set of closed points is dense in spectrum of $A$.

Question

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 $\rm{Spec}(A)$.

Answer 1

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.

Answer 2

Judging by the phrasing, I guess this is from Vakil's notes. Here's a proof using the version of Nullstellensatz that he's stated (3.2.5):

(HN) If $\mathsf{k}$ is any field, every maximal ideal of $\mathsf{k}[x_1, \ldots, x_n]$ has residue field a finite extension of $\mathsf{k}.$

Instead of using the theorem about the intersection of maximal ideals, I only use the fact that the intersection of all primes is the nilradical. This is true for any ring. In particular, if $f \in A$, then $D(f)$ is empty iff $f$ is a nilpotent.

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It suffices to show that every nonempty $D(f)$ contains a closed point. (Since $\{D(f)\}$ forms a base.)

To this end, let $f \in A$ be such that $D(f) \neq \emptyset.$ Then, $f$ is not nilpotent, and thus, $A_f$ is not the zero-ring. In turn, $A_f$ has a maximal ideal. This is of the form $\mathfrak{p}_f$ for some prime $\mathfrak{p} \subset A$ not containing $f.$ In other words, $[\mathfrak{p}] \in D(f).$ We show that $[\mathfrak{p}]$ is a closed point by showing that $\mathfrak{p}$ is maximal. (Recall that the maximal ideals are the closed points.)

Note that $A/\mathfrak{p}$ is an integral domain. To show that $\mathfrak{p}$ is maximal, we must show that $A/\mathfrak{p}$ is a field. If we can show that $A/\mathfrak{p}$ is finite-dimensional as a $\mathsf{k}$-vector space, then we are done, by (3.2.G).

Note that $(A/\mathfrak{p})_f \cong A_f/\mathfrak{p}_f$ as $\mathsf{k}$-vector spaces and thus, the problem is reduced to showing that $A_f/\mathfrak{p}_f$ is a finite $\mathsf{k}$-vector space.

(This is where we use the fact that $A$ is a finitely generated $\mathsf{k}$-algebra.) Note that $A_f$ is also a finitely generated $\mathsf{k}$-algebra as it is the image of $A[x]$ under $x \mapsto \frac{1}{f}.$
Now, invoking (HN) tells us that $A_f/\mathfrak{p}_f$ is a finite $\mathsf{k}$-vector space, as desired.

Edit: The above only shows that $(A/\mathfrak{p})_f$ is a finite $\mathsf{k}$-vector space. I forgot that we had to show that $(A/\mathfrak{p})$ is also a finite-dimensional $\mathsf{k}$-vector space.

To see that, note that $A/\mathfrak{p}$ is an integral domain and thus, the canonical map $A/\mathfrak{p} \to (A/\mathfrak{p})_f$ is injective. Since this is also a $\mathsf{k}$-linear map, we are done.

Answer 3

I'm wondering if Vakil had a simpler solution in mind—please correct me if I've made a mistake.

We take our non-empty $D(f)$, which, corresponds to $A_f$, and take any maximal ideal $\mathfrak{p}A_f$ of $A_f$. Note that since $\mathfrak{p}\in D(f)$, $f\notin \mathfrak{p}$, so $(A_f)_{\mathfrak{p}} = A_\mathfrak{p}$. In particular, the residue field at $\mathfrak{p}$, as a point of $\operatorname{Spec}(A)$, which is $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$, is the same as the residue field at $\mathfrak{p}A_f$ viewed as a point of $\operatorname{Spec}(A_f)$. Thus we see that the question of $\mathfrak{p}$ being a a closed point in the two rings (both of which are finitely generated $k$-algebras) by the Nullstellensatz reduces to a question of $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ being a finite extension of $k$ or not, and this is really the same question for both rings. So the maximality of $\mathfrak{p}A_f\triangleleft A_f$ implies the closedness of $\mathfrak{p}A_f\in \operatorname{Spec}(A_f)$, and hence that $\kappa(\mathfrak{p}A_f) = A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is a finite extension of $k$, so $\mathfrak{p}$ is a closed point of $\operatorname{Spec}(A)$ contained in $D(f)$.