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

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4
votes
2answers
198 views

Set of associated primes of direct sum

Let $M$ be a module over a ring $R$. Let $\operatorname{Ass}(M)$ be the set of annihilator ideals $\operatorname{Ann}(x)$, which are prime, so $$\operatorname{Ass}(M) = \{\operatorname{Ann}(x) \mid ...
5
votes
2answers
1k views

Do the zero divisors form an ideal?

The problem is to prove that, in a commutative ring with identity, the set of ideals in which every element is a zero divisor has a maximal element with respect the order of inclusion, and that every ...
10
votes
2answers
768 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 ...
4
votes
1answer
275 views

Counting number of maximal ideals with given property?

There's a small theorem in algebra that I've been toying with. Let $R$ be integrally closed in its quotient field, and $E$ a splitting extension of the quotient field, and $S$ be the integral closure ...
3
votes
2answers
894 views

Finitely-generated $k$-algebras and their relationship with affine coordinate rings

Let $k$ be an algebraically closed field, $A = k[x_1, ... , x_n]$. For $Y \subseteq \mathbb A^n$, define $I(Y) = \{f \in A| f(P) = 0 \ \forall P \in Y\}$ Hartshorne's Algebraic Geometry, p. 4-5, says ...
5
votes
1answer
2k views

Prime ideals and irreducible ideals

I have the following definitions: An ideal $I$ is prime if, whenever $ab \in I$, either $a \in I$ or $b \in I$. An ideal $J$ is irreducible if, whenever $J = I_1 \cap I_2$ for ideals $I_1 $ and $ ...
3
votes
1answer
87 views

Needing a nudge with a Commutative Algebra Question

I have a commutative ring with identity $R$, and an $R$-module $M$. Next I have an $R$-submodule $N$ of $M$. Finally, I have a multiplicatively closed subset $S$ of $R$. I am asked to show that ...
4
votes
2answers
105 views

Show that $(A[X])^\times=A^\times+nil(A[X])$

I am currently reading through the book Basic Commutative Algebra, by Balwant Singh, wherein the exercise I.XVI reads like: Show that $(A[X])^\times=A^\times+nil(A[X])$. Here, for a ring ...
1
vote
4answers
152 views

Divisors in a Euclidean Domain

Let $R$ be a Euclidean Domain. If $(a,b)=1$, and $a$ divides $bc$, I would like a hint on how to prove that $a$ divides $c$.
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 ...
2
votes
3answers
806 views

How to determine whether a unique factorization domain is a principal ideal domain?

Could someone please provide an example of a unique factorization domain that is not a principal ideal domain? Furthermore, is there some way to determine whether a UFD is a PID?
5
votes
2answers
551 views

Prime ideals a very basic question

Let $A$ be a non zero commutative ring with identity. Show that the set of prime ideals of $A$ has minimal elements with respect to inclusion. I don´t know how to prove that, I can suppose that ...
3
votes
1answer
83 views

Integrality of quotient rings.

Suppose $A\subseteq B$ are commutative rings, $B$ integral over $A$. Let $\mathfrak{b}$ be an ideal of $B$, and set $\mathfrak{a}=A\cap\mathfrak{b}$. Apparently, $B/\mathfrak{b}$ is integral over ...
6
votes
1answer
206 views

A Definition in Eisenbud's Commutative Algebra

In Eisenbud's commutative algebra book, he defines a Weil divisor to be an element of the free abelian group generated by codimension 1 prime ideals of a ring $R$. He then goes on to define a Cartier ...
1
vote
2answers
78 views

If $f:A\to B$ is integral, then $S^{-1}f:S^{-1}A\to S^{-1}B$ is integral?

I'm trying to understand a proof that $f:A\to B$ is integral implies $S^{-1}f:S^{-1}A\to S^{-1}B$ is integral. Here $S$ is a multiplicative subset of $A$. Take $\alpha\in B$, since $B$ is integral ...
4
votes
2answers
212 views

If $B \ (\supseteq A)$ is a finitely-generated $A$-module, then $B$ is integral over $A$.

I'm going through a proof of the statement: Let $A$ and $B$ be commutative rings. If $A \subseteq B$ and $B$ is a finitely generated $A$-module, then all $b \in B$ are integral over $A$. Proof: ...
29
votes
6answers
2k views

Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$

This seems to be one of those tricky examples. I only know one proof which is quite complicated and follows by localizing $\mathbb{Z}[\sqrt[3]{2}]$ at different primes and then showing it's a DVR. ...
1
vote
2answers
240 views

a very basic question about commutative algebra

Sorry for asking this simple things... First I will put the results that are used. I have a stupid question about the proof of the corollary. I don´t know totally how to use the proposition 1.1, ...
4
votes
1answer
644 views

Is $(XY - 1)$ a maximal ideal in $k[[X]][Y]$?

Is $(XY - 1)$ a maximal ideal in $k[[X]][Y]$, and if so, how can I see it? It is at least prime because the generator is irreducible, and by the same argument it is maximal among all principal ...
3
votes
1answer
123 views

still torsion-free after nilredution?

This seems like it ought to be true and easy, but somehow I'm stymied. Let $A$ be a commutative ring (Noetherian if you like) and let $M$ be a finitely generated $A$-module. Suppose that $M$ is ...
2
votes
1answer
1k views

Krull dimension and localization

If we have a commutative ring with 1, what can we say about $\operatorname{dim} S^{-1}A$ for some multiplicative subset $S$, or more specifically, what happens if $S = A \backslash \mathfrak{p}$ for a ...
0
votes
4answers
199 views

Divisibility question for non UFD rings

Let $p$ be a prime element. I need an example of a domain in which $p^n$ divides $ab$ and $p^n$ does not divide $a$ and $p$ does not divide $b$. Obviously, the domain I'm looking for is not a UFD. ...
6
votes
1answer
288 views

Elementary Question About Hopf Algebras

Let $A$ be a smooth Hopf Algebra over a field $k$ with comultiplication map $\Delta$ and Augmentation Ideal $I$. Then I can regard the composition of $\Delta$ with the natural projection of $A ...
9
votes
1answer
316 views

Are projective rings over $\mathbb{Z}$ free?

Let $R$ be a ring. Say that an $R$-algebra $A$ is $R$-projective if it has the left lifting property with respect to surjections of $R$-algebras: that is, whenever $B \to C$ is a surjection of ...
2
votes
2answers
301 views

How can one interpret a module as a vector space? (in a specific circumstance)

I'm reading a remark in Atiyah, Macdonald - Introduction to Commutative Algebra and I can't really see what is going on. After giving the definition and a few examples of $R$-Modules (for a ring ...
3
votes
1answer
2k views

If $I$ is a maximal ideal of $R$, why is $R/I$ a field?

If $I$ is a maximal ideal of $R$, why is $R/I$ a field? I'm trying to use the fact that $I$ is maximal to show that $R/I$ only have ideals $\{0\}$ and $R/I$. Can anyone help me with this method. Many ...
3
votes
1answer
185 views

Extending an ideal of a polynomial ring to a polynomial ring with more indeterminates. Is it a tensor product?

Let $\mathbb{k}$ be a field, let $S'=\mathbb{k}[x_1,x_2,\dots,x_m]$, and let $I'\subseteq S'$ be an ideal. For some $n>m$, let $$S=\mathbb{k}[x_1,x_2,\dots,x_n]\ \ \ ...
6
votes
3answers
851 views

Dimension of $\mathbb Z[\sqrt 5]$

I know that every non-zero prime ideal in a Dedekind Domain is maximal. Since $\mathbb Z[\sqrt 5]$ is not integrally closed in its field of fractions $\mathbb Q[\sqrt 5]$, I wonder whether we can ...
4
votes
3answers
2k views

irreducibility of a polynomial in several variables over ANY field

The irreeducibility of a polynomial $f\!\in\!K[x_1,\ldots,x_n]$ in general depends on what the field $K$ is (for example, if $K=\mathbb{R}$, then $f=x_1^2+1$ is irreducible, but if $K=\mathbb{C}$, it ...
7
votes
1answer
194 views

Integral closure in the total ring of fractions

My question is linked with normalization of reduced algebraic curves that are not necessarily irreducible. Let $(A,\mathfrak{m})$ be a local reduced noetherian ring with Krull dimension $1$, let ...
1
vote
1answer
159 views

Queries on proof that every PID is a factorisation domain

I'm reading a proof from C. Musili's Rings and Modules that every PID is a factorisation domain. The author defines a factorisation domain as a commutative integral domain $R$ with a unit such that ...
2
votes
1answer
340 views

Finitely presented modules

I know that one can compute Fitting ideals of a finitely presented module (over a commutative ring with identity). However, are they the only invariants of such a module? In other words, my question ...
5
votes
1answer
358 views

definition of Krull dimension of a module

Let $R$ be a commutative ring with $1$. We know that the Krull dimension of $R$ is by definition the length of the longest chain of prime ideals of $R$. Now if $M$ is a $R$-module, the Krull ...
3
votes
2answers
449 views

$\operatorname{MaxSpec}(A)$ closed

If $A$ is an arbitrary commutative ring, is $\operatorname{MaxSpec}(A)$ closed as a subset of $\operatorname{Spec}(A)$? I wanted to think of a counterexample, but so far without success. I tried to ...
11
votes
1answer
739 views

Are all subrings of the rationals Euclidean domains?

This is a purely recreational question -- I came up with it when setting an undergraduate example sheet. Let's go with Wikipedia's definition of a Euclidean domain. So an ID $R$ is a Euclidean domain ...
7
votes
5answers
2k views

Integral domain that is not a factorization domain

I am looking for rings that are integral domains but not factorization domains, that is, rings in which it is not possible to express a nonzero nonunit element as a product of irreducible elements. ...
4
votes
3answers
1k views

Height and minimal number of generators of an ideal

How can I determine the height and the least number of generators of the ideal $ I=(xz-y^2,x^3-yz,z^2-x^2y) \subset K[x,y,z] $? I tried to calculate the dimension of the vector space ...
0
votes
2answers
142 views

integral cohomology groups of an aspherical manifold are isomorphic to the integral cohomology groups of it's fundemental group

(of corresponding dimensions). how can I prove this? I think my main stumbling block is my general ignorance of group cohomology.
0
votes
1answer
631 views

Quotient of a local regular ring

How can I prove this: Let $A$ be a local regular ring with maximal ideal $\mathfrak m$ and $x \in \mathfrak m-\mathfrak m^2$. Then $A/(x)$ is a regular ring. Prove also that if $x\in\mathfrak ...
6
votes
2answers
199 views

Computing contractions of ideals in Macaulay2

Does Macaulay2 compute contractions of ideals under ring homomorphisms. Specifically, if $R\subseteq S$ is a ring extension (say polynomial rings over $\mathbb{Q}$ which can be specified in M2) and ...
6
votes
2answers
1k views

$\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain, and its field of fractions is isomorphic to $\mathbb Q(t)$

How can I prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain? Also, I need to prove that its field of fractions is isomorphic to the field of rational functions ...
6
votes
1answer
456 views

Localizing a quotient ring $A/ \mathfrak{p}$

Let's assume $A$ is a commutative ring with $1$ and $\mathfrak{p} \subset A$ is a prime ideal. We shall consider $A/ \mathfrak{p}$ as an $A$-module, so there is a localization $(A/ ...
1
vote
1answer
181 views

Prime ideals in coordinate rings

Is there a way to characterise prime ideals in affine coordinate rings (i.e. quotients of polynomial rings). To be more specific, how can I say if principal ideals in such rings are prime or not in an ...
6
votes
2answers
748 views

Finite length modules over local rings

Let $A$ be a noetherian local ring and $M$ be an artinian and noetherian module over $A$. Does one know a priori anything about the structure of $M$? Furthermore: if one knows that the length of $M$ ...
2
votes
1answer
413 views

concrete isomorphisms of polynomial rings

Question 1: In A Singular Introduction to Commutative Algebra, page 222, there is written: How can I check that this isomorphism actually holds? I would really prefer a computational proof (using a ...
8
votes
5answers
1k views

Commutative rings without assuming identity

I was going through Exercises in Dummit&Foote, which does not assume identity in the definition of a ring, and reached the following exercise: Prove that in a Boolean ring ($a^2 = a$ for all ...
6
votes
1answer
630 views

The ideal $(x,y)$ is not a free $K[x,y]$-module

Given a field $K$ we have the polynomial ring $K[x,y]$ in $2$ variables, which is also a left module (over itself). How can we prove that the ideal $(x,y)$ is not a free module?
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vote
2answers
73 views

a simple question about a local ring, and modules.

I put the paragraph to clarify because it is a vector space. I have a question with the proposition, I don´t know why he concludes the red line assertion, only knowing that there exist a surjective ...
12
votes
1answer
228 views

An inverse limit

Let $k$ be a field. Consider the inverse limit $\varprojlim k[x,y]/(y\cdot x^n)$. I wonder if there is a nice description of this ring? Geometrically, we look at the union of the line $y=0$ ...
0
votes
4answers
114 views

there exist an extension such that this element is a zerodivisor?

Everyone knows that if in a ring A a unit a $\in$ A can´t be a zerodivisor. But could also be possible that "a" not be a zero divisor ( i.e does not exist a nonzero x $\in$ A , such that $ax=0$) but ...