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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

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.

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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."


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!!!!) .

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