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This is exercise 3.6.J from the most recent Vakil's notes.

Suppose that k is a field, and A is a finitely generated k-algebra. Show that closed points of Spec A are dense, by showing that if f ∈ A, and D(f) is a nonempty (distinguished) open subset of Spec A, then D(f) contains a closed point of Spec A. Hint: note that $A_f$ is also a finitely generated k-algebra. Use the Nullstellensatz 3.2.5 to recognize closed points of Spec of a finitely generated k-algebra B as those for which the residue field is a finite extension of k. Apply this to both B = A and B = $A_f$.

I know we have $A_f/pA_f=(A/p)_f$, then if p is a prime ideal of $A_f$, then the left side is a field and is a finite extension of k. Then how should I get the conclusion that $A/p$ is also a field?

hardmath
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    As an irrelevant comment, this is actually false for schemes in general! It's possible to have a scheme with no closed points at all. – Mark Jan 14 '16 at 13:51

1 Answers1

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We know that $(A/\mathfrak p)_f$ is finite-dimensional. Since $A/\mathfrak p$ is a domain and the image of $f$ in $A/\mathfrak p$ is nonzero, the map $A/\mathfrak p \to (A\mathfrak p)_f$ is injective. Hence $A/\mathfrak p$ is a domain that is a finite-dimensional $k$-algebra. Therefore, it is a field (exercise).

Remy
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  • Tnx, I did conclude everything by myself, except the hint that a domain that is a finite vector space over a field, must be also a field (extension). – Tal Oct 15 '22 at 18:44