Linked Questions
1
vote
1answer
377 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
164 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
28 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 ...
0
votes
0answers
23 views
$A$ is a ring, $M$ a left noetherian $A$-module, $f:M \to M$ is a left homomorphism of modules. 2 part question. [duplicate]
$A$ is a ring, $M$ a left noetherian $A$-module, $f:M \to M$ is a left homomorphism of modules.
Part A -- Suppose $Ker f^n=Ker f^{2n}$, then prove $Ker f^n \cap Im f^n = \{0\}$.
Suppose that $a \in ...
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
984 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
748 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
674 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
691 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
122 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
767 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
304 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
386 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
165 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
185 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 ...