Converse to Hilbert basis theorem

Solution 1:

Let $I\subseteq R$ be an ideal. Then $J := I + X\cdot R[X]\subseteq R[X]$ is an ideal, hence finitely generated, say $J = \langle p_0, \ldots, p_n\rangle$. Now let $a_i = p_i(0) \in R$ for $0 \le i \le n$. We have $\langle a_0, \ldots, a_n\rangle\subseteq I$ by definition of $J$. Now let $a \in I$. Then for some $f_i \in R[X]$ we have $$ a = \sum_{i=0}^n f_i p_i $$ which, evaluated at $0$ gives $$ a = \sum_{i=0}^n f_i(0)a_i \in \langle a_0, \ldots, a_n \rangle $$ Hence $ I = \langle a_0,\ldots, a_n\rangle$ and $I$ is finitely generated.

Solution 2:

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!!!!) .