proof of Hilbert basis theorem

Let $R$ be a noetherian ring and let $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{\dots }+a_{1}x+a_0\in R[x]$ with $a_n\ne 0$. 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 $f_0,f_1,\mathrm{\dots },f_k$ have been chosen then choose $f_{k+1}$ from $I\setminus (f_0,f_1,\mathrm{\dots },f_k)$ 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 $J=(a_1,a_2,a_3,\mathrm{\dots })$ of initial coefficients. Since $R$ is noetherian, $J=(a_0,\mathrm{\dots },a_N)$ for some $N$.

Then $I=(f_0,f_1,\mathrm{\dots },f_N)$. For if not then $f_{N+1}\in I\setminus (f_0,f_1,\mathrm{\dots },f_N)$, and $a_{N+1}={\sum }_{k=0}^N{u}_k{a}_k$ for some $u_1,u_2,\mathrm{\dots },u_N\in R$. Let $g(x)={\sum }_{k=0}^N{u}_k{f}_k{x}^{{\nu }_k}$ where ${\nu }_k=\mathrm{deg}(f_{N+1})-\mathrm{deg}(f_k)$.

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

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

Title	proof of Hilbert basis theorem
Canonical name	ProofOfHilbertBasisTheorem
Date of creation	2013-03-22 12:59:27
Last modified on	2013-03-22 12:59:27
Owner	bwebste (988)
Last modified by	bwebste (988)
Numerical id	6
Author	bwebste (988)
Entry type	Proof
Classification	msc 13E05