TITLE: Converse to Hilbert basis theorem
QUESTION [14 upvotes]: Prove the converse to Hilbert basis theoren:

If the polynomial ring $R[x]$ is Noetherian, then $R$ is noetherian.

REPLY [4 votes]: Converse to Hilbert basis theorem has a trivial answer......
use the result, "Any homomorphic image of a Noetherian ring is Noetherian."
Proof:-
Let $f : M \to N$ be a homomorphism of $A$-modules, where $M$ is Noetherian.
Then $f(M)$ is isomorphic to $M/ \ker f$.(Use first isomorphisom theorem for rings)
Now to prove $f(M)$ is noetherian use the following result,
"Let $A$ be a ring, $M$ be an $A$-module and $N$ be an $A$-submodule
of $M$. Then $M$ is noetherian if and only if $N$ and $M/N$ are noetherian."
to prove this consider the following exact sequence....
$$0 \to N \to M \to M/N \to 0.\qquad \text{(This is an exact sequence).}$$
so by the above result, $f(M)$ is Noetherian.
Now,for the homomorphisom part of $T:R[x] \to R$
Consider, $T(f(x))=f(0)$.
This map is trivially a homomorphisom (Check!!!!) .