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### 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.
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### $A$ 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 ...
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### 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 ...
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### 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 ...
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### 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. ...
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
### 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 ...