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