Hilbert Basis Theorem

Statement

Let $R$ be a Noetherian ring. Then $R[x]$ is Noetherian.

Proof

Let $I\subset R[x]$ be an ideal, we aim to show that $I$ is finitely generated.

Let $f_i\in I$ be of minimal degree. For $n\ge 1$, if $(f_1,\dots ,f_n)\ne I$ then chose another polynomial $f_1\in I\backslash (f_1,\dots f_n)$ of minimal degree (if there isn’t another one then that means that $I=(f_1,\dots ,f_n)$ is already finitely generated.)

We can take the leading coefficient $c_i\in R$ of each $f_i$. Thus, the ideal generated by all coefficients $(c_1,c_2,\dots )$ is finitely generated (since $R$ is Noetherian) by some list of generators we can call $(r_1,\dots ,r_N)$.

We aim to prove that $(f_1,\dots ,f_N)=I$ for some $N$. For contradiction suppose that $I\ne (f_1,\dots ,f_N)$. Then, like above we can pick some $f_{N+1}$ of minimal degree such that $f_{N+1}\in I\backslash (f_1,\dots ,f_N)$.

We can write the leading coefficient $c_{N+1}\in (c_1,c_2,\dots )$ — which has $N$ finite generators — as

$$c_{N+1}=\sum _{i=1}^N{c}_i{r}_i{r}_i\in R.$$

We know that each of $f_i$ are of minimal degree at the time chosen so $\text{deg }{f}_{N+1}\ge \text{deg }{f}_j$ for $1\le j\le N$, then we can build the polynomial

$$g:=\sum _{i=1}{r}_i{x}^{\mathrm{deg}{f}_{N+1}-\mathrm{deg}{f}_i}{f}_i.$$

The leading term of $g$ is built to be that of $f_{N+1}$ by how we build $c_{N+1}$. Thus, we can take $f_{N+1}-g$ and the leading terms will cancel giving a polynomial of degree less than $f_{N+1}$. Importantly, $f_{N+1}-g\notin (f_1,\dots ,f_n)$ since $f_{N+1}\notin (f_1,\dots ,f_n)$. However, this is a contradiction since we assumed that $f_{N+1}$ is of minimal degree.

Therefore, $I$ is finitely generated and thus $R[x]$ is Noetherian.