Vakil 11.1.D: Fibers of integral morphisms have dimension 0

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I'm working through Ravi Vakil's "Rising Sea" notes for an independent study, and I'm working on Exercise 11.1.D in the most recent version of the notes.

I'm not sure I understand his hint for this exercise. He takes a $k$-algebra $A$ that is an integral extension of $k$, and mods $A$ out by a prime ideal $\mathfrak{p}$ of $A$, and relabels the result as $A$ in order to make $A$ an integral domain, and then proves that $A$ is a field so that $\dim A = 0$.

How is he able to do this? Wouldn't modding out $A$ by a prime ideal increase the dimension of the original $A$ by one?