# proof of Hilbert basis theorem

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.$\square$

Title proof of Hilbert basis theorem ProofOfHilbertBasisTheorem 2013-03-22 12:59:27 2013-03-22 12:59:27 bwebste (988) bwebste (988) 6 bwebste (988) Proof msc 13E05