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

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7
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1answer
685 views

Infinite coproduct of rings

I just learned from Wikipedia that coproduct of two (commutative) rings is given by tensor product over integers, and that coproduct of a family of rings is given by a "construction analogous to the ...
1
vote
2answers
105 views

What does it mean when elements act as units on a set?

I'm reading Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, and I am confused by one of his proofs. The setup is that $R$ is a commutative ring, $U$ is a multiplicatively closed ...
2
votes
1answer
62 views

Some questions about rings

All rings are commutative and unital Q1: what means notation $$A\cong A_1\times\ldots\times A_n?$$ Is it true that elements of $A_1\times\ldots\times A_n$ are collection of elements of $A_1,\ldots ...
3
votes
2answers
553 views

Being maximal ideal follows from being a kernel

The ideal of all polynomials in $k[x_1,\ldots,x_n]$ with zero constant term is maximal (since it is the kernel of the homomorphism $k[x_1,\ldots,x_n]\to k$ which maps $f\mapsto f(0)$). I ...
2
votes
1answer
83 views

Behaviour of conductor ideal

Let $f : A \to B$ be a homomorphism of finitely generated $k$-algebras, where $k$ is a field. Let $J_A$ and $J_B$ denote the conductor ideals of $A$ and $B$ respectively for the corresponding ...
2
votes
0answers
125 views

Clarifications on Noether Normalization

I finished reading Noether Normalization but given that I have almost no prior algebra training I am concerned that my understanding is wrong. (Starting masters in Mathematics but previously was an ...
2
votes
0answers
339 views

Field of Fractions for Commutative Ring with Identity

I'm trying to complete a problem that is walking me through creating a field of fractions $F$ from a commutative ring with identity (and no non-zero divisors) $R$. The construction first asks me to ...
1
vote
1answer
442 views

Modules of finite type over Noetherian rings

Let M be a unitary module of finite type over a commutative Noetherian ring R with a unit. Can M then always be represented as a quotient of a pair of free R-modules of finite-type?
0
votes
2answers
76 views

Elementary Question about Torsion Subgroups

Let $G$ be an abelian group which is killed by multiplication with the integer $n\geq 1$. Let $n=a\cdot b$ with $a,b \geq 1$ and relatively prime. Denote by $G[a]$ resp. $G[b]$ the $a$-resp. ...
1
vote
1answer
102 views

Relation between DVR's of a local domain and localizations of its integral closure.

$\textbf{1.}\,\,\,\,\,\,\,\,$ Let $(A,\mathfrak m_A)$ be a one dimensional local domain and let $B$ be its integral closure in the fraction field $L=\textrm{Frac}\,A$. Assume that $B$ is finitely ...
6
votes
1answer
329 views

polynomials over a local Artinian (or finite) ring

In this question " Zero-divisors and units in $\mathbb Z_4[x]$ " it looks like it has been shown that the set of zero divisors of $\mathbb{Z}_4[x]$ coincides with its nilpotent elements. Since the ...
1
vote
1answer
78 views

Some question on filtrations

Let $S$ be a noetherian ring and $M$ a finitely generated $S$-module. There exists a filtration by submodules $$0=M_0 \subseteq M_1\subseteq \cdots \subseteq M_r=M.$$ I want to show that for any ...
5
votes
1answer
556 views

Generators for the intersection of two ideals

Let $I=\langle a_1,\dots, a_s\rangle, J=\langle b_1,\dots, b_t\rangle$ be ideals of arbitrary commutative ring. Then we know that $I+J=\langle a_1,\dots, a_s, b_1,\dots, b_t\rangle, ...
1
vote
1answer
95 views

Flat module over A implies flat module over B

Let $A,B$ be commutative rings with 1, $\varphi:B\rightarrow A$ a ring homomorphism and M an $A$-module. If M is flat as an $A$-module, is it also flat as a $B$-module? (The structure of $B$-module is ...
3
votes
1answer
629 views

Tensor product of projective modules

(All rings are commutative) Let $A$ be a noetherian ring. Let $B$ be a noetherian $A$-algebra (not nessecerily f.g!) Suppose $M$ and $N$ are finitely generated projective $B$-modules (for my ...
4
votes
2answers
184 views

Galois theorem for ideals?

Let $R$ be an integrally closed integral domain with fraction field $K$. Let $L$ be a finite Galois extension of $K$ and let $\sigma_1,\dots,\sigma_n$ be the elements of $Gal(L/K)$. Let $S$ be $R$'s ...
5
votes
2answers
201 views

When does $\mathfrak{a}B\cap A = \mathfrak{a}$?

Let $A\subset B$ be rings, and let $\mathfrak{a}$ be an ideal of $A$. Under what circumstances does $\mathfrak{a}B\cap A = \mathfrak{a}$? More precisely, are there conditions on $A,B$ that guarantee ...
1
vote
1answer
209 views

relation of annihilators on exact sequence

Let $0\to M' \to M \rightarrow M^{''} \to 0$ be an exact sequence of modules. I want to show that ${\rm Ann}(M)= {\rm Ann}(M')\cap {\rm Ann}(M^{''})$. The "$\subset$" case I have shown, but I can't ...
0
votes
1answer
204 views

Semicontinuity of fiber dimension

The following two on commutative algebra are true? Let $S$ be a f.g. algebra over a field $k$. Let $e$ be an integer. Then (1) There is an ideal $I\subset S$ such that if $Q$ is a maximal ideal of ...
2
votes
1answer
310 views

Definition of a universal example

I'm not sure how the term is being used here: Let $R$ be a commutative ring and $X_1,\ldots, X_n$ indeterminates over $R$. Set $P = R[X_1, \ldots, X_n]$. Given a ring homomorphism $\phi: R ...
8
votes
1answer
506 views

Is the dualizing functor $\mathcal{Hom}( \cdot, \mathcal{O}_{X})$ exact?

In Hartshorne's Algebraic Geometry II.8.20.1 (page 182), he takes the dual of Euler sequence $$0 \rightarrow \Omega_{X/k} \rightarrow \mathcal{O}_{X}(-1)^{n+1} \rightarrow \mathcal{O}_{X} \rightarrow ...
1
vote
1answer
107 views

Filtrations of graded modules

Let $S$ be a graded noetherian ring and $M$ a finitely generated graded $S$- module. Then I know that there exists a filtration $$0=M_0 \subseteq M_1\subseteq \cdots \subseteq M_r=M$$ by graded ...
3
votes
1answer
74 views

Reference Request: Finite Length Modules

Where is a good place to read about the properties of the length function on modules over a commutative ring (in particular, quotients of the ring)? I'm looking mainly for basic properties. I've ...
1
vote
1answer
293 views

Dimension of a tensor product of affine rings

The dimension of a ring is defined as the length of a longest prime chain as usual. Let $A,B$ be affine rings over a field $k$. Then $$\dim A\otimes_k B = \dim A + \dim B.$$ How can we prove or ...
2
votes
1answer
87 views

Points lying on more than one irreducible component

Let $X$ be a scheme over a field $k$. Let $X = X_1 \cup \cdots \cup X_n$ be its decomposition into irreducible components. If a point $x \in X$ lies in more than one component, is it necessarily ...
1
vote
1answer
110 views

Extension of rings of integers always locally free

In his answer to this question, Andrea claims that if $A \subset B$ is an extension of rings of integers of number fields, $B$ is locally free over $A$. How can one prove this? Furthermore, I am ...
2
votes
0answers
91 views

Is there any other way to prove this statement?

Suppose that $v_1, ..., v_n \in R^m$, where $R^m = M_{m1}(R)$ for $R$ a commutative ring and $R^m$ a vector space. Let the matrix $A= [v_1 |...| v_n]$, the matrix whose $i$th column $= v_i$. I want to ...
2
votes
1answer
196 views

Some question on Hilbert polynomial

Let $S=k[X_1,\cdots,X_n]$ and $\{f_1,\cdots f_q\}$ be a $S$-regular sequence with ${\rm deg}(f_i)=a_i$. What is Hilbert polynomial of $S/ \langle f_1,\cdots,f_q\rangle$?
4
votes
3answers
138 views

Product of fractional ideals

Let $R$ be a Noetherian commutative ring. Let $I,J\subset K(R)$ be fractional ideals where $K(R)$ is the total quotient ring. Define $I^{-1}:=\{s\in K(R) : sI\subset R\}.$ Further suppose that $I$ is ...
0
votes
1answer
123 views

Example of a ring with zero dimension that is not Artinian (Atiyah - MacDonald)

Let $K$ be a field and consider the ring $A=K[x_1,x_2,\cdots]$ in countably infinite indeterminates and its ideal $\alpha=(x_1,x_2^2,\cdots,x_n^n,\cdots)$. Then Atiya and MacDonald in their ...
3
votes
2answers
358 views

A quotient field is finitely generated as a module?

When is the following true and why? If a commutative ring $A$ is a finitely generated module over a commutative ring $B$, then its quotient field $K(A)$ is a finitely generated module over $K(B)$. ...
3
votes
1answer
108 views

If $S$ is integral over integrally closed $R$, can $\mathfrak{a}S$ be principal without $\mathfrak{a}\triangleleft R$ being principal?

Let $R\subset S$ be integral domains, with $R$ integrally closed in its field of fractions, and $S$ integral over $R$. Suppose that the fraction field of $S$ is a finite Galois extension of the ...
8
votes
2answers
422 views

Classgroup of $\mathbb{Q}(\sqrt{2},\sqrt{-13})$

How would you compute the classgroup of the biquadratic number field $\mathbb{Q}(\sqrt{2},\sqrt{-13})$? I would prefer a method as "from scratch" as possible. Please avoid, if possible, quoting ...
2
votes
1answer
148 views

Is the integral closure of $k[[t]]$ in a finite extension of $k((x))$ necessarily a free module?

In Milne Prop 2.29, it is said that the integral closure $B$ of a PID $A$ in a separable finite extension of its fraction field is a free $A$-module. On the other hand, I have read here that if the ...
3
votes
0answers
152 views

Primary decomposition of large ideals

Short version: I'd like to do a primary decomposition of an ideal with 38 generators in a polynomial ring with 44 generators. However, the ideal seems far to large to naively decompose in, say, ...
3
votes
1answer
222 views

Proving something is the basis of a quotient space

Let $k$ be a field which does not have characteristic 2. Let $M$ be the free $k$-vector space generated by two elements $\{ c, x \}$. Let $T(M)$ be the tensor algebra of $M$ and let $I$ be the ideal ...
3
votes
3answers
123 views

Modules which are not free

I am trying to figure out the following: Given a principal ideal domain $R$ which is not a field, does there necessarily exist a module over $A$ which is not free? $A$ is a PID, so taking submodules ...
5
votes
5answers
485 views

Localizations of Dedekind Domains are Discrete Valuation Rings

I am trying to prove the following implication, and can't seem to find my way around all the equivalent definitions of Dedekind domains and DVRs: I have a ring $R$ with the following properties: 1) ...
1
vote
1answer
163 views

Projective dimension of tensor product $M\otimes M$

If the projective dimension of an $R$-module $M$ is finite, then can we say that projective dimension of tensor product $M\otimes M$ (as an $R\otimes R$-module) is finite?
10
votes
2answers
568 views

A non-nilpotent formal power series with nilpotent coefficients

Does anyone have an example of a formal power series $$p=a_0+a_1x+ a_2x^2 + \cdots \in R[[x]]$$ ($R$ is a commutative ring) all of whose coefficients $a_i$ are nilpotent in $R$ such that $p$ is not ...
0
votes
1answer
52 views

Binomial/Tensor Identity

Let $k$ be a a field and consider the space $k[x] \otimes_k k[x]$. I would like to verify the equation $$ \sum_{k=0}^{m+n} {m+n \choose k} x^k \otimes x^{(n+m)-k}= \sum_{i=0}^n \sum_{j=0}^m{n \choose ...
0
votes
2answers
100 views

What do I need to know to understand the completion of the field of rational functions of a non-singular projective curve?

So the title gives the jist of my question. Specifically, let $X$ be a non-singular projective curve, $P$ a point on $X$, $v_P$ the discrete valuation associated to the ring $\mathcal{O}_P$. Then I ...
1
vote
2answers
116 views

Syzygies and Free Resolutions

I am going to attend a workshop on Syzygies and Free Resolutions and want to prepare for that. I haven't had introduction to the subject but I studied first course in commutative algebra. I request ...
5
votes
2answers
725 views

Maximal ideal in a quotient ring

Consider the ring $A=\mathbb{C}[x,y]/(y^2-x^4+5x-4)$, and consider the ideal $\mathfrak{m}=(y, x+2)$. Is $\mathfrak{m}$ maximal? My sketched solution: consider an arbitrary element of $A$ and a ...
5
votes
2answers
376 views

Can the completion of a domain be a non-domain?

Suppose $R$ is a domain, finitely generated over an algebraically closed field, and $\mathfrak{m}\subset R$ is a maximal ideal. Is $\underleftarrow{\lim} R/\mathfrak{m}^n$ necessarily a domain?
6
votes
2answers
139 views

Which elements of a ring have zero differential?

Let $A$ be an algebra over $B$ (all rings commutative with a unit) and let $\Omega_{A/B}$ be the module of differentials of $A$ over $B$ with $d:A\to \Omega_{A/B}$ the universal derivation. Is there a ...
4
votes
2answers
167 views

Tensor product of faithful modules

In commutative algebra, is it true that the tensor product of two faithful modules is a faithful module? I have written for myself a proof for the case of finitely generated modules over reduced ...
3
votes
1answer
224 views

An $(R,S)$-bimodule is a left $R \otimes_k S^{\text{op}}$-module

Let $k$ be a commutative ring, and let $R,S$ be $k$-algebras. To me "$R$ is a $k$-algebra" means that $R$ is a $k$-module such that $a(rs)=(ar)s=r(as)$ for all $a\in k$ and $r,s \in R$. Let $M$ be a ...
2
votes
0answers
200 views

“$A$ is a ring with zero ideal the product of a finite number of maximal ideals . Then $A$ is Noetherian if and only if $A$ is Artinian.”

"Let $A$ be a ring in which the zero ideal is the product of the maximal ideals $\mathfrak{m}_i, \, i=1, \cdots, r$. Then $A$ is Noetherian if and only if $A$ is Artinian." This is Corollary 6.11, ...
3
votes
4answers
188 views

Why is $\mathbb{Z}_2 \ast \mathbb{Z}_2$ not a free group?

I recently start reading Hatcher's book for self-study. On page $46$ it gives such an example that is a free product and not a free group. I don't quite understand the explanation given in the book. ...