# Noetherian Rings

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.

## Answers and Replies

Hurkyl
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Suppose A is a Noetherian ring, phi:A->A any surjective ring homomorphism. Show that phi is also injective.
That's surely not true!

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.

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 $\phi(A)=A$.

morphism
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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
Staff Emeritus
Science Advisor
Gold Member
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.