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?

  • 3
    $\begingroup$ 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. $\endgroup$ – 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

$F \subset D \subset E; \tag 1$


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

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

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


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

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

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


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

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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.