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Noetherian Rings

  1. Oct 24, 2007 #1
    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. jcsd
  3. Oct 24, 2007 #2

    Hurkyl

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    That's surely not true!
     
  4. Oct 24, 2007 #3
    If all prime ideals are finitely generated then the ring is noetherian, theorem of Cohen.
     
  5. Oct 25, 2007 #4
    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].
     
  6. Oct 25, 2007 #5

    morphism

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    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
  7. Oct 25, 2007 #6

    Hurkyl

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    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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