# Noetherian ring, any empimorphism is also injective,

• December 9th 2012, 04:34 AM
rayman
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
• December 9th 2012, 10:04 AM
Deveno
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.
• December 10th 2012, 12:02 AM
rayman
Re: Noetherian ring, any empimorphism is also injective,
thank you, your reformulation of the problem helped me to finally get this :)