Hilbert’s Basis Theorem implies that the polynomial ring ${\mathbb{K}}[{\pmb{x}}]$ is a Noetherian ring, i.e.,

\[thm:asc-chain\]For any ascending chain of ideals $$I_1\subseteq I_2\subseteq \cdots\subseteq I_n \subseteq \cdots$$ of ${\mathbb{K}}[{\pmb{x}}]$, there exists an $N\in\mathbb{N}$ such that $I_n=I_N$ for all $n\geq N$.

Algebraic sets and varieties
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By assigning values in ${\mathbb{K}}^d$ to ${\pmb{x}}$, a polynomial $f\in{\mathbb{K}}[{\pmb{x}}]$ can be regarded as a function from the affine space ${\mathbb{K}}^d$ to ${\mathbb{K}}$. Then the set of zeros of a polynomial set $P\subset{\mathbb{K}}[{\pmb{x}}]$ can be defined as $Z(P)\triangleq\{{\pmb{x}}\in{\mathbb{K}}^d\mid\forall f\in P.\, f({\pmb{x}})=0\}$. It is easy to verify that $Z(P)=Z(\langle P\rangle)$.

A subset $X\subseteq{\mathbb{K}}^d$

1.  is algebraic, if there exists some $P\subseteq{\mathbb{K}}[{\pmb{x}}]$ such that $X=Z(P)$, and $P$ is called a set of generating polynomials of $X$;

2.  is reducible, if it has two algebraic proper subsets $X_1$ and $X_2$ such that $X=X_1\cup X_2$; otherwise it is called irreducible;

3.  is a variety, if it is a nonempty irreducible algebraic set.

The following properties on algebraic and variety can be easily verified: the union of two algebraic sets is an algebraic set, and the intersection of any family of algebraic sets is still algebraic; suppose $X_1,X_2,\cdots,X_n$ are algebraic sets and $X$ is a variety, then $$X\subseteq X_1\cup\cdots\cup X_n\Rightarrow X\subseteq 