The following is an exercise by Qing Liu Algebraic geometry and arithmetic curves,
Exercise 1.2.
Let φ: A → B be a homomorphism of the finitely generated algebras over a field. Show that the image of a closed point under Spec φ is a closed point.
The following is the solution of Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.
Write $ k $ for the field below. Let us analyze the statement. A closed point in $ operatorname {Spec} B $ means a maximum ideal $ n $ from $ B $, And $ operatorname {Spec} (φ) (n) = φ ^ {- 1} (n) $, So let's show that $ p: = φ {-1} (n) $ is a maximum ideal in $ A $, First, $ p $ is definitely an ideal ideal of $ A $ and $ φ $ descends to an injectable $ k $Algebra homomorphism $ ψ: A / p to B / n $, But the card $ k to B / n $ Defines a finite field extension of $ k $ from Corollary 1.12. So the holistic domain $ A / p $ is included between a finite field extension. Such domains are necessarily fields $ p $ is maximum in $ A $,
In the penultimate sentence, the author says that the integral area $ A / p $ is included between a finite field extension. I do not know exactly what it means, but I think there are two injective ring homomorphisms $ f: k to A / p $ and $ g: A / p to B / n $ so that $ g circ f $ makes $ B / n $ a finite field extension of $ k $, But why is that? $ A / p $ is a field?
