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

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 $\rm{Spec}(A)$.

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.

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.

• So you have shown that $A_f/\mathfrak p_f$ is finite-dimensional. How to conclude $A/\mathfrak p$ is finite-dimensional? $\ddot\smile$ (Note also that the notions finite $k$-algebra and finitely generated $k$-algebra are totally different!)
– cqfd
Commented Jun 3, 2021 at 6:44
• @ShiveringSoldier: Oops, thank you. I'll fix that. Also, what is the difference you're referring to? That the former means finite-dimensional as $k$-vector space and the latter means that it's finitely generated as a $k$-algebra? (That is, it is of the form $k[r_1, \ldots, r_n]$?) (Hopefully, you didn't mean that the former means that it is finite as a set. I thought it was standard terminology to use "finite vector space" to mean "finite dimensional".) Commented Jun 4, 2021 at 10:34
• Yes, that's correct. The latter is also referred to as finite type, I guess.
– cqfd
Commented Jun 4, 2021 at 10:38
• Why aren't we saying $A_f$ is a finitely generated $k$-algebra here? since it seems like that's what we need to apply HN? or am I confused? Commented Jan 30, 2022 at 21:57
• @BelowAverageIntelligence: I did say that in my solution. Commented Jan 31, 2022 at 14:39

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.

• As I have mentioned this is not what I am expecting...
– user312648
Commented Feb 9, 2017 at 14:17
• Use the fact that $p=\sqrt{p}=\bigcap_{p\subset M \text{ maximal}}M$, As $f\not\in p$, then there exist $M$ maximal containing $p$ wich does not contain $f$. Commented Feb 9, 2017 at 14:24
• This then mean that closed points of spec A is dense irrespective of what $A$ is..
– user312648
Commented Feb 9, 2017 at 14:31
• The equality $p=\bigcap_{p\subset M\text{ maxiamal}}M$ hold in a finitely generated algebra over a field. Commented Feb 9, 2017 at 14:37
• Ok OK.. I understand
– user312648
Commented Feb 9, 2017 at 14:51

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 $$D(f)$$.