# 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?

• This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion. – Jyrki Lahtonen Mar 25 '19 at 6:06

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 $$g = \psi$$.

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.

• I'm confused about the last step. Why is $p_0^{-1} \in D$? – Vincent Mar 25 '19 at 9:34
• @Vincent: From (4) and (6), $p_0 \in F \subset D$; since $F$ is a field, $p_0^{-1} \in F \subset D$. – Robert Lewis Mar 25 '19 at 16:00
• You are right, this is obvious, I was mixing up $F$ and $E$. Thanks for the clarification! – Vincent Mar 25 '19 at 21:19
• @Vincent: glad to be of service! Cheers! – Robert Lewis Mar 25 '19 at 21:21

$$A$$ 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.