Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - Noetherian ring, any empimorphism is also injective,

  1. #1
    Member
    Joined
    Sep 2010
    From
    Germany
    Posts
    112
    Thanks
    4

    Noetherian ring, any empimorphism is also injective,

    Let A be a Noetherian ring, then show that any ring empimorphism \varphi A\to A is also injective.
    Hint: consider a chain of ideals ker\varphi\subset ker\varphi^2....

    Solution:
    As A is Noetherian, the chain of ideals becomes stationary, i.e ker\varphi\subset ker\varphi^2\subset....\subset ker\varphi^n=ker\varphi^{n+1} for some n
    Moreover we know that \varphi^{n}:A\to A is also onto.

    Suppose a\in ker\varphi, then a\in A equals \varphi^{n}(b) for some b\in A

    Now \varphi(a)=0 as a\in ker\varphi\Rightarrow \varphi^{n+1}(b)=0\Rightarrow \varphi^{n}(b)=0 i.e a=0\Rightarrow ker\varphi=0

    My question is why a\in A equals \varphi^{n}(b) for some b\in A?????? how do we get this?
    and this line \varphi(a)=0 as a\in ker\varphi\Rightarrow \varphi^{n+1}(b)=0\Rightarrow \varphi^{n}(b)=0 i.e a=0\Rightarrow ker\varphi=0
    Thanks
    Last edited by rayman; December 9th 2012 at 04:44 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,384
    Thanks
    750

    Re: Noetherian ring, any empimorphism is also injective,

    we know that φn is onto.

    for suppose a is any element of A. since φ is onto, there exists b1 with φ(b1) = a.

    since b1 is likewise an element of A, there exists b2 with φ(b2) = b1, therefore φ2(b2) = φ(φ(b2)) = φ(b1) = a.

    a short inductive proof then shows that φn is onto, for any n.

    so if a is any element of the kernel of φ (so that φ(a) = 0), we know there exists b in A with φn(b) = a, no matter what n we pick.

    in particular, if we pick n = N (the integer at which the kernels stabilize), then a = φN(b).

    in which case φN+1(b) = φ(φN(b)) = φ(a) = 0.

    so if a is in ker(φ), then b is in ker(φN+1) = ker(φN). which means φN(b) = 0, that is: a = 0, so φ is injective.
    Thanks from rayman
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    From
    Germany
    Posts
    112
    Thanks
    4

    Re: Noetherian ring, any empimorphism is also injective,

    thank you, your reformulation of the problem helped me to finally get this
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 8th 2012, 05:06 AM
  2. Noetherian ring
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: September 24th 2012, 06:29 AM
  3. Noetherian Ring
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: September 7th 2010, 11:05 AM
  4. Noetherian ring, finitely generated module
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 5th 2009, 12:09 PM
  5. Noetherian Ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 22nd 2008, 09:03 PM

Search Tags


/mathhelpforum @mathhelpforum