Let $A$ be a finitely generated $k$-algebra.
I would like to show that the set of closed points of $\operatorname{Spec}A$ is dense, but without using the fact that the nilradical of a finitely generated $k$-algebra is the intersection of all maximal ideals.
I know that the closed points of $\operatorname{Spec}A$ are the maximal ideals of $A$.
I also know that if $\frak m$ is a maximal ideal of $A$, then $A/\frak m$ is a finite extension of $k$, by Hilbert's Nullstellensatz.
I would like to show that if $f \in A$ and $D(f) \ne \varnothing$, then $D(f)$ contains a maximal ideal of $A$.
I know that $A_f$ is also a finitely generated $k$-algebra as well.
Best Answer
Write $A\cong k[x_1,\cdots,x_n]/I$ as a quotient of a polynomial ring. Then $D(f)$ is the spectrum of $A_f\cong k[x_1,\cdots,x_n,z]/(I,zf-1)$, and by the assumption that $D(f)\neq\emptyset$, we have that $(I,zf-1)$ is a proper ideal of $k[x_1,\cdots,x_n,z]$ and is hence contained in a maximal ideal $\mathfrak{m}$. This gives a closed point in $D(f)$.
To see this is also a closed point of $\operatorname{Spec} A$, we use the characterization the closed points in the spectrum of a finitely generated $k$-algebra have residue field a finite extension of $k$. As $(A/I)_f\cong A_f/I_f$, the residue field of the closed point of $D(f)=\operatorname{Spec} A_f$ is the same as the residue field of that same point considered in $\operatorname{Spec} A$, so it is again a closed point of $\operatorname{Spec} A$.