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

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4
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242 views

Converse to Chinese Remainder Theorem

So as seen on this question Converse of the Chinese Remainder Theorem, we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, ...
4
votes
2answers
1k views

A characterization of invertible fractional ideals of an integral domain

Let $A$ be an integral domain, $K$ its field of fractions. Let $M$ be a fractional ideal of $A$. I'd like to prove that $M$ is invertible if and only if $MA_P$ is a principal fractional ideal of $A_P$ ...
4
votes
1answer
248 views

“Instructive” proof of “If I is maximal among ideals not …, then I is prime”

In this question all rings are commutative with identity. Consider the following well-known statement: (*) Let $R$ be a ring and $S$ a multiplicatively closed subset of $R$. Suppose $I$ is an ...
4
votes
1answer
893 views

The total ring of fractions of a reduced Noetherian ring is a direct product of fields

This is question 6.5 in Matsumura's "Commutative ring theory": How can I prove that the total ring of fractions of a reduced Noetherian ring is a direct product of fields?
8
votes
3answers
268 views

sufficient conditions for being a PID

Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, ...
7
votes
2answers
713 views

Lying-over theorem without Axiom of Choice

This question is motivated by this and this. Can the following proposition be proved without Axiom of Choice? Proposition: Let $k$ be a field. Let $A$ and $B$ be commutative algebras without ...
6
votes
2answers
980 views

$x$ not nilpotent implies that there is a prime ideal not containing $x$.

Let $\mathscr{N}(R)$ denote the set of all nilpotent elements in a ring $R$. I have actually done an exercise which states that if $x \in \mathscr{N}(R)$, then $x$ is contained in every prime ideal ...
5
votes
1answer
252 views

Commutative integral domain with d.c.c. is a field [duplicate]

If $R$ is a commutative integral domain and it also satisfies descending chain condition on its ideals then how will we show that such ring $R$ will be a field?
5
votes
1answer
654 views

Constructing Idempotent Generator of Idempotent Ideal

Exercise 2.1 in Matsumura's Commutative Ring Theory reads as follows: "Let $A$ be a commutative ring and $I$ an ideal that is finitely generated and $I=I^2$. Then $I$ is generated by an idempotent." ...
4
votes
3answers
809 views

Dedekind's theorem on the factorisation of rational primes

Let $K$ be an algebraic number field, and suppose its ring of integers is $\mathcal{O}_K = \mathbb{Z}[\theta]$ for some $\theta \in \mathcal{O}_K$. Let $f \in \mathbb{Z}[X]$ be the minimal polynomial ...
3
votes
1answer
409 views

Use Nakayama's Lemma to show that $I$ is principal, generated by an idempotent. [duplicate]

Let $I$ be a finitely generated ideal in $R$, such that $I^2 = I$. Using the fact there exists $x\in R$ such that $e = 1 - x\in I$ and $xI = 0$, use Nakayama's Lemma to show that $I$ is principal, ...
3
votes
2answers
628 views

Principal prime ideals are minimal among prime ideals in a UFD

Fulton, "Algebraic Curves," Exercise 1.39(a): Let $R$ be a UFD, and $P = (t)$ a principal, proper, prime ideal. Show there is no prime ideal $Q$ with $0 \subset Q \subset P$. After being ...
8
votes
2answers
623 views

How to prove that this subring is not noetherian?

Consider the subring $R=k[x,xy,xy^2,\ldots]$ of $k[x,y]$. I want to prove that $R$ is not noetherian. An ascending chain of ideals is the following ...
3
votes
2answers
262 views

If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).

If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat). The converse, although not immediately apparent, can be proven quite easily. But ...
0
votes
2answers
482 views

Every finitely generated algebra over a field is a Jacobson ring

Knowing that the polynomial ring in $n$ variables over a field $k$ is a Jacobson ring, how can we prove from it that every finitely generated $k$-algebra is a Jacobson ring? EDIT: We define a ...
82
votes
4answers
3k views

Does $R[x] \cong S[x]$ imply $R \cong S$?

This is a very simple question but I believe it's nontrivial. I would like to know if the following is true: If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and ...
31
votes
5answers
3k views

Why should I care about adjoint functors

I am comfortable with the definition of adjoint functors. I have done a few exercises proving that certain pairs of functors are adjoint (tensor and hom, sheafification and forgetful, direct image and ...
29
votes
3answers
3k views

Reference request: introduction to commutative algebra

My goal is to pick up some commutative algebra, ultimately in order to be able to understand algebraic geometry texts like Hartshorne's. Three popular texts are Atiyah-Macdonald, Matsumura ...
23
votes
2answers
3k views

Tensor products commute with direct limits

This is Exercise 2.20 in Atiyah-Macdonald. How can we prove that $\varinjlim (M_i \bigotimes N) \cong (\varinjlim M_i) \bigotimes N$ ? Atiyah gives a suggestion, he says that one should obtain a map ...
29
votes
3answers
1k views

Ideals of $\mathbb{Z}[X]$

Is it possible to classify all ideals of $\mathbb{Z}[X]$? By this I mean a preferably short enumerable list which contains every ideal exactly once, preferably specified by generators. The prime ...
12
votes
1answer
1k views

Does localisation commute with Hom for finitely-generated modules?

Question. Let $R$ be a ring, $\mathfrak{p}$ a prime, $M$ a finitely-generated $R$-module, and $N$ any $R$-module. Is the natural map $$\textrm{Hom}_R(M, N)_\mathfrak{p} \to ...
11
votes
1answer
893 views

Every maximal ideal is principal. Is $R$ principal?

Let $R$ be a commutative ring with 1. If every maximal ideal of $R$ is principal, is $R$ a principal ideal ring?
13
votes
2answers
880 views

Fields finitely generated as $\mathbb Z$-algebras are finite?

Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra. (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite?
12
votes
2answers
654 views

Hom and tensor with a flat module

Let $A$ be a commutative noetherian ring. Let $M, N$ be $A$-modules, and assume that $M$ is finite over $A$. Let $P$ be a flat $A$-module. Is it true that there is an isomorphism ...
12
votes
3answers
840 views

Examples of rings with ideal lattice isomorphic to $M_3$, $N_5$

In thinking about this recent question, I was reading about distributive lattices, and the Wikipedia article includes a very interesting characterization: A lattice is distributive if and only if ...
8
votes
1answer
144 views

Infinite linear independent family in a finitely generated $A$-module

So I'm stuck with this problem. Let $A$ be a commutative ring (with unit). I have several questions that are really close to each other. 1) Let $M$ be a finitely generated module over $A$. Can we ...
7
votes
3answers
714 views

Verifying that the ideal $(x^3-y^2)$ is prime

How to prove that the ideal $I=(x^3-y^2)$ in $k[x,y]$ is prime? I have constructed a map from $k[x,y]$ to $k[t]$, which maps $x$ to $t^2$, and $y$ to $t^3$. Then, I want to show that the kernel ...
9
votes
3answers
2k views

If a ring is Noetherian, then every subring is finitely generated?

Let $R$ be a commutative ring with $1$, and let $K$ be a field. We know that $R$ is Noetherian iff every ideal of $R$ is finitely generated as an ideal. Question 1: If $R$ is Noetherian, is every ...
11
votes
2answers
817 views

Must $k$-subalgebra of $k[x]$ be finitely generated?

Suppose $k$ is a field, $A$ is a $k$-subalgebra of the polynomial ring $k[x]$. Must $A$ be a finitely generated $k$-algebra? Thanks.
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votes
2answers
1k views

When is a tensor product of two commutative rings noetherian?

In particular, I'm told if $k$ is commutative (ring), $R$ and $S$ are commutative $k$-algebras such that $R$ is noetherian, and $S$ is a finitely generated $k$-algebra, then the tensor product ...
6
votes
1answer
285 views

Normalization of a quotient ring of polynomial rings (Reid, Exercise 4.6)

I solved all parts of Exercise 4.6 of the book Undergraduate Commutative Algebra of Miles Reid except the last one. Let $A=k[X]$ and $f\in A$ has a square factor but it is not a square polynomial ...
8
votes
3answers
2k views

If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. [duplicate]

Possible Duplicate: Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field I need to prove this result, but the only starting point I think of is to ...
6
votes
2answers
618 views

Does every Noetherian domain have finitely many height 1 prime ideals?

Let $A$ be a Noetherian domain. Is the set $\{P\subset A \mid P \mbox{ prime ideal, } \dim A_P=1\}$ always finite? I can prove for $f \neq 0, f\in A$, the set $\{P\subset A \mid \dim A_P=1, f\in ...
4
votes
1answer
273 views

Multidimensional Hensel lifting

I have a question about a practical application of (some) generalised form of Hensel's Lemma. I cannot find it stated in an appropriate form in Bourbaki or anywhere else, so here goes ... Let $p$ be ...
2
votes
1answer
269 views

A proposition on a Dedekind domain

I need a proof of the following proposition(?). Actually I think I came up with a proof. But it's nice to confirm it and/or to know other proofs. Thanks. Proposition Let $A$ be a Dedekind domain. Let ...
9
votes
1answer
580 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 ...
5
votes
4answers
540 views

Examples of Non-Noetherian Valuation Rings

For valuation rings I know examples which are Noetherian. I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind? I am very ...
3
votes
1answer
97 views

In arbitrary commutative rings, what is the accepted definition of “associates”?

In an integral domain, the following are equivalent: $r \mid s$ and $s \mid r$ $r=us$ for some unit $u$ However in arbitrary commutative rings this is no longer the case; in particular, (2) ...
3
votes
1answer
347 views

Primary decomposition of a monomial ideal

Can anyone give me an idea about the primary decomposition of the ideal $I=(x^3y,xy^4)$ of the ring $R=k[x,y]$? I am trying to connect the primary decomposition with the set Ass(R/I) which i ...
3
votes
4answers
685 views

Radical ideal of $(x,y^2)$

How does one show that the radical of $(x,y^2)$ is $(x,y)$ over $\mathbb{Q}[x,y]$? I have no idea how to do, please help me.
2
votes
3answers
209 views

If $a,b$ is an $R$-sequence, then $ax-b$ is prime (Eisenbud, Exercise 10.4)

This is the exercise mentioned above: Let $a,b$ be regular sequence over a domain $R$. Prove that $ax-b$ is a prime of $R[x]$. Thank you for your answer!
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votes
4answers
271 views

$\mathbb{C}[x,y]/(f,g)$ is an artinian ring, if $\gcd(f,g)=1$. [closed]

This problem extends the fact that $\mathbb{C}[x,y]/(x^n,y^m)$ is artinian ring. Let $f,g \in \mathbb{C}[x,y]$ such that $\gcd(f,g)=1$. Show that $\mathbb{C}[x,y]/(f,g)$ is an artinian ring.
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vote
3answers
353 views

Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice

Let $A$ be an integral domain which is finitely generated over a field $k$. Let $f \neq 0$ be a non-invertible element of $A$. Can one prove that there exists a prime ideal of $A$ containing $f$ ...
1
vote
1answer
736 views

A maximal ideal among those avoiding a multiplicative set is prime

Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime. If $R$ is an integral domain, ...
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votes
2answers
415 views

Is quotient of a ring by a power of a maximal ideal local?

Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring? This is a well ...
3
votes
2answers
226 views

Atiyah and Macdonald, Proposition 2.9

The following simple claim is used without proof in Proposition 2.9 of Atiyah and MacDonald (p.23). Although I believe I can prove it with a fairly involved argument, the claim is treated by the ...
2
votes
2answers
1k views

Roots of a polynomial in an integral domain

Let $R$ be a ring and $f(X) \in R[x]$ be a non-constant polynomial. We know that the number of roots, of $f(X)$ in $R$ has no relation, to its degree if $R$ is not commutative, or commutative but not ...
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vote
1answer
134 views

A relation involving an endomorphism of a finitely generated module

Suppose $R$ is a commutative ring, $I$ is an ideal of $R$, and $M$ a finitely generated $R$-module. (Usually in this problem $R$ includes $1_R$.) Let $\phi : M \to M$ be an $R$-homomorphism, and ...
0
votes
1answer
155 views

Primary decomposition of $I = (x^2, y^2, xy)$

I want to find a primary decomposition of the ideal $$ I = (x^2,y^2,xy) \subset k[x,y]$$ where $k$ is a field. How to proceed? Are there algorithms to find such decompositions? Where can I find ...
0
votes
1answer
411 views

Quotient ring of a localization of a ring

Let $A$ be a commutative ring. Let $P$ be a prime ideal of $A$. Let $I$ be an ideal of $A$ such that $I \subset P$. Let $\bar A = A/I$. Let $\bar P = P/I$. Is $\bar A_{\bar P}$ isomorphic to ...