# Hilbert's Basis Theorem/Corollary

## Theorem

Let $A$ be a Noetherian ring.

Let $n \geq 1$ be an integer.

Let $A\left[{ x_1,\ldots,x_n }\right]$ be the ring of polynomial forms over $A$ in the indeterminates $x_1,\ldots,x_n$.

Then $A\left[{ x_1,\ldots,x_n }\right]$ is also a Noetherian ring.

## Proof

We proceed by induction over $n \geq 1$.

### Basis for the induction

Suppose that $n = 1$.

Then $A\left[{ x_1,\ldots,x_n }\right] = A\left[{ x_1}\right]$ is the ring of polynomial forms in a single indeterminate $x_1$.

Thus by Hilbert's Basis Theorem, $A\left[{ x_1}\right]$ is noetherian.

### Induction step

Suppose that $A\left[{ x_1,\ldots,x_{n-1} }\right]$ is noetherian.

We wish to show that $A\left[{ x_1,\ldots,x_{n} }\right]$ is noetherian.

$A\left[{ x_1,\ldots,x_{n} }\right] = A\left[{ x_1,\ldots,x_{n-1} }\right]\left[{ x_n }\right]$

is the polynomial ring in a single indeterminate $x_n$ over the noetherian ring $A\left[{ x_1,\ldots,x_{n-1} }\right]$.

Thus by Hilbert's Basis Theorem, $A\left[{ x_1,\ldots,x_{n} }\right]$ is noetherian.

$\blacksquare$