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

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8
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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 ...
4
votes
1answer
237 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
449 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
258 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 ...
4
votes
1answer
131 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?
4
votes
3answers
531 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 ...
1
vote
3answers
329 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 $f$ ...
4
votes
1answer
194 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 ...
3
votes
2answers
189 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 ...
3
votes
2answers
574 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 $A_P$ ...
0
votes
2answers
264 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 ...
69
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 $S$ ...
26
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 ...
11
votes
2answers
2k views

Prove that a UFD $R$ is a PID if and only if every nonzero prime ideal in $R$ is maximal

Prove that a UFD $R$ is a PID if and only if every nonzero prime ideal in $R$ is maximal. The forward direction is standard, and the reverse direction is giving me trouble. In particular, I can prove ...
12
votes
5answers
1k views

A non-noetherian ring with $\text{Spec}(R)$ noetherian

Question 1: Does such a ring can be found? Note: The definition of a noetherian topological space is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite ...
9
votes
3answers
778 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 ...
5
votes
2answers
752 views

I want a proof without using Nakayama's lemma

I am trying to understand Nakayama's lemma. It looks like some "fixed point theorem". Using Nakayama's lemma , I can easily solve the following question. I want another proof. Thanks. Let $A$ be a ...
8
votes
4answers
835 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 ...
11
votes
2answers
351 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 ...
10
votes
2answers
778 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
2answers
589 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.
5
votes
1answer
278 views

Field of fractions of $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ [duplicate]

This problem goes as follows: Prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ is an integral domain and that its field of fractions is isomorphic to the ring of rational functions ...
5
votes
3answers
189 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 ...
5
votes
1answer
204 views

Finite injective dimension of the residue field implies that the ring is regular

Let $(R,\mathfrak m,k)$ be a noetherian local ring. If $\operatorname{inj dim}_R k$ is finite, then $R$ is regular. This is exercise 3.1.26 from Bruns and Herzog, Cohen-Macaulay Rings. I don't ...
5
votes
2answers
333 views

Where is the Axiom of choice used?

In Reid's commutative algebra, there is a proof of equivalent conditions of Noetherian rings, especially (1) The set of ideals of $A$ has the a.c.c. $\Rightarrow$ (2) Every ideal in $A$ is finitely ...
3
votes
3answers
302 views

$R \otimes_R M \cong M$

Let $R$ be a commutative unital ring and $M$ an $R$-module. I'm trying to prove $R \otimes_R M \cong M$ but I'm stuck. If $(R \otimes M, b)$ is the tensor product then I thought I could construct an ...
10
votes
2answers
314 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 the primary decomposition of $I$. I tried to draw the graph of the variety of $I$ and get a decomposition of ...
7
votes
2answers
670 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 ...
3
votes
1answer
250 views

Working out the normalization of $\mathbb C[X,Y]/(X^2-Y^3)$.

I'm trying to identify the normalization of the ring $A := \mathbb C[X,Y]/\langle X^2-Y^3 \rangle$ with something more concrete. First, $X^2-Y^3$ is irreducible in $\mathbb C[X,Y]$, making $\langle ...
3
votes
2answers
165 views

Showing that if $R$ is local and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$.

Let $R$ be a local ring, and let $\mathfrak m$ be the maximal ideal of $R$. Let $M$ be an $R$-module. I understand that $M \otimes_R (R / \mathfrak m)$ is isomorphic to $M / \mathfrak m M$, but I ...
8
votes
2answers
864 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 ...
6
votes
1answer
121 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 subset of ...
6
votes
2answers
483 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 ...
5
votes
1answer
431 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." ...
4
votes
3answers
452 views

Does any integral domain contain an irreducible element?

Let $R$ be an integral domain which is not a field. Does $R$ necessarily have an irreducible element? I suspect the answer is no, but I couldn't find an example showing that...
4
votes
2answers
842 views

Nilpotency of the Jacobson radical of an Artinian ring without Axiom of Choice

Let $A$ be a commutative ring. Suppose $A$ has a composition series as an $A$-module. EDIT Since $A$ has a composition series, $A$ has a maximal ideal. Let $J$ be the intersection of all the maximal ...
4
votes
1answer
446 views

Is $(XY - 1)$ a maximal ideal in $k[[X]][Y]$?

Is $(XY - 1)$ a maximal ideal in $k[[X]][Y]$, and if so, how can I see it? It is at least prime because the generator is irreducible, and by the same argument it is maximal among all principal ...
3
votes
1answer
104 views

Proof that $K\otimes_F L$ is not noetherian

Let $F$ be a field and $K$ and $L$ be extension fields of $F$ such that $\mathrm{tr.deg}_F(K) = \infty$ and $\mathrm{tr.deg}_F(L) = \infty$. It seems to be proved that $K\otimes_F L$ is not ...
3
votes
1answer
302 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, ...
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vote
5answers
282 views

Maximal ideal of Dedekind domain

Let $\Lambda$ be a Dedekind domain and $\mathcal{m}$ be a maximal ideal of $\Lambda$. Is it possible that $\mathcal{m}=\mathcal{m}^2$? If not, how can I prove it?
0
votes
1answer
333 views

Krull dimension of some quotient rings

I have difficulties in doing some calculations of heights and Krull dimensions; I hope that somebody could help me unveil the "tricks of the trade". In the following $\alpha,\beta,\gamma$ denote ...
5
votes
1answer
95 views

Height unmixed ideal and a non-zero divisor

Let $R$ be a commutative Noetherian ring with unit and $I$ an unmixed ideal of $R$. Let $x\in R$ be an $R/I$-regular element. Can we conclude that $x+I$ is an unmixed ideal? Background: A ...
5
votes
1answer
207 views

Injective Cogenerators in the Category of Modules over a Noetherian Ring

Let $R$ be a Noetherian ring and let $\mathcal{A}$ be an injective $R$-module. The injectivity of $\mathcal{A}$ is equivalent to the exactness of the functor $Hom_R(-,\mathcal{A})$, i.e. whenever we ...
4
votes
1answer
327 views

There exists a unique isomorphism $M \otimes N \to N \otimes M$

I want to show that there is a unique isomorphism $M \otimes N \to N \otimes M$ such that $x\otimes y\mapsto y\otimes x$. (Prop. 2.14, i), Atiyah-Macdonald) My proof idea is to take a bilinear $f: M ...
4
votes
3answers
687 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 ...
3
votes
4answers
447 views

Noetherian module implies Noetherian ring?

I know that a finitely generated $R$-module $M$ over a Noetherian ring $R$ is Noetherian. I wonder about the converse? I believe it to be false and I am looking for counterexamples. Also I wonder if ...
3
votes
4answers
540 views

Radical ideal of $(x,y^2)$

How does one show that the radical of $(x,y^2)$ is $(x,y)$ over $\mathbb{Q}[x,y]$? I have no idea how to do, please help me.
2
votes
4answers
197 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.
2
votes
4answers
178 views

Consider $R[x]$ and let $S$ be the subring generated by $rx$, where $r \in R$ is some non-invertible element. Then $x$ is not integral over $S$

Consider $R[x]$ and let $S$ be the subring generated by $rx$, where $r \in R$ is some non-invertible element. Then I want to show that $x$ is not integral over $S$ I'm not seeing why this is the ...
2
votes
1answer
198 views

Ways to prove that an extension of Noetherian modules is Noetherian

Let $M$ be a module and $N$ a submodule of $M$. If $N$ is Noetherian and $M/N$ is Noetherian, so is $M$. This is usually proven like this: Let $(A_n)$ be an ascending series of submodules of $M$. ...