A nontrivial principal section has pure codimension one

Write $A=R\mathrm{/}I(X)$, where R is a finite-variable polynomial ring over k. The field k has only two ideals, so repeated F6 makes R Noetherian; lifting ideals makes its quotient A Noetherian. A is a nonzero domain: if the product of two coordinate functions vanishes on X, their closed zero sets cover the irreducible X, forcing one function to vanish everywhere.

Lift the proper ideal (f) to an ideal J of R. If $V_X(f)=V(J)$ were empty, F5 would give $\sqrt{J}=R$, and then $1\in J$, contradicting properness of (f). Hence $V_X(f)$ is nonempty. This invocation and the prime/closed-set dictionary F1 are the exact inherited Nullstellensatz uses of AC.

Let Z be any irreducible component of $V_X(f)$. By F1 its vanishing prime $p\subset A$ contains f. It is minimal among primes containing (f): a strictly smaller such prime would correspond, by F1, to a strictly larger irreducible closed subset of $V_X(f)$, contrary to the maximality defining a component. The same dictionary identifies $k[Z]=A\mathrm{/}p$.

F3 applies to A and p, so $\mathrm{ht}p\le 1$. Since A is a domain and $f\mathrm{\ne }0$ lies in p, the strict chain $(0)⊊p$ shows the reverse inequality. Thus $\mathrm{ht}p=1$.

F4 now gives $\dim (A\mathrm{/}p)=\dim A-1$. Apply F1 to X and Z, then F2, to obtain $\dim Z=\dim X-1$ and ${\mathrm{codim}}_{X}Z=1$ for every component. All these dimensions are finite because the rings are finite-type domains.