Why does the integral domain "being trapped between a finite field extension" implies that it is a field? The following is an exercise from Qing Liu's Algebraic Geometry and Arithmetic Curves.

Exercise 1.2.
Let $\varphi : A \to B$ be a homomorphism of finitely generated algebras over a field. Show that the image of a closed point under $\operatorname{Spec} \varphi$ is a closed point.

The following is the solution from Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.

Write $k$ for the underlying field. Let’s parse the statement. A closed point in $\operatorname{Spec} B$ means a maximal ideal $n$ of $B$. And $\operatorname{Spec}(\varphi)(n) = \varphi^{−1}(n)$. So we want to show that $p := \varphi{−1}(n)$ is a maximal ideal in $A$. First of all, $p$ is definitely a prime ideal of $A$ and $\varphi$ descends to an injective $k$-algebra homomorphism $ψ : A/p \to B/n$. But the map $k \to B/n$ defines a finite field extension of $k$ by Corollary 1.12. So the integral domain $A/p$ is trapped between a finite field extension. Such domains are necessarily fields, thus $p$ is maximal in $A$.

In the second last sentence, the writer says that the integral domain $A/p$ is trapped between a finite field extension. I don't exactly know what it means, but I think it means that there are two injective ring homomorphisms $f:k\to A/p$ and $g:A/p\to B/n$ such that $g\circ f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?

$A$ and $B$ be finitely generated algebras over $k$. Let $\mathfrak m $ be maximal ideal of $B$. We have an injective map $A/\phi ^{-1}(\mathfrak m) \rightarrow B/\mathfrak m $. Identify $A/\phi ^{-1}(\mathfrak m)$ to its image via this map. Let $T\in A/\phi ^{-1}(\mathfrak m) $, then $1/T \in B/ \mathfrak m $- which is algebraic extension of the field $k$. So $1/T $ is there is a monic polynomial over $k$ which $1/T$ satisfies, multiplying this by $T^{n-1}$ you get that $1/T \in A/\phi ^{-1}(\mathfrak m) $ and you are done.
