Converse to Hilbert basis theorem

Question

Prove the converse to Hilbert basis theoren:

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

Answer 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.