Proving Injectivity of Surjective Ring Homomorphism in Noetherian Rings

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In summary: 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?If all prime ideals are finitely generated, then the ring is noetherian, theorem of Cohen.
  • #1
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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  • #2
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!
 
  • #3
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.
 
  • #4
Hurkyl said:
That's surely not true!

Consider the chain of ideals:

[tex]\ker\phi\subset\ker\phi^2\subset\ker\phi^3\subset\cdots[/tex]

And the fact that [itex]\phi(A)=A[/itex].
 
  • #5
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!?)
 
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  • #6
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:
 

Related to Proving Injectivity of Surjective Ring Homomorphism in Noetherian Rings

1. What is injectivity and surjectivity in ring homomorphisms?

Injectivity and surjectivity are properties of a function between two rings. A homomorphism is injective if it maps distinct elements in the domain to distinct elements in the codomain, and surjective if every element in the codomain has at least one preimage in the domain.

2. Why is it important to prove injectivity of a surjective ring homomorphism in Noetherian rings?

Noetherian rings have a special property called the ascending chain condition, which states that every increasing chain of ideals eventually stabilizes. Proving injectivity of a surjective ring homomorphism in Noetherian rings ensures that the homomorphism preserves this property, which is important for many algebraic and geometric applications.

3. How can one prove that a ring homomorphism is injective and surjective?

To prove injectivity, one can show that the kernel of the homomorphism is trivial, i.e. it only contains the identity element. To prove surjectivity, one can show that every element in the codomain has at least one preimage in the domain. This can be done by explicitly constructing the preimages or by using the fact that the homomorphism is surjective if and only if its image is equal to the entire codomain.

4. Can a ring homomorphism be both injective and surjective?

Yes, a ring homomorphism can be both injective and surjective. Such a homomorphism is called an isomorphism, and it essentially means that the two rings are structurally identical.

5. Are there any other important properties of ring homomorphisms that should be considered when proving injectivity and surjectivity?

Yes, there are other important properties such as bijectivity, which means the homomorphism is both injective and surjective, and the fact that it preserves the ring structure, i.e. it preserves addition and multiplication operations. These properties are also important to consider when proving injectivity and surjectivity.

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