Proving Injectivity of Surjective Ring Homomorphism in Noetherian Rings

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Homework Help Overview

The discussion revolves around the properties of surjective ring homomorphisms in the context of Noetherian rings, specifically questioning whether a surjective ring homomorphism is also injective. Additionally, there is a query about the implications of having finitely generated prime ideals in a ring and whether this condition guarantees that the ring is Noetherian.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between surjectivity and injectivity of ring homomorphisms, with some expressing skepticism about the original poster's assertion. There are also discussions about the implications of finitely generated prime ideals and references to theorems in commutative algebra.

Discussion Status

The discussion is active, with participants questioning the validity of the original claim regarding surjective homomorphisms and providing counterexamples. There is also a suggestion that the problem about prime ideals relates to established theorems, indicating a potential direction for further exploration.

Contextual Notes

Some participants mention specific theorems and concepts from commutative algebra, indicating a background in the subject. There is a reference to Zorn's lemma in the context of finding maximal ideals, which suggests a level of complexity in the discussion.

ZioX
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Suppose A is a Noetherian ring, phi:A->A any surjective ring homomorphism. Show that phi is also injective.

Also, if all the prime ideals of a ring A are finitely generated then is A noetherian?

I'm pretty sure it is. I figure I can take all of the ideals that are not finitely generated and find a maximal prime ideal that contains these ideals.

I've just started reading a book on commutative algebra with the hopes of moving on to algebraic geometry.
 
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ZioX said:
Suppose A is a Noetherian ring, phi:A->A any surjective ring homomorphism. Show that phi is also injective.
That's surely not true!
 
ZioX said:
Suppose A is a Noetherian ring, phi:A->A any surjective ring homomorphism. Show that phi is also injective.

Also, if all the prime ideals of a ring A are finitely generated then is A noetherian?

I'm pretty sure it is. I figure I can take all of the ideals that are not finitely generated and find a maximal prime ideal that contains these ideals.

I've just started reading a book on commutative algebra with the hopes of moving on to algebraic geometry.

If all prime ideals are finitely generated then the ring is noetherian, theorem of Cohen.
 
Hurkyl said:
That's surely not true!

Consider the chain of ideals:

\ker\phi\subset\ker\phi^2\subset\ker\phi^3\subset\cdots

And the fact that \phi(A)=A.
 
Also, if all the prime ideals of a ring A are finitely generated then is A noetherian?

I'm pretty sure it is. I figure I can take all of the ideals that are not finitely generated and find a maximal prime ideal that contains these ideals.
I'm not sure if I'm reading what you're saying correctly, but what you want to do is take all of the ideals that are not finitely generated, get a maximal such ideal (Zorn), and then show it's prime.

Come to think of it -- this is an exercise in Eisenbud, a book that might be very suitable for what you want. (Maybe you already aknow this, and this is where this problem is from!?)
 
Last edited:
Hurkyl said:
That's surely not true!
Well, what I saw is surely not true. But now that I look again, I can see clear as day that you didn't write phi:A->B. :frown:
 

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