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

learn more… | top users | synonyms (1)

11
votes
2answers
391 views

Minimal systems of generators for finitely generated algebras over commutative (graded) rings

Let $S$ be some base ring (a commutative ring or even just a field), and $R$ a commutative ring containing $S$ which is finitely generated (as an algebra) over $S$. What conditions guarantee that any ...
11
votes
2answers
2k views

Show $\mathbb{Z}[\sqrt{6}]$ is a Euclidean domain

I'm attempting to modify the proof the $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain to prove a similar result for $\mathbb{Z}[\sqrt{6}]$. The idea is to prove that $\mathbb{Q}[\sqrt{6}]$ is Euclidean ...
11
votes
2answers
90 views

When is $X_1^{a_1} \cdots X_n^{a_n}-1$ irreducible?

Let $F$ be a field, and $a_1, ... , a_n \geq 1$ integers. When is the polynomial $$f = X_1^{a_1} \cdots X_n^{a_n}-1$$ irreducible in $F[X_1, ... ,X_n]$? I believe this should be the case if and ...
11
votes
1answer
161 views

Examples of rings whose polynomial rings have large dimension

If $A$ is a commutative ring with unity, then a fact proved in most commutative algebra textbooks is: $$\dim A + 1\leq\dim A[X] \leq 2\dim A + 1$$ Idea of proof: each prime of $A$ in a chain can ...
11
votes
1answer
1k views

Intuition behind Hilbert's Nullstellensatz

maybe that's a pointless question, however I'm having problems in "understanding" (accepting) the Hilbert's Nullstellensatz. I understand the proof, however I cannot understand the concept in a more ...
11
votes
1answer
740 views

Homomorphism of local rings

Let $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ be local Noetherian rings. Suppose that $\phi : A\rightarrow B$ is a map such that $\phi(\mathfrak{m}) \subset \mathfrak{n}$ and suppose $A/\mathfrak{m}...
11
votes
1answer
151 views

DVR, power series expansion.

Let $A$ be a discrete valuation ring with quotient field $K$, maximal ideal $\mathfrak{m}$, uniformizing parameter $t$. Let $k = A/\mathfrak{m}$, so $k$ is a field. How do I show that there is a ...
11
votes
0answers
134 views

Study of rings of the form $R+I$

In my life I saw lots of ways of constructing rings: polynomial rings, quotient rings, localizations, endomorphism rings, rings of fractions, integral closure of a ring, center of a ring, etc... These ...
10
votes
2answers
1k views

How does Hilbert's Nullstellensatz generalize the “fundamental theorem of algebra”?

What is Hilbert's Nullstellensatz in the sense of the generalization of "fundamental theorem of algebra"? I've seen that in some texts it was referred to as the generalization of the fundamental ...
10
votes
3answers
1k 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 ...
10
votes
3answers
2k views

Nil-Radical equals Jacobson Radical even though not every prime ideal is maximal?

Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)? Are ...
10
votes
1answer
690 views

Prove that $\mathbb{C}[x,y] \ncong \mathbb{C}[x]\oplus\mathbb{C}[y]$

Prove that $\mathbb{C}[x,y] \ncong \mathbb{C}[x]\oplus\mathbb{C}[y]$ $\mathbb{C}[x,y]$ is the polynomial ring of two variables over $\mathbb{C}$. I guess that we can consider images of $xy$ and $x+...
10
votes
2answers
589 views

A proof using Yoneda lemma

Martin Brandenburg pointed out elsewhere in the comments that he could give a one line proof, using the Yoneda lemma, of $$\frac{\mathbf{C}[x_1,\ldots,x_{n+m}]}{I(X)^e+I(Y)^e} \cong \frac{\mathbf{C}[...
10
votes
2answers
1k views

Direct summand of a free module

Let $M$, $L$, $N$ be $A$-modules and $M=N\oplus L$. If $M$ and $N$ are free, is $L$ necessarily free?
10
votes
2answers
1k views

Projective module over a PID is free? [duplicate]

A common result is that finitely generated modules over a PID $R$ are projective iff they are free. Is the same true that an arbitrary projective module over a PID is free? I can't find this fact ...
10
votes
2answers
397 views

Is every prime element of a commutative ring “veryprime”?

Let $R$ denote a commutative ring. Define a function $$\| : R \times R \rightarrow \mathbb{N} \cup \{\infty\}$$ such that $a \| b$ is the number of times $a$ divides $b$ (and include $0$ in $\mathbb{...
10
votes
3answers
2k views

What are the integers $n$ such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed?

I was recently reading about integral ring extensions. One of the first examples given is that $\mathbb{Z}$ is integrally closed in its quotient field $\mathbb{Q}$. Another is that $\mathbb{Z}[\sqrt{5}...
10
votes
1answer
1k views

Finitely generated algebra

I am getting the confusion with the definition of algebra. When we say $A$ is a finitely generated $R$- algebra then is that mean $A$ has a ring structure and finitely generated as an $R$-module. ...
10
votes
2answers
486 views

How to understand “tensor” in commutative algebra?

Tensor is sure an important concept in commutative algebra, but the definition is kind of abstract, so is there any way to understand it which is easier? Thanks advance! The definition I see is the ...
10
votes
2answers
1k views

When is the integral closure of a local ring also a local ring?

Suppose $A$ is a normal local domain contained in a field $K$. Suppose $B$ is the integral closure of $A$ in $K$. Under what conditions on $A$ is $B$ local?
10
votes
2answers
2k views

Prerequisites for Atiyah Macdonald

I am currently doing a one semester course on groups and rings where we have learned about (so far): Definitions of groups, subgroups, cyclic and normal subgroups, the symmetric group, homomorphisms, ...
10
votes
3answers
2k views

Commutative property of ring addition

I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring: Let $R\neq \...
10
votes
1answer
918 views

Affine scheme $X$ with $\dim(X)=0$ but infinitely many points

As the title says, I'm looking for an affine scheme of dimension zero, but with infinitely many points. At first I doubted that something like this could exist, and I still can't think of an example, ...
10
votes
2answers
790 views

Hartshorne Exercise 1.1 (a)

(see bottom for apology) Let $Y$ be the plane curve $y = x^2$ (i.e., $Y$ is the zero set of the polynomial $f = y - x^2$). Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $...
10
votes
1answer
574 views

Is a regular ring a domain

A regular local ring is a domain. Is a regular ring (a ring whose localization at every prime ideal is regular) also a domain? I am unable to find/construct a proof or a counterexample. Any help would ...
10
votes
3answers
103 views

Can we recover a compact smooth manifold from its ring of smooth functions?

It is well-known that if $X$ is a reasonably nice topological space (compact Hausdorff, say) then we can recover $X$ from the ring $C(X)$ of continuous functions $X\to\mathbb R$; see this MSE question ...
10
votes
2answers
913 views

Inverse image of the sheaf associated to a module

In Hartshorne, Algebraic geometry it's written, that for every scheme morphism $f: Spec B \to Spec A$ and $A$-module $M$ $f^*(\tilde M) = \tilde {(M \otimes_A B)}$. And that it immediately follows ...
10
votes
1answer
370 views

Question about whether axiom of choice is needed in this proof

Do I need axiom of choice in this proof here? I think not: at each step we choose one element from a set $N - \langle g_1, \dots, g_k \rangle $. So while there is indeed a countable number of sets ...
10
votes
2answers
928 views

Ideal class group of a one-dimensional Noetherian domain

Let $A$ be a one-dimensional Noetherian domain. Let $K$ be its field of fractions. Let $B$ be the integral closure of $A$ in $K$. Suppose $B$ is finitely generated $A$-module. It is well-known that B ...
10
votes
2answers
997 views

Infinite product of fields

The main source of inspiration for this question is this excerpt Recall: An ultrafilter on the set X gives you a maximal ideal in the ring of all real-valued functions, and these are the only ...
10
votes
2answers
197 views

Can an ideal in a commutative integral domain be its own square?

If $I$ is a non-zero proper ideal of a commutative integral domain, is it possible for $I$ to be its own square?
10
votes
1answer
688 views

A question on faithfully flat extension

This question arose while reading page 116 of Red Book by Mumford. Let $B$ be a faithfully flat extension of $A$. Can I claim that $b \otimes 1 = 1 \otimes b$ in $B\otimes_A B$ if and only if $b\...
10
votes
1answer
459 views

Primes in a Power series ring

Let $\mathbb Z$ be the ring of rational integers. Consider the power series ring $\mathbb Z[[x]]$. It is known that $\mathbb Z[[x]]$ is unique factorization domain. What are the primes in $\mathbb Z[[...
10
votes
1answer
1k views

A question about the tensor product of $\mathbb{Q}$

I'm reading this blog post about $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$ and I have two questions about it: Is a simple tensor a tensor that cannot be written as a sum of tensors? On the first ...
10
votes
1answer
1k views

(Ir)reducibility criteria for homogeneous polynomials

Suppose I have a homogeneous polynomial in at least 3 variables over some algebraically closed field (of characteristic 0, if need be). Question: How may I test — by hand — whether it is irreducible? ...
10
votes
2answers
508 views

Existence of prime ideals in rings without identity

Let $R$ be a commutative ring (not necessarily containing $1$). Say that $R$ is the trivial ring if it has trivial (zero) multiplication. If $R$ is the trivial ring, then $R$ has no prime ideals (as ...
10
votes
3answers
347 views

Deducing results in linear algebra from results in commutative algebra

Here are two examples of results which can be deduced from commutative algebra: Any $n\times n$ complex matrix is conjugate to a Jordan canonical matrix (can be proven using the structure theorem ...
10
votes
2answers
2k views

Finitely generated projective module

Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free? What I know about a finitely generated projective module $M$ over a PID $R$ is ...
10
votes
2answers
879 views

Rings whose spectrum is Hausdorff

Let $A$ be a commutative ring with $1$ and consider the Zariski topology on $\operatorname{Spec}(A)$. When will $\operatorname{Spec}(A)$ be a Hausdorff space? If $A$ has positive or infinite Krull ...
10
votes
1answer
414 views

Finitely generated modules over PID

Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus $ $B$ $\cong$ $C\oplus $ $D$ and $A\oplus $ $D$ $\cong$ $C\oplus $ $B$ . Prove that $A$ $\cong$ $C$ and $B$ $\...
10
votes
1answer
329 views

Conditions for $\sqrt{\mathfrak{a + b}} = \sqrt{\mathfrak{a}} + \sqrt{\mathfrak{b}}$

Let $A$ be a commutative ring with identity and, $\mathfrak{a}$ and $\mathfrak{b}$ ideals.I'm trying to find sufficient and necessary conditions for $\sqrt{\mathfrak{a + b}} = \sqrt{\mathfrak{a}} + \...
10
votes
1answer
498 views

Why does the structure theorem for finitely generated modules over PIDs fail for arbitrary modules over a PID?

The proof that I know of the theorem goes like this: Any module $M$ is a quotient of a free module $F$ (over any ring). Any submodule $K$ of a free module $F$ over a PID $R$ is a free module, so in ...
10
votes
1answer
272 views

If $I$ is a finitely generated ideal of $A[X]$, is $I\cap A$ necessarily finitely generated for a commutative unital ring $A$?

Let $A$ be a commutative ring with $1$ and $A[X]$ the ring of polynomials in one variable over $A$. Assume $I$ is a finitely generated ideal of $A[X]$. My question is Is $I\cap A$ necessarily ...
10
votes
2answers
763 views

Intersection of finitely generated ideals

Let $I$, $J$ be finitely generated ideals in a ring $A$ (commutative with identity). I know that the intersection need not be finitely generated: can somebody give me an example? Thanks.
10
votes
1answer
258 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 ...
10
votes
1answer
850 views

What conditions guarantee that all maximal ideals have the same height?

It fails in general that all maximal ideals in a commutative ring with unity have the same height. It's easy to construct a counter-example when the ring is NOT an integral domain (consider the ...
10
votes
2answers
613 views

Exactness of sequences of modules is a local property, isn't it?

It's well known, that passing to modules of fractions is exact, i.e. if $M'\xrightarrow{f} M\xrightarrow{g} M''$ is an exact sequence of $A$-modules ($A$ being a commutative ring with ...
10
votes
1answer
655 views

Construct ideals in $\mathbb Z[x]$ with a given least number of generators

How do you construct, for each $n\geq 1$, an ideal in $\mathbb Z[x]$ of the form $(a_1,a_2,\dots,a_n)$ with $a_i\in \mathbb Z[x]$ such that it is impossible to have $(b_1,b_2,\dots,b_m)=(a_1,a_2,\dots,...
10
votes
1answer
1k views

Adjointness of Hom and Tensor

Could someone provide me a link to the proof of the adjointness of Hom and Tensor. I did an extensive google search but could not find anything self contained that presented the proof in full ...
10
votes
2answers
274 views

A commutative group structure on $R\times R$ for a ring $R$

Let $R$ be a commutative ring. The Cartesian square $A=R\times R$ is endowed with the operation $(a_1,b_1)\circ(a_2,b_2)=(a_1+a_2,b_1+b_2+a_1a_2^2+a_1^2a_2)$ which turns $A$ into a commutative group....