Field of fractions of a finite $\mathbb{Z}$-module is finite extension of $\mathbb{Q}$

Let $A$ be a ring which is also a finitely generated $\mathbb{Z}$-module. If $A$ is an integral domain and $K$ is its field of fractions and $K$ has characteristic zero, then why is $K$ a finite dimensional vector space over $\mathbb{Q}$?

I see that $K$ contains $\mathbb{Q}$ but why is the extension finite?

Let $a_1,\ldots,a_n$ generate $A$ as a $\mathbf{Z}$-module, and let $F=\mathbf{Q}(a_1,\ldots,a_n)\subseteq K$. Note that $\mathbf{Q}(a_1,\ldots,a_n)$ contains $A$, so in fact we must have $\mathbf{Q}(a_1,\ldots,a_n)=K$. Now, because $A$ is finite over $\mathbf{Z}$, it is in particular integral over $\mathbf{Z}$, which means that every element of $A$ is the root of a monic polynomial with coefficients in $\mathbf{Z}$. In particular, this is true of the $a_i$, and so each $a_i$ is algebraic over $\mathbf{Q}$ (integrality over a field is the same as algebraicity). So $K$ is generated over $\mathbf{Q}$ by finitely many algebraic elements, and therefore is a finite extension of $\mathbf{Q}$.

Show first that $K$ is an algebraic extension by considering the minimal polynomial over $\mathbb{Z}$ satisfied by any element of $A$. Then show that any finite generating set for $A$ as a $\mathbb{Z}$-module generates $K$ as a field. A finitely generated algebraic field extension is a finite field extension.