# Tagged Questions

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

666 views

### Quotient of polynomials, PID but not Euclidean domain?

While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$ is ...
779 views

### Is the radical of an irreducible ideal irreducible?

Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is irreducible if $I = J_1 \cap J_2$ for ideals $J_1$ and $J_2$ only when either $I = J_1$ or $I = J_2$. Question : Assume that $I$ is an ...
381 views

### Bound on nilpotency index of endomorphisms

Let $A$ be a Noetherian ring (commutative with $1$) and $M$ a finitely generated $A$-module. I want to show that there exists a bound $n$ such that for every nilpotent endomorphism $T : M \to M$ we ...
993 views

### $\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$

Let $A$ be an integral domain of finite Krull dimension. Let $\mathfrak{p}$ be a prime ideal. Is it true that $$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$$ where $\dim$ ...
2k views

### $A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$

I'm trying to prove that if $A\neq 0$ is a commutative ring and there is an injective $A$-module homomorphism $A^m\hookrightarrow A^n$ then $m\leq n$ must necessarily hold. This is exercise 2.11 ...
832 views

### Structure of Finite Commutative Rings

Is every finite commutative ring $A$ a direct product of finite algebras over $\mathbb Z/p^n$?
488 views

### Basic counterexample re: preimages of ideals

I'm trying to think of an example of a homomorphism of commutative rings $f:A\rightarrow B$ and ideals $I,J$ of $B$ such that $f^{-1}(I)+f^{-1}(J)$ is not a preimage of any ideal of $B$. I can't seem ...
387 views

### Is $\mathbb{Z}$ the only totally-ordered PID that is “special”?

(All my rings are commutative and unital.) Definition. Call a totally-ordered ring $R$ special iff for all non-zero $b \in R,$ every coset of $bR$ has a unique element in the interval $[0,|b|).$ ...
481 views

### Modules with $m \otimes n = n \otimes m$

Let $R$ be a commutative ring. Which $R$-modules $M$ have the property that the symmetry map $$M \otimes_R M \to M \otimes_R M, ~m \otimes n \mapsto n \otimes m$$ equals the identity? In other words,...
416 views

### Wanted: A purely algebraic proof of the Frobenius theorem on distributions

Is there a purely algebraic proof of the Frobenius theorem? Here's a rough sketch of what i'm looking for: Let $Der(R)$ denote the $R$-module of ($R$-valued) derivations of the algebra $R$ endowed ...
4k views

### Why is the localization at a prime ideal a local ring?

I would like to know, why $\mathfrak{p} A_{\mathfrak{p}}$ is the maximal ideal of the local ring $A_{\mathfrak{p}}$, where $\mathfrak{p}$ is a prime ideal of $A$ and $A_{\mathfrak{p}}$ is ...
12k views

### A ring is a field iff the only ideals are $(0)$ and $(1)$

Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$. I can't figure out where to start other that I ...
2k views

### A subring of the field of fractions of a PID is a PID as well.

Let $A$ be a PID and $R$ a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the field of fractions of $A$. How to show $R$ is also a PID? Any hints?
2k views

### Complement of maximal multiplicative set is a prime ideal

Let $R$ be a commutative ring with identity. I've been trying to prove the following: If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$. I have spent ...
543 views

### An exercise with Zariski topology

I read this exercise: Prove that the set $S = \{ (n, 2^n, 3^n ) \mid n \in \mathbb{N} \}$ is dense in $\mathbb{C}^3$ with Zariski topology. I have seriously thought about it, but I do not manage to ...
2k views

### What does a zero tensor product imply?

I'm trying to prove that for two finitely generated $A$-modules $M,N$ ($A$ being any cmmutative ring), the tensor product $M\otimes_A N$ is zero iff $\operatorname{Ann}(M)+\operatorname{Ann}(N)=A$. ...
1k views

334 views

2k views

### Intuitive explanation of Nakayama's Lemma

Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$. I've read the proof, ...
713 views

### Direct way to show: $\operatorname{Spec}(A)$ is $T_1$ $\Rightarrow$ $\operatorname{Spec}(A)$ is Hausdorff

In the book of Atiyah and MacDonald, I was doing exercise 3.11. One has to show that for a ring $A$, the following are equivalent: $A/\mathfrak{N}$ is absolute flat, where $\mathfrak{N}$ is the ...
4k views

### One-to-one correspondence of ideals in the quotient also extends to prime ideals?

I'm beginning to learn some grothendieck's algebraic geometry and I have a doubt about a property of commutative algebra. For a comm. ring $A$ and an ideal $I$ of $A$, does the one-to-one ...
889 views

### Localization at a prime ideal is a reduced ring

Here is the question that I came up with, which I am having trouble proving or disproving: Let $A$ be a ring (commutative). Let $p \in Spec(A)$ such that $A_p$ is reduced. Then there exists an open ...
I want to know some examples with the following properies. Let $R$ be a domain such that every non unit element $x$ is a product of finite irreducible elements,but $R$ is not a UFD, and there is ...