Corollary to the Hilbert Basis Theorem

# Corollary to the Hilbert Basis Theorem

Recall from The Hilbert Basis Theorem page that if $R$ is a Noetherian ring then $R[x]$ is a Noetherian ring. We will now state a few very important corollaries to the Hilbert basis theorem with regards to affine algebraic sets.

Corollary 1: Let $R$ be a Noetherian ring. Then for all $n \in \mathbb{N}$, $R[x_1, x_2, ..., x_n]$ is a Noetherian ring. |

**Proof:**Let $R$ be a Noetherian ring. By the Hilbert basis theorem we have that $R[x_1]$ is a Noetherian ring. Applying the Hilbert basis theorem again we have that $R[x_1][x_2] = R[x_1, x_2]$ is a Noetherian ring. Inductively we see that $R[x_1, x_2, ..., x_n]$ is a Noetherian ring. $\blacksquare$

Corollary 2: Let $K$ be a field and let $X \subseteq \mathbb{A}^n(K)$ be an affine algebraic set. Then $X$ is the zero locus of a finite set of polynomials in $K[x_1, x_2, ..., x_n]$, that is, there exists $F_1, F_2, ..., F_s \in K[x_1, x_2, ..., x_n]$ such that $X = V(F_1, F_2, ..., F_s)$. |

**Proof:**Since $K$ is a field, the only ideals in $K$ are $(0)$ and $(1)$ (where $0$ is the additive identity in $K$ and $1$ is the multiplicative identity in $K$). So every ideal in $K$ is finitely generated which implies that $K$ is a Noetherian ring.

- By Corollary 1, $K[x_1, x_2, ..., x_n]$ is a Noetherian ring.

- Now since $X$ is an affine algebraic set there exists a set $S \subseteq K[x_1, x_2, ..., x_n]$ such that $X = V(S)$. Since $(S)$ is an ideal in $K[x_1, x_2, ..., x_n]$ and $K[x_1, x_2, ..., x_n]$ is a Noetherian ring so there exists $F_1, F_2, ..., F_s \in K[x_1, x_2, ..., x_n]$ such that:

\begin{align} \quad (S) = (F_1, F_2, ..., F_s) \end{align}

- Observe that $V(S) = V((S))$. Therefore:

\begin{align} \quad X = V(S) = V(F_1, F_2, ..., F_s) \quad \blacksquare \end{align}