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Prove the converse to Hilbert basis theoren:

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

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Any factor ring of a noetherian ring is noetherian. Since $x$ generates a two-sided ideal of $R[x]$ and $R[x]/(x)$ is isomorphic to $R$, then $R$ is noetherian. –  J. Gaddis Nov 19 '12 at 13:55
    
tank you ${{{{}}}}$ –  mshj Nov 19 '12 at 17:24

2 Answers 2

Let $I\subseteq R$ an ideal. Then $J := I + X\cdot R[X]\subseteq R[X]$ is an ideal, hence finally generated, say $J = \langle p_0, \ldots, p_n\rangle$. Now let $a_i = p_i(0) \in R$ for $1 \le i \le n$. 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.

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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 → 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 → N→ M → M/N → 0.(This is an exact sequence).

so by the above result, f(M) is noetherian.

Now,for the homeomorphisom part of T:R[x]→ R

Consider, T(f(x))=f(0) .

This map is trivially a homeomorphisom(Cheek!!!!) .

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