This is my proof of the Hilber's Basis Theorem. I think it is incorrect. Because it is easier than other proofs. But I can't find out the mistake in my proof. Can anyone help me? Thanks!
Claim If $R$ is a Noetherian ring, then $R[x]$ is also a Noetherian ring.
Let $R$ be a Noetherian ring and $I[x]$ be an ideal in $R[x]$, where $I$ is an ideal of $R$. Since $R$ is Noetherian, $I$ is finitely generated. Suppose that $I=\langle b_1, b_2, ..., b_n\rangle=\{r_1 b_1+r_2 b_2+\cdots+r_n b_n\mid r_1, r_2, ..., r_n\in R\}$. For any $f(x)\in I[x]$, suppose that $f(x)=a_m x^m+a_{m-1}x^{m-1}+\cdots +a_1 x+a_0\in I[x]$ and $$ \left( \begin{array}{c} a_m \\ \vdots \\ a_1 \\ a_0 \\ \end{array} \right) = \left( \begin{array}{ccccc} r_{m,1} & r_{m,2} & \cdots & r_{m,n} \\ r_{m-1,1} & r_{m-1,2} & \cdots & r_{m-1,n} \\ \vdots & \vdots & \ddots & \vdots \\ r_{1,1} & r_{1,2} & \cdots & r_{1,n} \\ r_{0,1} & r_{0,2} & \cdots & r_{0,n} \\ \end{array} \right) \left( \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \\ \end{array} \right) $$ and $$ f(x)= \left( \begin{array}{cccc} x^m & \cdots & x & 1 \\ \end{array} \right) \left( \begin{array}{c} a_m \\ \vdots \\ a_1 \\ a_0 \\ \end{array} \right) = \left( \begin{array}{cccc} x^m & \cdots & x & 1 \\ \end{array} \right) \left( \begin{array}{ccccc} r_{m,1} & r_{m,2} & \cdots & r_{m,n} \\ r_{m-1,1} & r_{m-1,2} & \cdots & r_{m-1,n} \\ \vdots & \vdots & \ddots & \vdots \\ r_{1,1} & r_{1,2} & \cdots & r_{1,n} \\ r_{0,1} & r_{0,2} & \cdots & r_{0,n} \\ \end{array} \right) \left( \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \\ \end{array} \right). $$ Therefore, $I[x]$ is also generated by $\{b_1, b_2, ..., b_n\}$ and $R[x]$ is Noetherian.
Not every ideal of $R[x]$ is of the form $I[x]$ for some ideal $I\subset R$, so your proof is lacking.
Moreover, your argument doesn't go through for the ideal $(X)\subset\Bbb{Z}[X]$, for example.