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 $f \in A$, 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 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 fail to understand why do we have to consider that finitely generated algebra and all. All that we want is $D(f)\neq \emptyset$ and then show that it contains a closed point.
Let $f\in A$ and consider localization $A_f$. This ring has a maximal ideal. This corresponds to a prime ideal in $A$ say $\mathfrak{p}$ that does not contain $f$. This says $\mathfrak{p}\in D(f)$. I am stuck here though. I am not sure that i can choose a maximal ideal (which is a closed point) containing $\mathfrak{p}$ and not containing $f$. I am sure this can be solved. Any suggestions are welcome.
I am still not convinced how this set of closed points is different from set of all maximal ideals in case of $\rm{Spec}(A)$.
Best Answer
Let $f\in A$, suppose $D(f)$ does not contain a closed point. Hence for any maximal ideal $M$ in $A$, $f\in M$ , so
$$f\in \displaystyle\bigcap_{M\text{ maximal}}M$$ But by nullstellenzats $\bigcap_{M\text{ maximal}}M=\sqrt{0}$, it follow that $f$ is nilpotent and $D(f)$ is empty.