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 $a_{n}\eq 0$ . Thencall $a_{n}$ the initial coefficient of $f$ .

Let $I$ be an ideal in $R[x]$ . We will show $I$ is finitelygenerated, so that $R[x]$ is noetherian. Now let $f_{0}$ be apolynomial 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 inductivelygives a sequence $(f_{k})$ of elements of $I$ .

Let $a_{k}$ be the initial coefficient of $f_{k}$ , andconsider the ideal $J=(a_{1},a_{2},a_{3},\\ldots)$ of initialcoefficients. Since $R$ is noetherian, $J=(a_{0},\\ldots,a_{N})$ forsome $N$ .

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 $g(x)=\\sum_{k=0}^{N}u_{k}f_{k}x^{\u_{k}}$ where $\u_{k}=\\operatorname{deg}(f_{N+1})-\\operatorname{deg}(f_{k})$ .

Then $\\operatorname{deg}(f_{N+1}-g)<\\operatorname{deg}(f_{N+1})$ , and $f_{N+1}-g\\in I$ and $f_{N+1}-g\otin(f_{0},f_{1},\\ldots,f_{N})$ . But this contradictsminimality of $\\operatorname{deg}(f_{N+1})$ .

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

单词	ProofOfHilbertBasisTheorem
释义	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 $a_{n}\eq 0$ . Thencall $a_{n}$ the initial coefficient of $f$ . Let $I$ be an ideal in $R[x]$ . We will show $I$ is finitelygenerated, so that $R[x]$ is noetherian. Now let $f_{0}$ be apolynomial 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 inductivelygives a sequence $(f_{k})$ of elements of $I$ . Let $a_{k}$ be the initial coefficient of $f_{k}$ , andconsider the ideal $J=(a_{1},a_{2},a_{3},\\ldots)$ of initialcoefficients. Since $R$ is noetherian, $J=(a_{0},\\ldots,a_{N})$ forsome $N$ . 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 $g(x)=\\sum_{k=0}^{N}u_{k}f_{k}x^{\u_{k}}$ where $\u_{k}=\\operatorname{deg}(f_{N+1})-\\operatorname{deg}(f_{k})$ . Then $\\operatorname{deg}(f_{N+1}-g)<\\operatorname{deg}(f_{N+1})$ , and $f_{N+1}-g\\in I$ and $f_{N+1}-g\otin(f_{0},f_{1},\\ldots,f_{N})$ . But this contradictsminimality of $\\operatorname{deg}(f_{N+1})$ . Hence, $R[x]$ is noetherian. $\\square$
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