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Thread: Noetherian ring, any empimorphism is also injective,

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    Noetherian ring, any empimorphism is also injective,

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

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

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

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

    My question is why $\displaystyle a\in A$ equals $\displaystyle \varphi^{n}(b)$ for some $\displaystyle b\in A$?????? how do we get this?
    and this line $\displaystyle \varphi(a)=0$ as $\displaystyle a\in ker\varphi\Rightarrow \varphi^{n+1}(b)=0\Rightarrow \varphi^{n}(b)=0$ i.e $\displaystyle a=0\Rightarrow ker\varphi=0$
    Thanks
    Last edited by rayman; Dec 9th 2012 at 04:44 AM.
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    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
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    Re: Noetherian ring, any empimorphism is also injective,

    thank you, your reformulation of the problem helped me to finally get this
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