Let $R$ be a noetherian ring and let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \in R[x]$ with $a_n\neq 0$ Then call $a_n$ the initial coefficient of $f$

Let $I$ be an ideal in $R[x]$ We will show $I$ is finitely generated, so that $R[x]$ is noetherian. Now let $f_0$ be a polynomial of least degree in $I$ and if $f_0, f_1, \ldots , f_k$ have been chosen then choose $f_{k+1}$ from $I\smallsetminus (f_0, f_1, \ldots , f_k)$ of minimal degree. Continuing inductively gives a sequence $(f_k)$ of elements of $I$

Let $a_k$ be the initial coefficient of $f_k$ and consider the ideal $J=(a_1, a_2, a_3, \ldots )$ of initial coefficients. Since $R$ is noetherian, $J=(a_0, \ldots , a_N)$ for some $N$

Then $I=(f_0, f_1, \ldots , f_N)$ For if not then $f_{N+1}\in I\smallsetminus (f_0, f_1, \ldots , f_N)$ and $a_{N+1} = \sum_{k=0}^N u_k a_k$ for some $u_1, u_2, \ldots , u_N\in R$ Let $g(x)=\sum_{k=0}^N u_k f_k x^{\nu_k}$ where $\nu_k = \operatorname{deg}(f_{N+1})-\operatorname{deg}(f_k)$

Then $\operatorname{deg}(f_{N+1} - g) < \operatorname{deg}(f_{N+1})$ and $f_{N+1} - g \in I$ and $f_{N+1}-g\notin (f_0, f_1, \ldots , f_N)$ But this contradicts minimality of $\operatorname{deg}(f_{N+1})$

Hence, $R[x]$ is noetherian.$\square$