# Hilbert's Basis Theorem/Corollary

## Theorem

Let $A$ be a Noetherian ring.

Let $n \ge 1$ be an integer.

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

Then $A \sqbrk {x_1, \ldots, x_n}$ is also a Noetherian ring.

## Proof

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

### Basis for the induction

Suppose that $n = 1$.

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

Thus by Hilbert's Basis Theorem, $A \sqbrk {x_1}$ is Noetherian.

### Induction step

Suppose that $A \sqbrk {x_1, \ldots, x_{n - 1} }$ is Noetherian.

We wish to show that $A \sqbrk {x_1, \ldots, x_n}$ is Noetherian.

$A \sqbrk {x_1, \ldots, x_n} = A \sqbrk {x_1, \ldots, x_{n - 1} } \sqbrk {x_n}$

is the polynomial ring in a single indeterminate $x_n$ over the Noetherian ring $A \sqbrk {x_1, \ldots, x_{n - 1} }$.

Thus by Hilbert's Basis Theorem, $A \sqbrk {x_1, \ldots, x_n}$ is Noetherian.

$\blacksquare$