Linked Questions

1
vote
1answer
253 views

Ascending chain of ideals [duplicate]

Let $R$ be a commutative ring with identity such that every ascending chain of ideals terminate. Let $f:R \to R$ be a surjective homomorphism. Prove that it is an isomorphism.
2
votes
1answer
68 views

$A$ Noetherian and $f:A \rightarrow A$ suryective then $f$ inyective [duplicate]

I think this must have been questioned before, but after searching, I couldn't find it. I thought of considering a set of $\{x_1, ..., x_n\}$ such that $\lt x_1, ... x_n\gt = A$. The hypotesis shows ...
0
votes
0answers
24 views

Hopfian module over commutative ring [duplicate]

I have run into a problem which reads: "Any finitely generated module $M$ over a commutative ring $R$ is Hopfian, i.e., any $R$-epimorphism from $M$ to itself is one-to one." I think the problem ...
6
votes
2answers
2k views

Can you always find a surjective endomorphism of groups such that it is not injective?

If we take the following endomorphism, $\phi:R[t] \to R[t]$ by $\sum_{i = 0}^n a_it^i \mapsto \sum_{i = 0}^{\lfloor n/2 \rfloor} a_{2i} t^i$, it is surjective but not injective. (It just removes odd ...
4
votes
2answers
712 views

Invertible matrices in commutative rings

Let $A$ be a square matrix over a commutative ring $R$. Then $A$ has a left inverse iff it is invertible. Does there exist a elementary proof of this fact? (i.e. without using the determinant!)
0
votes
2answers
499 views

Surjective endomorphism of abelian group is isomorphism

Let $A$ be a finitely generated abelian group and $f:A\rightarrow A$ a surjective homomorphism. How do I prove that $f$ is an isomorphism? And if $f$ were injective instead of surjective would the ...
4
votes
1answer
542 views

Confusion about chain of subspaces

This is part of Problem 11.1.6 of Dummit and Foote. The problem reads Let $V$ be a vector space of finite dimension. If $\phi$ is any linear transformation from $V$ to $V$ prove that there is ...
1
vote
1answer
582 views

Surjective endomorphisms of Noetherian modules are isomorphisms.

I'm trying to solve this question: I didn't understand why the hint is true and how to apply it. I really need help, because it's my first question on this subject and my experience on this field is ...
2
votes
2answers
94 views

$\phi: A^n \to A^n$ be any injective $A$ linear map,will $\phi$ be surjective?

Let $A$ be a commutative ring with $1$ and $\phi: A^n \to A^n$ be any injective $A$ linear map. Can I say $\phi $ is surjective ? We know about the converse that surjectiveness implies injectivitness,...
0
votes
1answer
451 views

Surjective endomorphism of an $R$-module is injective.

I know this is a duplicate question. However, I haven't seen anything that invokes the isomorphism theorem. Here's my idea: By the isomorphism theorem we have that $M/\ker\varphi \cong \...
0
votes
2answers
141 views

Big list of undergraduate exercises in module theory [closed]

I would like to write down a big list of exercises in module theory addressed to a course of introduction to this subject. The background of the average student will be the definition of ring, and ...
0
votes
1answer
275 views

Surjective Homomorphisms of Isomorphic Abelian Groups [duplicate]

Is a surjective homomorphism between two (abstractly) isomorphic finitely generated abelian groups necessarily an isomorphism? I know this is true if the groups are torsion (finite) or torsion-free. ...
3
votes
1answer
120 views

Neither Noetherian nor Artinian

Let $R$ be a ring with $1$ and $M$ a non zero left $R$-modulesuch that $M\cong M\oplus M$ Why $M$ is neither Noetherian nor Artinian? Thanks
5
votes
1answer
120 views

Groups $G$ for which every finitely generated $\mathbb{Z}G$-module is Hopfian

Let $\mathbb{Z}G$ be the group ring over a group $G$. It is a well-known fact that every finitely generated module over a commutative ring is Hopfian. Hence if $G$ is abelian, then every finitely ...