Let $A$ and $B$ be integral domains, such that $f:A\to B$ is an injective finite-type homomorphism. Then there is an induced map $\text{Frac}(A)\to\text{Frac}(B)$ by the universal property of localisation (since $f$ is injective).
Why is $\text{Frac}(B)/\text{Frac}(A)$ a finite field extension? What is its degree?
I was thinking that since $A\to B$ is finite type, $B\cong A[x_1,\dots,x_n]/I$ where $I$ does not contain any non-zero element of $A$, and localisation commutes with quotients, so that $$\text{Frac}(B)\cong A[x_1,\dots,x_n]_I/I A[x_1,\dots,x_n]_I,$$
and that $\text{Frac}(A)\to A[x_1,\dots,x_n]_I\twoheadrightarrow \text{Frac}(B)$ had first map factoring through $\text{Frac}(A)[x_1,\dots,x_n]\to A[x_1,\dots,x_n]_I$, and if this was itself surjective, then $\text{Frac}(A)[x_1,\dots,x_n]\to \text{Frac}(B)$ would at least make $\text{Frac}(B)$ finite dimensional over $\text{Frac}(A)$, but it wouldn't tell me what the degree is.
This is not true. For a very simple example, if $B=A[x]$, then the fraction field of $B$ is $K(x)$ where $K$ is the fraction field of $A$, which is not a finite extension of $K$. It is true iff $B$ is algebraic over $A$, since a finitely generated field extension is finite iff it is algebraic.