Thread: Noetherian ring, any empimorphism is also injective,

Let A be a Noetherian ring, then show that any ring empimorphism is also injective.
Hint: consider a chain of ideals

Solution:
As A is Noetherian, the chain of ideals becomes stationary, i.e for some n
Moreover we know that is also onto.

Suppose , then equals for some

Now as i.e

My question is why equals for some ?????? how do we get this?
and this line as i.e
Thanks

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 = φ^N(b).

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

so if a is in ker(φ), then b is in ker(φ^N+1) = ker(φ^N). which means φ^N(b) = 0, that is: a = 0, so φ is injective.

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