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

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3
votes
1answer
172 views

How to find a generating set for the lattice ideal when one has a lattice basis?

This is an example from Miller, Sturmfels "Combinatorial Commutative Algebra". Let $\pi:\mathbb{Z}^3 \rightarrow \mathbb{Z}$ be the group homomorphism defined by the matrix $\left(\begin{array}{ccc} ...
4
votes
2answers
115 views

If $P$ is a prime ideal of $R$, $\sqrt{P^{n}}=P\ \forall n\in\mathbb{N}$?

Let $R$ be a commutative ring. If $P$ is a prime ideal of $R$, $\sqrt{P^{n}}=P\ \forall n\in\mathbb{N}$?
2
votes
1answer
84 views

Modules over $\mathbb{C}[x]$

Suppose I have a module $M$ over $\mathbb{C}[x]$ such that $M\otimes_{\mathbb{C}[x]}\mathbb{C}[x]/(x-\lambda)\cong\mathbb{C}$ for all $\lambda \in \mathbb{C}$. What kind of general hypotheses could I ...
0
votes
1answer
92 views

Poincaré series of an $A$-module

I'm studying Atiyah and Macdonald's "Introduction to Commutative Algebra" and I'm having some problems computing the Poincaré series of an $A$-module. Even the Example after $11.3$, which is dealt ...
1
vote
1answer
43 views

What is the name of this factor-algebra?

In the polynomial algebra $k[x_1,x_2,\ldots, x_n]$ consider an ideal $I$ generated by the polynomials of the form $x_i^k-x_i$, $i=1 \ldots n$ and $k=2,3,\ldots.$ Consider the quotient algebra ...
0
votes
1answer
43 views

Recovering ring information from localizations [closed]

I'm curious if such a statement is true: Let $R$ be a commutative ring with unity. Then if $x \in R$ and $x = 0$ in $R_p$ for all primes $p$, where $R_p$ is the localization of $R$ at prime $p$, ...
0
votes
2answers
51 views

Showing that $x$ is algebraically independent in $k[x,y]/(xy-1)$

I am studying Noether's normalization lemma, and I couldn't find a way to prove that $\bar{x}$ is algebraically independent in $k[x,y]/(xy-1)$. Can anyone give me a hint or help me please?
1
vote
2answers
699 views

Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring. [duplicate]

Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring. Thanks for any insight.
4
votes
2answers
145 views

If $M$ is finitely generated then $M/N$ is finitely generated.

Let $N$ be a submodule of $R$-module $M$. Prove that if $M$ is finitely generated then $M/N$ is finitely generated. Help me some hints.
1
vote
0answers
55 views

Normal domain is equivalent to integrally closed domain. Is it true?

Normal domain is equivalent to integrally closed domain. Is it true? Can anyone tell me?
6
votes
1answer
84 views

Does $1 \otimes_A b= b\otimes_A 1$ imply $b \in A$?

Supppse $B$ is a faithfully flat $A$-algebra, and $b$ an element in $B$. Does $1 \otimes_A b= b\otimes_A 1$ in $B\otimes_A B$ imply $b \in A$?
0
votes
0answers
52 views

What are some important examples of algebras which are not free modules?

Obviously I can take any commutative ring and regard it as an algebra over any of its subrings, and in general this won't result in the original ring being a free module. However, in cases where we ...
0
votes
1answer
34 views

$x\in R$ is a unit iff $(x+RadR)$ is a unit in $R/RadR$

Let $R$ be a commutative ring. Prove that the element $x\in R$ is a unit iff $(x+RadR)$ is a unit in $R/RadR$. ($RadR$ is Jacobson radical of $R$) Thanks in advanced.
1
vote
2answers
63 views

Local algebras over algebraically closed fields

How can one prove that every element $x$ of a finitely generated local commutative algebra $A$ with identity over an algebraically closed field $K$ is unit or nilpotent? Of course, this is ...
1
vote
1answer
51 views

Is the image of $\mathrm{Spec}(\varphi)$, where $\varphi$ is a localization, always open?

Let $R$ be a commutative ring and $S \subseteq R$ multiplicatively closed subset. Consider $\varphi: R \rightarrow S^{-1}R$ the localization of $R$ by $S$. Then it can be shown that the corresponding ...
2
votes
2answers
102 views

Is the ring of polynomial invariants of a finite perfect group an UFD?

Let $G$ be a finite group. $G$ acts on $\mathbb K[x_1,...,x_n]$ by automorphisms fixing $K$. $\mathbb K[x_1,...,x_n]^G=\{ T\in \mathbb K[x_1,...,x_n],\forall \sigma \in G, T^{\sigma}=T\}$ is the ring ...
1
vote
3answers
40 views

$\mathbb{Z}_{p}\left[x\right]/\left\langle f\right\rangle $ is a semilocal ring.

Let $p$ be a prime, $f$ be a nonconstant polynomial that is contained in $\mathbb{Z}_{p}\left[x\right]$. Prove that $\mathbb{Z}_{p}\left[x\right]/\left\langle f\right\rangle $ is a semilocal ring.
1
vote
0answers
36 views

Explicit display of the contraction of an ideal in polynomial ring extensions

$K$ is a finite Galois extension of $k$, $I=(f_1,\dots,f_m)$ an prime ideal in $K[X_1,\dots,X_n]$, then what is $I\cap k[X_1,\dots,X_n]$? Is it ...
0
votes
1answer
122 views

definition of the grade of an ideal

Definition: Let $A$ be a Noetherian ring and $M$ a non-zero finite $A$-module. Then the grade of $M$ is defined as $grade(M) = \inf_i \left\{Ext^i(M,A) \neq 0\right\}$. Matsumura (CRT) p.132, ...
1
vote
1answer
135 views

height of ideal generated by regular sequence

Let $(A,m)$ be a local Noetherian ring, $M$ a finitely generated non-zero $A$-module and $a_1,\cdots,a_r$ an $M$-sequence. If $M=A$, then by using the Hauptidealsatz we can prove that ...
5
votes
1answer
254 views

Ring of formal Laurent series over an Euclidean domain.

If $R$ is a commutative Euclidean domain, then the ring $R((x))$ of formal Laurent series will also be an Euclidean domain. (Here we can assume that $R((x))$ is a commutative integral domain.) ...
2
votes
1answer
149 views

About localization of a finitely generated $k$-algebra

I've some questions about localization. My problem is the following. Let $A$ be a finitely generated domain over a field $k$. We know by hypothesis that its fraction field $K(A)$ is a finite ...
1
vote
1answer
84 views

Ideals in commutative noetherian rings with unique prime ideal

Let $R$ be a commutative noetherian ring with $1$ having only one prime ideal $\mathfrak{P}$. It follows that $\mathfrak{P}^n = 0$ for some integer $n$. Can we say that every proper ideal in $R$ is a ...
4
votes
1answer
105 views

Prime ideals not containing the conductor of a finite extension of a domain

Let $A\subseteq B$ be integral domains, where $B$ is contained in the quotient field $K$ of $A$, and $B$ is finitely generated as an $A$-module. Let $\mathfrak{f}=\{a\in A\mid aB\subseteq A\}$ denote ...
7
votes
1answer
105 views

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

So I'm stuck with this problem. Let $A$ be a commutative ring (with unit). I have several questions that are really close to each other. 1) Let $M$ be a finitely generated module over $A$. Can we ...
0
votes
0answers
40 views

cartesian product of integral $A$-algebras

Given two $A$-algebras $B_1$ and $B_2$ by the homomorphisms $f_i: A \to B_i$. Suppose that $B_1$ and $B_2$ are integral $A$-algebras. Then $B:=B_1\times B_2$ is an $A$-algebra by ...
2
votes
2answers
541 views

Prime ideal in a polynomial ring over an integrally closed domain

Let $R$ be an integrally closed domain. Then $R[x]$ is also integrally closed. Let $P$ be a prime ideal of $R[x]$ with the property that $P\cap R=\left\{ 0\right\} $ and $P$ contains a monic ...
6
votes
1answer
106 views

Principal ideal of an integrally closed domain

Let $R$ be an integrally closed domain and $S$ be an integral domain that contains $R$. Assume that $a\in S$ is integral over $R$. Prove that $I=\left\{ f\left(x\right)\in R\left[x\right]\mid ...
0
votes
1answer
330 views

An irreducible element is not a zero divisor

Let $R$ be a commutative ring, and let $a \in R$ be an irreducible element. Prove that $a$ is not a zero divisor. I need help proving this. I know that $b \in R$ is a zero divisor if there is ...
5
votes
1answer
204 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?
2
votes
3answers
177 views

A noetherian ring $R$ which is commutative integral domain but not a PID?

I am looking for an example of a ring $R$ which is a commutative and Noetherian integral domain but not a PID. Thanks.
2
votes
1answer
80 views

Primary decomposition in a quotient ring

Suppose $I$ and $J$ are two ideals in a polynomial ring $R=\mathbb{Q}[x_1,\dots,x_n]$, what's the relation between the primary decomposition of $I$ in the quotient ring $R/J$ and the primary ...
2
votes
1answer
234 views

Noetherian module implies Noetherian ring

Let $M$ be a Noetherian $A$-module where $A$ is a commutative ring with identity. Let $I_1\leq I_2\leq\cdots$ be an ascending chain of ideals of $A$. Then $I_1M\leq I_2M\leq\cdots$ is an ascending ...
-3
votes
2answers
77 views

What is the Jacobson radical of $\mathbb{Z}_{(p)}$?

Let $\mathbb{Z}_{(p)}$ be the ring of integers localized at a prime ideal $(p)$. Could anyone tell me what its Jacobson radical is?
2
votes
1answer
110 views

Polynomial ring with uncountable indeterminates

In Rotman's Advanced Modern Algebra second edition (2010), on page 883 (or on page 905 in its first edition (2002)), in the proof of the existence of localization of a commutative ring $R$ on its ...
2
votes
2answers
101 views

A Noetherian module annihilated by a power of maximal ideal must has finite length.

Let $M$ be a Noetherian $R$-module and $P^kM=0$ from some maximal ideal $P$ of $R$ and some integer $k$. How to show that $M$ has finite length? The length of a module is defined to be the ...
3
votes
2answers
132 views

Subrings of rationals are noetherian

Let $R$ be a subring of $\mathbb{Q}$. Then $R$ is noetherian. Since $1 \in R$ it follows, that $\mathbb{Z} \subseteq R \subseteq \mathbb{Q}$. Consider an ideal $I \subseteq R$. Then $I':=I ...
0
votes
0answers
33 views

Noetherianty of ring of continuous functions [duplicate]

Let $X$ be a topological space. Is there any topological property on $X$ that be equivalent to $C(X,\mathbb R)$ be noetherian ring?
1
vote
1answer
53 views

Zero divisor graph of commutative ring

Let $R$ be a commutative ring with unity and let $Zd(R)=P_1∪P_2$, where $P_1$ and $P_2$ are maximal (prime) ideals in $Zd(R)$. Let $P_1∩P_2≠{0}$. Show that the diameter of the zero divisor graph ...
3
votes
0answers
75 views

A scheme $X$ finite type over $R$, but $\Gamma(X,O_X)$ is not finitely generated $R$-algebra

A scheme $X$ over $R$ is of finite type if $X$ is quasi-compact and for all affine subsets $U \subset X$, $\Gamma(U,O_X)$ is a finitely generated $R$-algebra. Is there an example of scheme $X$ ...
0
votes
1answer
144 views

When is the ring of continuous functions Noetherian?

Let $X$ be a topological space. Is there any topological property on $X$ that be equivalent to $C(X,\mathbb R)$ being a noetherian ring?
0
votes
0answers
75 views

System of congruences that do not satisfy CRT assumptions (via algorithm)

Let $x_i,a_i\!\in\!\mathbb{Z}$. The following procedure solves a system of congruences $$x \equiv x_i\pmod{a_i}\;\;\text{ for }i\!=\!1,\ldots,n$$ when $a_i$ are pairwise coprime. Assume that ...
2
votes
0answers
37 views

Why does taking the residue commute with the discriminant if $B$ is free over $A$ and not in general?

Let $K$ be a number field, $A$ its ring of integers, $L/K$ be a finite extension, and $B$ the integral closure of $A$ in $L$. Lemma (residue of the discriminant): Assume $B$ is free over $A$, let $a$ ...
3
votes
1answer
156 views

Module has finitely generated projective resolution

Let $M$ be a finitely generated module (over a local noetherian ring $(R,\mathfrak m))$ such that the projective dimension of $M$ is finite $(pd\ M=n<\infty)$. We know that i) There is a free ...
3
votes
2answers
991 views

The twisted cubic is an affine variety.

Let $Y=\{(t,t^2,t^3)\mid t\in k\}$ be the twisted cubic curve. I'm trying to prove this curve is a variety, i.e., it's irreducible and affine algebraic set. The easier part is to prove the ...
2
votes
1answer
89 views

If $X$ is a cone, show that $I(X)$ is homogeneous.

The exercise is 1.3(3) from HP Kraft, "Appendix A: Basics from Algebraic Geometry." If a closed subset $X\subseteq \mathbb C^n$ is a cone, show that $I(X)$ is generated by homogeneous functions. ...
4
votes
1answer
123 views

If $A[[x]]$ is Noetherian, will $A$ be Noetherian?

Let $A$ be a commutative ring with unit and $A[x]$ be the polynomial ring with coefficients in $A$. Then $A$ is Noetherian if and only if $A[x]$ is Noetherian. This is obtained by Hilbert Basis ...
2
votes
2answers
194 views

How can I prove a coordinate ring is not isomorphic to a polynomial ring

Let $Z$ be the plane curve $xy=1$. I would like to prove that $A(Z)$ is not isomorphic to a polynomial ring in one variable over $k$. I'm already prove that the coordinate ring is ...
1
vote
0answers
59 views

Diagonalization and integrality over power/Laurent series rings

The question itself might seem overly specialized and technical (and by this, I mean boring), and it is quite difficult to explain the real motivation for it (but there is!), so I will try to give ...
3
votes
1answer
76 views

Is every prime ideal in $\Pi_{n=1}^{\infty}{k}$ maximal?

Suppose k is a algebraic closed field, is every prime ideal $\mathfrak{p}$ in the product ring $\Pi_{n=1}^{\infty}{k}$ maximal?