# 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 |