A classical affine algebraic set has a unique finite irredundant decomposition

Every ideal of the field $k$ is either 0 or $k$: a nonzero member is invertible, putting 1 in the ideal. These ideals are generated by 0 and 1, respectively, so F2 makes $k$ Noetherian and F1 makes $R=k[x_1,\dots ,x_n]$ Noetherian, including $n=0$. Apply F3 to the radical ideal $I(X)$, with its stipulated DC cost, to obtain its distinct minimal primes $P_1,\dots ,P_m$ with $I(X)={\bigcap }_i{P}_i$.

Put $X_i=V(P_i)$. F4 makes each nonempty irreducible. A point outside every $X_i$ admits $f_i\in P_i$ nonzero there; the finite product belongs to every $P_i$ and is nonzero at the point. Thus $V({\bigcap }_i{P}_i)\subseteq {\bigcup }_i{X}_i$; the reverse inclusion follows because every element of the intersection vanishes on each $X_i$. So $X={\bigcup }_i{X}_i$. For $m=0$, the intersection is $R$ and the union empty. Distinct minimal primes are incomparable; F4 therefore makes the $X_i$ incomparable.

An irreducible nonempty closed set $Z$ covered by finitely many closed sets must be contained in one: repeatedly split the cover as the first member and the union of the others; if $Z$ is not contained in the first, irreducibility puts it in the remaining union. For a one-member cover this ends immediately; a zero-member cover cannot cover nonempty $Z$. Thus every irreducible closed subset of $X$ is contained in some $X_i$, and incomparability makes the $X_i$ precisely the maximal ones. It also prevents removal of an $X_i$ from the cover.

If $X={\bigcup }_j{Y}_j$ is another finite irredundant irreducible closed decomposition, step 3.1 puts each $Y_j$ inside an $X_i$ and that $X_i$ inside some $Y_h$. Irredundancy forces $Y_j=Y_h$ (otherwise $Y_j$ could be removed), so $Y_j=X_i$. Reversing the two covers shows their members coincide. After removal of duplicate indexing, this is uniqueness up to permutation.