Classical varieties have finite irreducible decompositions

Given: The objects and hypotheses in the statement.

Fix an algebraically closed field $k$. A classical algebraic prevariety over $k$ is a quasi-compact locally ringed space over $k$ that is covered by open subspaces isomorphic, as locally ringed spaces over $k$, to affine algebraic sets with their regular-function sheaves. Equivalently, it admits a finite such affine cover. A map is regular when it is a morphism of these locally ringed spaces over $k$; equivalently, this can be checked on affine charts. Thus its maps on structure sheaves respect the fixed $k$-algebra structures. A prevariety is separated when the equalizer of every pair of regular maps into it is closed. A (classical) algebraic variety is a separated prevariety. In the comparison below, “irreducible classical variety” means a nonempty variety whose underlying topological space is irreducible. (Classical algebraic prevarieties, regular maps, and varieties).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space). The space $X$ is Noetherian when every ascending chain $U_0\subseteq U_1\subseteq U_2\subseteq \cdots$ of open subsets stabilizes. Equivalently, $X$ is Noetherian when every descending chain $F_0\supseteq F_1\supseteq F_2\supseteq \cdots$ of closed subsets stabilizes. (Noetherian topological spaces via ACC on opens or DCC on closed subsets).

Let $k$ be an algebraically closed field and let $X\subseteq {\mathbf{A}}_k^n$ be an affine algebraic set. Then there exist finitely many irreducible closed subsets $X_1,\dots ,X_m\subseteq X$ such that $X=X_1\cup \cdots \cup X_m,$ no $X_i$ is contained in the union of the others, and the family $\{X_1,\dots ,X_m\}$ is uniquely determined up to reordering. These sets are the irreducible components of $X$. (Every affine algebraic set has finitely many irreducible components).

Let $R$ be a Noetherian commutative ring. Then the polynomial ring $R[x]$ is a Noetherian commutative ring. No hypothesis beyond Noetherianity is placed on $R$: it may have zero divisors, and it may be the zero ring. (Hilbert basis theorem: if $R$ is Noetherian then $R[x]$ is Noetherian).