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

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8
votes
2answers
446 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
162 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
167 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
223 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
126 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
545 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
168 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?
11
votes
2answers
636 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
57 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
115 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
121 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
795 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
428 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
142 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
176 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
230 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
221 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
191 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. ...
10
votes
1answer
516 views

The ring of integers of a number field is finitely generated.

For a number field $K$, we define the ring of integers of $K$ to be $$\mathcal{O}_K:=\{x\in K\big|\ (\exists f\in\mathbb{Z}[X])(f\ \text{ is monic and } f(x)=0)\}.$$ Is there any easy way to see from ...
4
votes
4answers
180 views

Why Does Finitely Generated Mean A Different Thing For Algebras?

I've always wondered why finitely generated modules are of form $$M=Ra_1+\dots+Ra_n$$ while finitely generated algebras have form $$R=k[a_1,\dots, a_n]$$ and finite algebras have form ...
8
votes
4answers
801 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 ...
9
votes
4answers
637 views

Equality of two notions of tensor products over a commutative ring

Let $R$ be a ring (not necessarily commutative), let $M$ be a right $R$-module and let $N$ be a left $R$-module. Then the tensor product $M \otimes_R N$ is an abelian group satisfying the universal ...
6
votes
2answers
192 views

Noetherian domains with finitely many primes

For any domain $A$ let $A^\times$ be its group of units. Let $A$ be a noetherian domain with only finitely many prime ideals, and field of fractions $K$. Is the group $K^\times/A^\times$ finitely ...
1
vote
0answers
138 views

linear functors commute with finite direct sums (in R-mod)

I'm working on Exercise 8.16 from http://web.mit.edu/18.705/www/syl11f.html. In particular: Let F : ((R-mod)) → ((R-mod)) be a linear functor. Show that F always preserves finite direct sums. ... ...
5
votes
3answers
330 views

A question about Cohen-Kaplansky domains

Let $A$ be a domain. Recall that $A$ is Cohen-Kaplansky (or CK) if (CK) any nonzero nonunit of $A$ is a product of irreducible elements, and there are only finitely irreducible elements up to ...
2
votes
2answers
337 views

Discrete valuation ring associated with a prime ideal of a Dedekind domain

Let $A$ be a Dedekind domain. Let $K$ be the field of fractions of $A$. Let $P$ be a non-zero prime ideal of $A$. Let $v_P$ be the valuation of $K$ with respect to $P$. Then the localization $A_P$ of ...
4
votes
4answers
135 views

Need help to show $R/I$ is not necessarily flat over $R$

Let $R$ be a ring with unit and $I$ an ideal in $R$. I want to show that $R/I$ is need not be flat over $R$, but I do not know how to come up with a counter-example. Any hint is appreciated.
1
vote
2answers
84 views

About absolute convergence and completeness in rings

EDIT: Let R be a commutative ring with unit ring and $I$ a maximal ideal in R. The completion of R with respect to $I$ is the inverse limit of the factor rings $R / I^k$ under the usual quotient maps. ...
26
votes
4answers
3k views

Commutative non Noetherian rings in which all maximal ideals are finitely generated

In commutative rings we have the following Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated. From this Theorem I am looking for commutative rings $R$ in which ...
5
votes
1answer
162 views

Original article on the Grothendieck group

Is there someone who knows the title of the original publication of Grothendieck on the construction of the Grothendieck group? Thanks in advance.
1
vote
1answer
482 views

Annihilator of a simple module

Let $R$ be a finitely generated commutative ring and $C$ an $R$-algebra ($C$ is not necessarily commutative). Assume that $C$ is a finitely generated $R$-module. If $S$ is a simple $C$-module, then ...
3
votes
1answer
111 views

Behavior of associated primes under inclusion

Let $A \subset B$ be an inclusion of commutative rings inducing $f: \text{Spec}(B) \rightarrow \text{Spec}(A)$. Must it be the case that $\text{Ass}(A) \subset f(\text{Ass}(B))$? If this isn't true ...
3
votes
1answer
797 views

On modules over polynomial rings

Let $\mathbb{A}$ be a polynomial ring in $n$ variables over an algebraically closed field $\mathbb F$. Given a maximal ideal $\mathfrak{m}$ of $\mathbb A$, consider the quotient ...
10
votes
1answer
246 views

When is $\mathbb{Z}$ a flat $\mathbb{Z}G$-module?

Suppose that $\mathbb{Z}$ is a flat $\mathbb{Z}G$-module for a group $G$. Question: Is $G$ the trivial group ? Nb. I know that the question can be answered affirmatively if $G$ is finitely ...
5
votes
2answers
137 views

Finite presentation of algebra of invariants

(1) Let $R$ be a ring, let $A$ be a finitely presented $R$-algebra, and let $G$ be a finite group of $R$-automorphisms of $A$. Is the algebra of invariant $A^G$ finitely presented over $R$? I can ...
2
votes
1answer
71 views

Extending $\phi: A \rightarrow \Omega$ to $A[x] \rightarrow \Omega$ where $A$ is integral domain and $x$ transcendental over $A$

Let $A \subseteq B$ be integral domains and let $\phi:A \rightarrow \Omega$ be a homomorphism of $A$ into the infinite algebraically closed field $\Omega$. Let $x \in B$ and suppose that $x$ is ...
0
votes
1answer
142 views

Irreducible polynomial over an algebraically closed field(2)

Suppose $k$ is algebraically closed field. And $p(x_1,\ldots,x_n)\in k[x_1,\ldots,x_n]$ is an irreducible polynomial. I wonder to show $p(x_1,\ldots,x_n)+z\in \overline{k(z)}[x_1,\ldots,x_n]$ is ...
1
vote
1answer
133 views

Artinian local algebras over a complete local noetherian ring.

So let $\Lambda$ be a complete local noetherian ring. The author claims that $\Lambda[t]/(t^i)$ is an Artinian local $\Lambda$-algebra with the same residue field as $\Lambda$. I don't see this. For ...
1
vote
1answer
79 views

Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal.

Let $R$ be an order, that is, a ring of which the additive group is finitely generated and torsion-free. As an exercise I am trying to prove that ...
2
votes
1answer
353 views

Tensor product of algebras

What is the tensor product $M_n(L)\otimes_K L$, where $L/K$ is a quadratic extension? Let $K$ be a field of characteristic $0$, $L/K$ a quadratic extension. Let $\rho\in \operatorname{Gal}(L/K)$ ...
13
votes
2answers
2k views

Tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$

I want to understand the tensor product $\mathbb C$-algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$. Of course it must be isomorphic to $\mathbb{C}\times\mathbb{C}.$ How can one construct an ...
7
votes
1answer
292 views

Extension of Homomorphisms (Lang, Atiyah and McDonald)

Let $A$ be a subring of a field $K$, and suppose that $A$ is a local ring with maximal ideal $\mathfrak{m}$. Let $x \in K, \, x \neq 0$. Let $\phi: A \rightarrow L$ be a homomorphism of $A$ into the ...
4
votes
4answers
1k views

Let $M$ be a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit. Show that $R$ is a local ring with maximal ideal $M$

So I'm trying to prove that if $M$ is a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit, then $R$ is a local ring with maximal ideal $M$, that is to say $R$ has a unique maximal ...
4
votes
0answers
213 views

Construction of graded rings and modules

In Algebraic Geometry and Homological Algebra - as far as I know - we often consider graded rings and modules so as to encode more information, say, some sort of (computational) complexity. For ...
11
votes
1answer
2k views

Inverse limit of modules and tensor product

Let $(M_n)_n$ be an inverse system of finitely generated modules over a commutative ring $A$ and $I\subset A$ an ideal. When is the canonical homomorphism $$\left(\varprojlim\nolimits_n ...
3
votes
1answer
549 views

The height of a principal prime ideal

A formal consequence of Krull's principal ideal theorem is the following: If $A$ is a Noetherian ring, and $I$ is an ideal generated by $r$ elements, then any prime ideal which is minimal among those ...
3
votes
0answers
101 views

Computation of determinant of a matrix with elements from an arbitrary commutative ring

The cofactor formula for computing the determinant of a matrix is applicable when elements of the matrix are from a commutative ring. However, the complexity of this method is extremely high and I ...
3
votes
2answers
450 views

Projective modules over polynomial rings

Let $M$ be a finitely generated right module over a polynomial ring $R$ (in any number of variables) over a field $\mathbb F$. Given a maximal ideal $\mathfrak{m}$ of $R$, consider ...
7
votes
1answer
187 views

Two corollaries in Lang's Algebraic Number Theory.

I'm having difficulty understanding the relationship between two corollaries in Lang's Algebraic Number Theory, on page 16 for those with the book. They can also be found in his Algebra. The first ...
3
votes
3answers
91 views

What can $\operatorname{Hom}\left(\prod_i M_i, N\right)$ look like?

It's easy to see that $\operatorname{Hom}\left(\bigoplus_i M_i, N\right) = \prod_i \operatorname{Hom}(M_i, N)$. However, there are a couple of ways this can conceivably fail if we replace the ...