variety with ...
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2answers
1k views

How to check whether an ideal is a prime (or maximal) ideal?

I have a ring $R$ which is known to be a Dedekind domain, but not necessarily a Euclidian domain, and a nonzero ideal generated by one or two elements in this ring. How can I check if this ideal is a ...
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107 views

Contents of Tor modules

I'm interested in knowing a concrete description of what elements of Tor modules $\mathrm{Tor}^i_R(M,N)$ "are". As it stands I have no real intuition for, say, maps between Tor modules induced by ...
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129 views

diagonalizing a matrix over the $\ell$-adics

Let $M$ be a $2 \times 2$ matrix with coefficients in $\mathbb{Z}_{\ell}$ whose characteristical polynomial is $$ P(T) = T^2- (a+d) T + (ad-bc). $$ I've encountered the following assertion: If ...
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409 views

Computing the “lying over”, “going up”, “going down” ideals.

For any commutative unital ring $R$ and an ideal $\mathfrak{a}$ of $R$, we shall denote $$\begin{align*} \mathrm{Spec}(R)&:=\{\text{prime ideals of }R\},\\ ...
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Grade of a maximal prime ideal in a Noetherian UFD

Here is an another problem in Commutative Rings by Kaplansky, p. 103, no. 15. Let $R$ be a Noetherian UFD. Let $(a,b) \not= R$ where $a,b \in R.$ Prove that any maximal prime of $(a,b)$ has grade ...
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143 views

$AB=z \mathrm{Id}_n$ implies $z^m BA = z^{m+1} \mathrm{Id}_n$ for what $m$?

This question builds on a series of questions looking for elementary proofs that $AB=\mathrm{Id}$ implies $BA=\mathrm{Id}$, for $A$ and $B$ both $n \times n$ matrices over a commutative ring. First ...
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1answer
165 views

Is there a characterization of integral domains in terms of the homomorphisms out of them?

In the $\mathbf{Set}$-concrete category of commutative rings, we can define that an object $A$ is a field iff for every homomorphism $f : A \rightarrow B$, precisely one of the following holds. $f$ ...
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3answers
255 views

How does one prove that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not Noetherian?

How does one prove that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not Noetherian? I let $(1, f_1, ..., f_n,...)$ be the $\mathbb{Z}$-basis of $\text{Int}(\mathbb{Z})$, ...
9
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1answer
187 views

Lifting isomorphisms between derived categories

Suppose $A$ and $B$ are commutative rings. Let $A\to B$ be a surjective ring homomorphism. I will denote by $D(A)$ and $D(B)$ the derived categories of unbounded complexes over $A$ and $B$. Suppose ...
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1answer
441 views

Exercise 11.5 from Atiyah-MacDonald: Hilbert-Serre theorem and Grothendieck group

I don't understand Exercise 11.5 of Atiyah & MacDonald, which demands one elaborate upon or rephrase the Hilbert–Serre Theorem (11.1) in terms of the Grothendieck group $K(A_0)$. Here's ...
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How to tell if an element of a quotient ring is a zero divisor

I am looking at Hartshorne Example III.9.8.4., p260. He says that $a$ is not a zero divisor in $k[a,x,y,z]/I$, where $$ I = (a^2(x+1) -z^2, ax(x+1)-yz, xz-ay,y^2-x^2(x+1)). $$ Is there a good way to ...
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343 views

Trivial intersection of algebraic sets?

The question came up while reading a bit more into the Hilbert-Zariski theorem I asked about the other week. Suppose $V$ is an algebraic variety over arbitrary field $k$. (For this situation, I'll ...
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3answers
2k views

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 ...
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A question on definition of field of fractions

Wikipedia defines the field of fractions of a domain as The field of fractions or field of quotients of an integral domain is the "smallest" field in which it can be embedded. What does ...
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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 ...
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3answers
433 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, $a$ ...
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2answers
602 views

How does Hilbert's Nullstellensatz generalize the “fundamental theorem of algebra”?

What is Hilbert's Nullstellensatz in the sense of the generalization of "fundamental theorem of algebra"? I've seen that in some texts it was referred to as the generalization of the fundamental ...
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218 views

Integral closure of $\mathbb{Q}[X]$ in $\mathbb{Q}(X)[Y]$

Consider the ring $\mathbb{Q}[X]$ of polynomials in $X$ with coefficients in the field of rational numbers. Consider the quotient field $\mathbb{Q}(X)$ and let $K$ be the finite extension of ...
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3answers
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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 ...
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4answers
329 views

Subrings of fraction fields

Let $R$ be an integral domain and let $S$ be a ring with $R \le S \le \text{Frac}(R)$ (fraction field). Question: Is there a multiplicatively closed subset $U \subseteq R\setminus \{0\}$ such that ...
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4answers
334 views

Let $R$ be a commutative ring with $1$. Suppose that every nonzero proper ideal of $R$ is maximal. Prove that there are at most two such ideals.

Let $R$ be a commutative ring with $1$. Suppose that every nonzero proper ideal of $R$ is maximal. Prove that there are at most two such ideals. Help me some hints. I have no idea to start. ...
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942 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 ...
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428 views

A fraction field is not finitely generated over its subdomain

I'm looking for proofs of the following fact. Suppose that $R$ is a domain which is not a field with fraction field $K$. Then $K$ is not finitely generated as $R$-module. I know this fact is ...
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2answers
864 views

Direct summand of a free module

Let $M$, $L$, $N$ be $A$-modules and $M=N\oplus L$. If $M$ and $N$ are free, is $L$ necessarily free?
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4answers
686 views

Non-Noetherian ring with a single prime ideal

My question: What are the most simple examples of a commutative ring R satisfying both of the following two properties: 1. R is not Noetherian. 2. R has exactly one prime ideal.
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Prerequisites for Atiyah Macdonald

I am currently doing a one semester course on groups and rings where we have learned about (so far): Definitions of groups, subgroups, cyclic and normal subgroups, the symmetric group, homomorphisms, ...
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Hartshorne Exercise 1.1 (a)

(see bottom for apology) Let $Y$ be the plane curve $y = x^2$ (i.e., $Y$ is the zero set of the polynomial $f = y - x^2$). Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over ...
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433 views

Is this ring Noetherian?

The subring of $\mathbb{C}[x,y]$ consisting of all polynomials $f(x,y)$ whose gradient vanishes at the point $x=y=0$. Is this ring Noetherian?
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1answer
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Is a regular ring a domain

A regular local ring is a domain. Is a regular ring (a ring whose localization at every prime ideal is regular) also a domain? I am unable to find/construct a proof or a counterexample. Any help would ...
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242 views

Is the ring of p-adic integers of finite type over the ring of integers?

Denote by $\mathbb{Z}_p$ the ring of $p$-adic integers. Is $\mathrm{Spec}(\mathbb{Z}_p)$ of finite type over $\mathrm{Spec}(\mathbb{Z})$?
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182 views

How does this step in the proof of the structure theorem for f.g. modules over a Dedekind domain work?

I am trying to show that every finitely generated projective module $P$ over a Dedekind domain $D$ is a direct sum of (fractional) ideals. May's notes on Dedekind domains claim the result can be ...
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507 views

Inverse image of the sheaf associated to a module

In Hartshorne, Algebraic geometry it's written, that for every scheme morphism $f: Spec B \to Spec A$ and $A$-module $M$ $f^*(\tilde M) = \tilde {(M \otimes_A B)}$. And that it immediately follows ...
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Commutative Algebra without the axiom of choice

It is well known that in a commutative ring with unit, every proper ideal is contained in a maximal ideal. The proof uses the axiom of choice. This fact, and others that are proved using essentially ...
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2answers
246 views

Exercise 3.15 [Atiyah/Macdonald]

I have a question regarding a claim in Atiyah, Macdonald. A is a commutative ring with $1$, $F$ is the free $A$-module $A^n$. Assume that $A$ is local with residue field $k = A/\mathfrak m$, and ...
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Commutative property of ring addition

I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring: Let $R\neq ...
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1answer
248 views

Irreducible polynomial over an algebraically closed field

Suppose $k$ is an algebraically closed field and $p(x,y)\in k[x,y]$ is an irreducible polynomial. Prove that there are only finite many $a\in k$ such that $p(x,y)+a$ is reducible, i.e. the set ...
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394 views

Approximation Lemma in Serre's Local Fields

Let $A$ be a Dedekind domain, and let $K$ be its field of fractions. In Serre's Local Fields, the following Lemma is stated. Approximation Lemma Let $k$ be a positive integer. For every $i$, ...
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678 views

When is the integral closure of a local ring also a local ring?

Suppose $A$ is a normal local domain contained in a field $K$. Suppose $B$ is the integral closure of $A$ in $K$. Under what conditions on $A$ is $B$ local?
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830 views

Finitely generated algebra

I am getting the confusion with the definition of algebra. When we say $A$ is a finitely generated $R$- algebra then is that mean $A$ has a ring structure and finitely generated as an $R$-module. ...
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4answers
634 views

Are bimodules over a commutative ring always modules?

Let $R$ be a commutative ring. It is true that every module over $R$ is an $(R,R)$-bimodule. Is the converse true? In other words is it possible that there is an $R$-module where left multiplication ...
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347 views

Exactness of a short sequence of quotient modules

Suppose R is a commutative ring with 1, I $\subset R$ is an ideal. We have R-Modules A, B and C with C being flat, as well as a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C ...
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Proposition 5.21 in Atiyah-MacDonald

There's just one step in this proof I can't see for the life of me. Set up: We have a field K and an algebraically closed field $\Omega$. $(B, g)$ is maximal in the set $\Sigma$ of ordered pairs ...
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Affine scheme $X$ with $\dim(X)=0$ but infinitely many points

As the title says, I'm looking for an affine scheme of dimension zero, but with infinitely many points. At first I doubted that something like this could exist, and I still can't think of an example, ...
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1answer
536 views

Do localization and completion commute?

Let A be a commutative Ring and $\mathfrak{p}$ be a prime ideal of A. Under which assumptions for A and $\mathfrak{p}$ does localization by $\mathfrak{p}$ and completion with respect to $\mathfrak{p}$ ...
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262 views

Is Nullstellensatz true for arbitrary fields if there aren't hidden points?

The ideals $I=(X,Y)$ and $J=(X^2+Y^2)$ in $\mathbb R[X,Y]$ are such that $V(I)=V(J)$ and their radicals aren't the same contradicting the Nullstellensatz (in case it was true for arbitrary fields). ...