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

learn more… | top users | synonyms (1)

2
votes
1answer
159 views

Prove that in a Noetherian ring, no invertible maximal ideal properly contains a nonzero prime ideal

Let $R$ be an integral domain which is Noetherian, let $P$ be an invertible maximal ideal, and let $Q<P$ be a prime ideal. How to show that $Q=0$? I have proved that $Q=QP$, and still haven't used ...
2
votes
1answer
162 views

$R\subseteq S$ integral extension and $S$ Noetherian implies $R$ Noetherian

The problem is as follows: Let $R\subseteq S$ be an integral extension and $S$ a Noetherian ring. Then show that for each $p\in \operatorname{Spec}R$, there are only finitely many $q\in ...
2
votes
0answers
103 views

Irreducible polynomials as formal power series

I'm studing the ring of formal series with complex coefficients $\mathbb{C}[[x]]$. I proved that the polynomial $y^2-x^3-x^2$ is irreducible in $\mathbb{C}[x,y]$ but reducible in $\mathbb{C}[[x,y]]$. ...
2
votes
1answer
116 views

for prime ideals, the intersection of the squares is the square of the intersection?

Here is something that i proved and i would appreciate feedback on my proof: Proposition: Let $A$ be a commutative Noetherian ring and $p,q \in \operatorname{Spec}(A)$. Then $p^2 \cap q^2 = (p\cap ...
2
votes
0answers
66 views

The action of a Galois group on a prime ideal in a Dedekind domain

Let $A$ be a commutative Dedekind domain and $K$ its field of fractions. Let $L/K$ be a finite Galois extension with Galois group $G$ and let $B$ be the integral closure of $A$ in $L$. If ...
2
votes
1answer
152 views

Extended ideals in power series ring

Let $A$ be a commutative ring with $1$ and consider the ring of formal power series $A[[X]]$. If $I \subseteq A$ is an ideal, let $I[[X]]$ denote the set of power series with coefficients in $I$. This ...
2
votes
0answers
244 views

Quick question on localization of tensor products

All rings are commutative with unit. Let $\rho:A\rightarrow B$ be a ring homomorphism. Suppose $\mathfrak q$ is a prime ideal of $B$, and let $\mathfrak p=\rho^{-1}(\mathfrak q)$. My question: Is ...
2
votes
1answer
204 views

Is a prime ideal in the polynomial ring over an algebraically closed field prime also in polynomial rings over extension fields?

Let $k \subsetneq K$ be algebraically closed fields, $I \subset k[x_1,\ldots, x_n]$ a prime ideal generated by the polynomials $f_1, \ldots, f_r$. Let $J \subset K[x_1, \ldots, x_n]$ be the ideal ...
2
votes
1answer
536 views

Finite morphisms of schemes are closed

I want to prove that finite morphisms of schemes are closed, but I cannot prove the affine case, namely: Given a finite morphism of rings $\varphi :B \to A$ prove that the induced morphism of ...
2
votes
2answers
520 views

Zero-dimensional ideals in polynomial rings

I have the following past exam paper question, a similar sort of question seems to come up every year. And I'm completely lost with it... Let $J$ denote the ideal in $\mathbb{Q}[x,y,z]$ generated ...
2
votes
1answer
127 views

Zero-dimensional ideals and finite-dimensional algebras [duplicate]

I encountered in the literature the term zero-dimensional ideal, however I can not find a relevant definition anywhere in Atiyah-MacDonald or Matsumura. In fact, I encounted the statement: $I$ is ...
2
votes
1answer
123 views

Constructing an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated [duplicate]

Let $R$ be a commutative ring and $M$ and $N$ be two finitely generated $R$-modules. I wanna construct an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated. It's well-known that if ...
2
votes
0answers
154 views

verifying if an ideal is prime

Let $R=\mathbb{Z}[\sqrt{-5}]$ and let $\mathfrak{i}=(2,1+\sqrt{-5})$ the ideal generated in $R$ by $2$ and $1+\sqrt{-5}$. I want to prove that $\mathfrak{i}$ is prime. So i considered the surjective ...
2
votes
2answers
304 views

Factorization of ideals in $\mathbb{Z}[\sqrt{5}]$

Consider the ring $R=\mathbb{Z}[\sqrt{5}]$. Let $I$ be the following ideal of $R$: $$I:=(3,1+\sqrt{5})$$ My teacher said that the following equation holds: $$I^2=(3)I,$$ but I actually can't ...
2
votes
1answer
203 views

Pure Submodules and Finitely Presented versus Finitely Generated Submodules

Let $A$ be a ring $M$ an $A$-module and $N$ a submodule. Definition: $N$ is called a pure submodule of $M$ if the sequence $0 \rightarrow N \otimes E \rightarrow M \otimes E$ is exact for every ...
2
votes
1answer
117 views

Injectivity is a local property

Let $R$ be a commutative noetherian ring, and let $M$ be an $R$-module. How can I show that if any localization $M_p$ at a prime ideal $p$ of the ring $R$ is injective over $R_p$ , then $M$ is ...
2
votes
2answers
179 views

Representation of an element of the field of fractions of a Dedekind domain as a fraction of elements which are relatively prime to a given ideal

This is a generalization of this question. Let $A$ be a Dedekind domain. Let $K$ be the field of fractions of $A$. Let $I$ be a non-zero ideal of $A$. Let $\alpha$ be a non-zero element of $K$ which ...
2
votes
1answer
141 views

Cohomology of Double Complex (extreme case)

Let $K^{p,q}$ be a double complex with ascending indices, i.e. the differentials are $d':K^{p,q} \rightarrow K^{p+1,q}$ and $d'':K^{p,q} \rightarrow K^{p,q+1}$. Suppose that $K^{p,q}=0$ if $p<0$ or ...
2
votes
3answers
254 views

Showing an ideal is a projective module via a split exact sequence

Let $R=\mathbb{Z}[\sqrt{-6}]$ and $I=(2,\sqrt{-6})$ the ideal generated by $2$ and $\sqrt{-6}$. I want to show that $I$ is a projective $R$-module by producing a short exact sequence that splits, ...
2
votes
1answer
192 views

Transcendence degree of $K[X_1,X_2,\ldots,X_n]$

Let $K$ be field. How do I proof that transcendence degree of $K[X_1,X_2,\ldots,X_n]$ is $n$? The set $\{X_1,X_2,\ldots,X_n\}$ is algebraically independent over $K$. So, I have to show that every ...
2
votes
3answers
229 views

Atiyah - Macdonald Exericse 9.7 via Localization

I am trying to show that the quotient of a Dedekind domain $A$ by an ideal $\mathfrak{a}$ is a PIR (principal ideal ring). Now by using the Chinese Remainder Theorem and the fact that a direct product ...
2
votes
1answer
106 views

How to make sense of the ideal norm in a localization?

This is a follow-up to this question: Localization of finite modules, or: compatibility of ideal norms with localization at a prime number Let $K$ be an algebraic number field and ...
2
votes
1answer
102 views

Making the fundamental theorem of Galois theory explicit

I encountered the present question when investigating that other recent question of mine. Let $x_1,x_2, \ldots, x_8$ be indeterminates. Let $s_1,s_2, \ldots s_n$ denote the elementary symmetric ...
2
votes
1answer
347 views

Hartshorne's proof of Proposition 2.5, Chapter II of his book Algebraic Geometry [duplicate]

Let $S = \sum_{n\ge 0} S_n$ be a graded commutative ring. Let $f$ be a homogeneous element of $S$ of degree $> 0$. Let $D_+(f) = \{\mathfrak{p} \in\operatorname{Proj} S\mid f \notin ...
2
votes
1answer
86 views

Describe invariant polynomials under action of commutative group of order eight.

I believe the question below should be fairly standard in invariant theory ; I hope someone more familiar with it than me can explain a bit more or point to a reference. Let $F$ be polynomial field ...
2
votes
1answer
141 views

The dual of a finitely generated module over a noetherian integral domain is reflexive.

As posted by navigetor23 in this question the dual of a finitely generated module over a noetherian integral domain is reflexive. Could you tell me how to prove it?
2
votes
0answers
111 views

Noetherian ring and primary decomposition result

I'm struggling with the following problem and I would appreciate some help if possible Let $R$ be Noetherian and let $I,J$ be ideals. Define $(I:J^{\infty}) = \bigcup_{n}(I:J^{n})$. (a) If $Q$ is ...
2
votes
3answers
367 views

Modules with projective dimension $n$ have not vanishing $\mathrm{Ext}^n$

Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with projective dimension $n$. Then for every finitely generated $R$-module $N$ we have $\mathrm{Ext}^n(M,N)\neq 0$. Why? By ...
2
votes
1answer
81 views

When the injective hull is indecomposable

Let $R$ be a ring and $M$ an $R$-module. Denote by $E(M)$ the injective hull of $M$. I was trying to prove that the following conditions are equivalent: 1) $(0)$ is meet-irreducible in $M$; 2) ...
2
votes
1answer
172 views

Integral Galois Extensions - Proposition 2.4 (Lang)

My question refers to the proof of Proposition 2.4, p. 341 in Lang's Algebra. Here is the context: Let $A$ be an integral domain, integrally closed in its field of quotients $K$ and let $L$ be a ...
2
votes
3answers
217 views

Isomorphism from $\widehat{G}$ to $\displaystyle \lim_{\longleftarrow} G/G_n$

Let $G$ be a topological abelian group and let $\widehat{G}$ denote its completion (i.e. equivalence classes of Cauchy sequences). Let $G_n$ be a descending sequence of subgroups, i.e. $G = G_0 ...
2
votes
1answer
177 views

Question about the $\mathrm{Tor}$ functor

Assume we want to define $\mathrm{Tor}_n (M,N)$ where $M,N$ are $R$-modules and $R$ is a commutative unital ring. We take a projective resolution of $M$: $$ \dots \to P_1 \to P_0 \to M \to 0$$ Now ...
2
votes
0answers
113 views

Generalized Hensel lifting in a Dedekind domain

The following theorem is known as generalized Hensel lifting(see here). Can we prove this without using $P$-adic completion? Theorem Let $A$ be a Dedekind domain. Let $P$ be a non-zero prime ideal of ...
2
votes
1answer
133 views

Example computation of $\operatorname{Tor_i}{(M,N)}$

Let $M = \mathbb Z / 284 \mathbb Z$ and $N = \mathbb Z / 2 \mathbb Z$. Can you tell me if my computation of $\operatorname{Tor_i}{(M,N)}$ is correct: (i) First we want a projective resolution of ...
2
votes
3answers
205 views

When is a module over $R$ and $S$ an $R \otimes S$-module?

Suppose $M$ is a module over $R$ and $S$, commutative rings with $1$. Under what conditions is $M$ also a $R \otimes S$-module? Also, a more general question: How to construct a structure of a $R ...
2
votes
0answers
152 views

Is an irreducible element still irreducible under localization?

Suppose $R$ is a domain. We say an element $x\in R$ is "irreducible" if $x=yz$ implies that $y$ or $z$ is a unit or both are units. I want to know if an irreducible element is still an irreducible ...
2
votes
2answers
317 views

Is there an analogue of the jordan normal form of an nilpotent linear transform of a polynomial ring?

Is there an analogue of the Jordan Normal Form for an infinite dimensional vector space? In general I think the answer is no. It's been awhile since I studied it, but I'm pretty sure something would ...
2
votes
1answer
104 views

Trouble with notation $I:J^{\infty}$

I am not sure I understand this notation correctly. The definition says, for a ring $R$ with $I,J$ ideals of $R$, we define $I:J^{\infty}=\cup_{i=1}^{\infty} I:J^i$. Now, $I:J$ is the set of elements ...
2
votes
1answer
437 views

characteristic of residue field and characteristic of the ring

I am looking for a proof of the result that appears at the end of page 418 here indicating the possibilities for equal characteristic and mixed characteristic. I could prove the following: Let ...
2
votes
2answers
154 views

Does a 1-dimensional noetherian domain obey cancellation law?

Someone calls this "order" which puzzles me, because I can't understand it's name. I was wondering whether these rings obey cancellation law, i.e. if $\mathfrak a\mathfrak b=\mathfrak a\mathfrak c$ ...
1
vote
1answer
56 views

Classifying complex conics up to isomorphism as quotient rings of $\mathbb{C}[x,y]$

This is a continuation of the question I asked here. The problem is now: Let $Q = ax^2 + bxy + cy^2 + dx + ey + f \in \mathbb{C}[x,y]$ be a general quadratic polynomial, that is, $a,b,c \not= 0$. ...
1
vote
1answer
69 views

Trouble showing flatness

Let $K$ be a field and $\pi: K[x]/(x^2) \to K$ be the ring homomorphism given by the valuation at $0$. I'm stuck in showing that $\pi^*(K)$ (the pullback) is not a flat module (over $K[x]/(x^2)$).
1
vote
2answers
114 views

Commutative ring is semisimple iff it's isomorphic to a finite direct product of fields.

I am trying to prove the following: Let $R$ be a commutative ring. Prove that $R$ is semisimple if and only if it is isomorphic to a direct product of a finite number of fields. Suppose $R$ is a ...
1
vote
2answers
53 views

Noether normalization for $k[x]_{x}$

According to the Noether normalization theorem, there exists a $k[t]$ where $t$ is an indeterminate and $k[t]\subseteq k[x]_{x}$ is a $k$-algebra extension so that $k[x]_{x}$ is a finitely generated ...
1
vote
0answers
59 views

Local complete intersection ring

Suppose $R$ is a local Noetherian complete intersection ring that is a finite $A$-algebra, where $A$ is a DVR. If the module of differentials of $R$ is free as an $R/\mathfrak a$-module for some ...
1
vote
0answers
66 views

$\operatorname{supp}(M) \subseteq \operatorname{supp}(N) \iff f_I(M)\subseteq f_I(N) $?

Let $ R $ be a commutative unital ring, $ I $ an ideal of $ R $, and $ M $ an $ R $-module. It has proven (here) that if $\operatorname{supp}(M) \subseteq \operatorname{supp}(N)$ then ...
1
vote
0answers
101 views

Surjectivity implies injectivity of finitely generated modules, localization?

The following problem is canonical: Suppose $A$ is a commutative unitary ring, and $M$ is a finitely generated module over $A$. If an endomorphism $f\colon M\to M$ is surjective, then it's also ...
1
vote
1answer
46 views

Both $R$ and $R/I$ are regular local rings

Let $R$ be a Noetherian local ring and $I$ is an ideal of $R$ such that both $R$ and $R/I$ are regular local rings. Could we deduce that $I$ is generated by an $R$-sequence? I know that a ...
1
vote
2answers
84 views

Grade of an ideal in a Noetherian ring

I want to prove that if $R$ is a Noetherian integral domain and $I$ is a nonzero ideal of $R$, then $I^{-1}=R$ if and only if $\operatorname{grade}(I)≥2$. For the "only if" part, I say ...
1
vote
2answers
134 views

Checking the maximality of an ideal

Let $R = \mathbb{Z}_{(2)}$ be the localization of $\mathbb{Z}$ at the prime ideal generated by $2$ in $\mathbb{Z}$. Then prove that the ideal generated by $(2x-1)$ is maximal in $R[x]$. Otherwise ...