I already got a proof for the fact that if a polynomial map is surjective then it is also injective. However, I used the invariant dimension of a ring and I want a simpler proof. Bravo for any try. For preciseness, the statement of the fact is as follows: 


Statement: Consider two polynomial rings $k[x\_1,...,x\_n], k[y\_1,...,y\_n]$. Let $\Phi: k[x\_1,...,x\_n] \rightarrow k[y\_1,...,y\_n]$ be a $k$-algebra homomorphism. If $\Phi$ is surjective then $\Phi$ is also injective.



A proof for a statement about polynomial automorphism
If $A$ is any Noetherian ring, then any surjective homomorphism $\varphi: A\to A$ is injective. One has the ascending chain of ideals $\ker \varphi\subseteq \ker \varphi^2\subseteq \cdots$. Thus $\ker \varphi^n=\ker \varphi^{n+1}$ for some $n$. Let $a\in \ker \varphi$. Since $\varphi^n$ is surjective, we can write $a=\varphi^n(b)$ for some $b\in A$. The $0=\varphi(a)=\varphi^{n+1}(b)$. So $b\in \ker \varphi^{n+1}=\ker \varphi^n$. Thus $a=\varphi^n(b)=0$ and so $\varphi$ is injective.


