Why does the integral domain "being trapped between a finite field extension" implies that it is a field?

Question

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?

Answer 1

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 $M_a\left(s\right) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_a\left(s\right) = as$, so that $as = M_a\left(s\right) = 1$. 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 $g = \psi$.

Answer 2

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 via (5),

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

and 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.

Answer 3

$A$ 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.