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A field that is a finite-type integer algebra has prime characteristic or zero. Prime characteristic and Zariski lemma make it a finite algebraic extension of a finite prime field. In characteristic zero, the same lemma makes it a number field; clearing finitely many coefficient denominators makes it integral over $\mathbb{Z}[1/N]$. The integral field extension forces the base domain to be a field, but a prime not dividing $N$ is not invertible there. This contradiction excludes characteristic zero.

The characteristic of $K$ is either a prime $p$ or zero. In characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$. The course's Zariski lemma says that a field finitely generated as an algebra over a field is a finite algebraic extension of that field. ThereforeIt remains to exclude characteristic zero. Write $K=\mathbb{Z}[a_1,\dots ,a_m]$. Because $K$ is then a field containing $\mathbb{Q}$, also $K=\mathbb{Q}[a_1,\dots ,a_m]$. Zariski lemma makes $K/\mathbb{Q}$ finite algebraic. Each $a_i$ has a monic equation with rational coefficients. Choose a positive integer $N$ clearing all their coefficient denominators. Each $a_i$ is then integral over $A=\mathbb{Z}[1/N]$, so $A[a_1,\dots ,a_m]$ is integral over $A$. This algebra equals $K$: it contains the given integer algebra $K$, and $1/N$ already lies in the field $K$.

An integral field extension forces the base domain to be a field. To see it here, for a nonzero $a\in A$ its inverse in $K$ satisfiesMultiplying by $a^{r-1}$ givesThus $A$ would be a field. But choose a prime $\ell \nmid N$. Reduction modulo $\ell$ is a well-defined map $\mathbb{Z}[1/N]\to {\mathbb{F}}_{\ell }$, so the nonzero element $\ell$ is not invertible in $A$. This contradiction rules out characteristic zero. Therefore the finite-field theorem for finitely generated integer algebras yieldsThis uses Zariski lemma and elementary integrality, not a claim that a field merely finitely generated as a field extension must be algebraic.