# Noetherian Rings

1. Oct 24, 2007

### ZioX

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.

2. Oct 24, 2007

### Hurkyl

Staff Emeritus
That's surely not true!

3. Oct 24, 2007

### Kummer

If all prime ideals are finitely generated then the ring is noetherian, theorem of Cohen.

4. Oct 25, 2007

### ZioX

Consider the chain of ideals:

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

And the fact that $\phi(A)=A$.

5. Oct 25, 2007

### morphism

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: Oct 25, 2007
6. Oct 25, 2007

### Hurkyl

Staff Emeritus
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.