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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 $f:kto A/p$ and $g:A/pto B/n$ such that $gcirc f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?

Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g : R to L$ be an injective $K$-linear map. Then, $R$ is a field.

Proof of Theorem 1. Since the $K$-linear map $g : R to L$ is injective, we have $dim R leq dim L$, where "$dim$" refers to the dimension of a $K$-vector space. But $dim L < infty$, since $L$ is finite-dimensional. Hence, $dim R leq dim L < infty$; thus, $R$ is a finite-dimensional $K$-vector space. Therefore, any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces (according to a well-known fact from linear algebra).

Now, let $a in R$ be nonzero. Let $M_a$ denote the map $R to R, r mapsto ar$. This map $M_a : R to R$ is $K$-linear and has kernel $0$ (because every $r in R$ satisfying $ar = 0$ must satisfy $r = 0$ (since $R$ is an integral domain and $a$ is nonzero)); thus, it is injective. Hence, it is an isomorphism of $K$-vector spaces (since any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces). Thus, it is surjective. Therefore, there exists some $s in R$ such that $M_aleft(sright) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_aleft(sright) = as$, so that $as = M_aleft(sright) = 1$. In other words, $s$ is a (multiplicative) inverse of $a$. Hence, $a$ has an inverse.

We have thus proven that every nonzero $a in R$ has an inverse. In other words, the ring $R$ is an integral domain. This proves Theorem 1. $blacksquare$

In your situation, you should apply Theorem 1 to $K = k$, $R = A/p$, $L = B/n$ and $g = psi$.

Suppose $F$ is any field, $E$ is a finite extension field of $F$, and $D$ is an integral domain such that

$F subset D subset E; tag 1$

since

$[E:F] = n < infty, tag 2$

every element of $D$ is algebraic over $F$; thus

$0 ne d in D tag 3$

satisfies some

$p(x) in F[x]; tag 4$

that is,

$p(d) = 0; tag 5$

we may write

$p(x) = displaystyle sum_0^{deg p} p_j x^j, ; p_j in F; tag 6$

then

$displaystyle sum_0^{deg p} p_j d^j = p(d) = 0; tag 7$

furthermore, we may assume $p(x)$ is of minimal degree of all polynomials in $F[x]$ satisfied by $d$. In this case, we must have

$p_0 ne 0; tag 8$

if not, then

$p(x) = displaystyle sum_1^{deg p} p_jx^j = x sum_1^{deg p} p_j x^{j - 1}; tag 9$

thus

$d displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{10}$

and via (4) this forces

$displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{11}$

since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial

$displaystyle sum_1^{deg p} p_j x^{j - 1} in F[x] tag{12}$

of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write

$displaystyle sum_1^{deg p}p_j d^j = -p_0, tag{13}$

or

$d left( -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} right ) = 1, tag{14}$

which shows that

$d^{-1} = -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} in D; tag{15}$

since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.