用户: 数学迷/General

Committee: Bhargav Bhatt (chair), Chenyang Xu, Ruobing Zhang.

Special topics: Algebraic Number Theory, Commutative Algebra.

I took my general on 02/21/2022, 1:30pm–3:00pm EST (UTC−5). It was on Ruobing Zhang’s office since Bhargav Bhatt has not got an office yet.

I try to report the whole process as faithfully as I can, but I cannot remember our words and sentences exactly. In the brackets are descriptions and my comments afterwards.

Xu: What are your special topics?

“Algebraic number theory and commutative algebra.”

Bhatt: Tell me about the Galois group of a local field.

“It contains 3 parts: the unramified part is $\hat{\mathbb{Z}}$, the tame part is ${\prod }_{\ell \ne p}{\mathbb{Z}}_{\ell }$, and the wild part is pro-$p$. Actually the tame part is ${\prod }_{\ell \ne p}{\mathbb{Z}}_{\ell }(1)$ if you take the conjugate action into account.”

Bhatt: You should also mention $\mathbb{R}$ but that’s fine. What about its $\ell$-adic representations?

“First, ${\mathrm{GL}}_n({\mathbb{Q}}_{\ell })$ is mainly pro-$\ell$ so the pro-$p$ part maps to a finite group; also only ${\mathbb{Z}}_{\ell }$ in the tame part and the unramified part can possibly map to an infinite group. But in fact the tame part also maps to a finite group, because say $\phi$ is a Frobenius and $a$ is a generator of the ${\mathbb{Z}}_{\ell }$ in the tame part, then $\phi a{\phi }^{-1}=a^q$ so the eigenvalues of $a$ must be in ${\mu }_{q-1}$.” (I made a mistake here; the eigenvalues don’t give finiteness.)

Bhatt: Wait. Why is it finite given that the eigenvalues are roots of unity?

“Ah! It might be not semisimple! OK. Then let’s raise $a$ to a power and then take log, and get something nilpotent. Let $N=\frac{1}{q-1}\log (a^{q-1})$. Then $N$ is nilpotent on the representation $V$, and it gives a filtration$$0=V_{-n-1}\subseteq \cdots \subseteq V_0\subseteq \cdots \subseteq V_n=V$$by Deligne such that $N{V}_i\subseteq V_{i-2}$. Is it called the weight filtration?” (In fact it’s the monodromy filtration.)

Bhatt: How does the Frobenius act on these?

“(After thinking for a while) I can’t remember.”

Bhatt: How is the filtration defined?

“(After thinking for a while) I can’t remember.”

Bhatt: That’s fine. I also can’t remember. (It’s Deligne’s magic. Let $n$ be the smallest number such that $N^{n+1}=0$. We use induction on $n$. For $n=0$ it’s obvious. For $n>0$ we just put $V_{n-1}=\ker (N^n)$ and $V_{-n}=\mathrm{im}(N^n)$, and since in $V_{n-1}/V_{-n}$ we have $N^n=0$, the induction gives us the rest. For this filtration we have $N^i$ induces an isomorphism $V_i/V_{i-1}\to V_{-i}/V_{-i-1}$. We call things in $V_{-i}$ primitive if they’re killed by $N$; then for representations that come from geometry, the weight monodromy conjecture predicts that $\phi$ acts by weight $i$ on the primitive elements in $V_{-i}$, which means that its eigenvalues are algebraic numbers with conjugates all of modulus $q^{i/2}$.)

Bhatt: Tell me about the Brauer groups of finite fields.

“Since finite division algebras are fields, we have $\mathrm{Br}(k)=0$ for $k$ finite.”

Bhatt: What about other fields?

“For $K$ non-Archimedean local we have $\mathrm{Br}(K)=\mathbb{Q}/\mathbb{Z}$ by the invariant map. For Archimedean, $\mathrm{Br}(\mathbb{R})=\frac{1}{2}\mathbb{Z}/\mathbb{Z}$ and $\mathrm{Br}(\mathbb{C})=0$. For $F$ global, we have a short exact sequence$$0\to \mathrm{Br}(F)\to \underset{v}{⨁}\mathrm{Br}(F_v)\to \mathbb{Q}/\mathbb{Z}\to 0$$where the second map is taking sum.”

Bhatt: Can you write the elements explicitly for $K$?

“Say $\frac{d}{n}\in \mathbb{Q}/\mathbb{Z}$. Let $\pi$ be a uniformizer of $K$. Then the division algebra should be something like $K⟨i,j⟩/⟨i^n={\pi }^d,j\cdots ,ij{i}^{-1}=\cdots ⟩$ (stuck for a while). OK, let $K_n$ be the degree-$n$ unramified extension of $K$. Then it should be $K_n⟨i⟩/⟨i^n={\pi }^d,ix{i}^{-1}=\sigma (x)⟩$ for $x\in K_n$, where $\sigma$ generates the Galois group $G(K_n/K)$.” (It’s actually the Frobenius but I forgot to say.)

Bhatt: How do you prove that these are the only elements?

“I can prove $H^2(G_K,{K^{\text{s}}}^{\times })=\mathbb{Q}/\mathbb{Z}$. We have an inflation map from $H^2(G(K^{\text{un}}/K),{K^{\text{un}}}^{\times })$ to it. We prove that this is an isomorphism and then directly compute the unramified one since $G(K^{\text{un}}/K)=\hat{\mathbb{Z}}$.”

Bhatt: Wait. $\hat{\mathbb{Z}}$ is procyclic. How can this $H^2$ be nonzero? (In fact the cohomological dimension $1$ only holds for torsion modules which I didn’t realize. Maybe he mistakenly thought that this holds for all modules.)

“Let’s compute it. We first compute $H^2(G(K_n/K),K_n^{\times })$ by the exact sequence$$1\to {\mathcal{O}}_{K_n}^{\times }\to K_n^{\times }\to k_n^{\times }\to 1$$...”

Bhatt: What is $k_n^{\times }$?

“It’s the multiplicative group of the residue field ...”

Bhatt: I mean it can’t be finite.

“Oh, it should be $\mathbb{Z}$, so it’s $$1\to {\mathcal{O}}_{K_n}^{\times }\to K_n^{\times }\to \mathbb{Z}\to 0$$OK. Now Tate periodicity gives us $H^2(G(K_n/K),\mathbb{Z})=H_{\text{T}}^0(G(K_n/K),\mathbb{Z})=\mathbb{Z}/n$, and the Herbrand quotient of ${\mathcal{O}}_{K_n}^{\times }$ is $1$, and the $H^1$ of $K_n^{\times }$ and $\mathbb{Z}$ are both $0$, so one can see that $|H^2(G(K_n/K),K_n^{\times })|=n$ ... but that’s not enough. How can we see that $H^0$ and $H^1$ of ${\mathcal{O}}_{K_n}^{\times }$ are $0$ ...”

Bhatt: How can you think of $H^1(G(K_n/K),{\mathcal{O}}_{K_n}^{\times })$ in an abstract way?

“It classifies the line bundles over ${\mathcal{O}}_K$ that are trivialized over ${\mathcal{O}}_{K_n}$ ... so it’s trivial! Now we have $H^2(G(K_n/K),K_n^{\times })=\mathbb{Z}/n$.”

Bhatt: But when you take the colimit where do the generators go?

“Yeah. The transition maps of the Tate resolutions are $$\cdots \mathbb{Z}[\mathbb{Z}/mn]\mathbb{Z}[\mathbb{Z}/mn]\mathbb{Z}[\mathbb{Z}/mn]\cdots \mathbb{Z}[\mathbb{Z}/n]\mathbb{Z}[\mathbb{Z}/n]\mathbb{Z}[\mathbb{Z}/n]$$and they vanish in the colimit ... You were right ... but $H^2(G(K^{\text{un}}/K),{K^{\text{un}}}^{\times })$ should be $\mathbb{Q}/\mathbb{Z}$ since every division algebra splits over $K^{\text{un}}$ ...” (In fact he was wrong; when you take $\mathrm{Hom}$ from this complex to a non-torsion module, you won’t necessarily get $0$ in the colimit of $H^2$.)

Bhatt: That’s fine. I’ll say that the class actually comes from some $H^1$. Chenyang, you should ask him about commutative algebra.

Xu: Do you know the definition of strongly $F$-regular?

“I don’t know.”

Xu: OK. What’s a Cohen–Macaulay ring?

“We say a $d$-dimensional Noetherian local ring $(A,\mathfrak{m},k)$ Cohen–Macaulay if there is a length-$d$ regular sequence inside $\mathfrak{m}$, or equivalently ${\mathrm{Ext}}^{<d}(k,A)=0$.”

Xu: What’s a Gorenstein ring?

“$A$ is Gorenstein if it is its own dualizing complex, up to some shift.”

Xu: Can you give another characterization?

“$A$ is of finite injective dimension over itself.”

Xu: Can you prove that they are equivalent?

“Let me first recall the definition of the dualizing complex. A dualizing complex $\omega$ is one that satisfies$$R\mathrm{Hom}(M,\omega {)}^{\wedge }=R\mathrm{Hom}(R{\Gamma }_{\mathfrak{m}}(M),E(k))$$... but this is not enough to characterize a complex ... Let’s say $\omega \in {\mathsf{D}}_{\mathsf{Coh}}^b(A)$; then it will be enough. Now a dualizing complex always has finite injective dimension, since the functor $R{\Gamma }_{\mathfrak{m}}$ has finite cohomological dimension and $E(k)$ is injective. For the other direction, I have no idea. (That’s because I forgot the true definition of the dualizing complex. As in stacks 0A7B, a dualizing complex is some $\omega \in {\mathsf{D}}_{\mathsf{Coh}}^b(A)$ with finite injective dimension such that $A=R\mathrm{Hom}(\omega ,\omega )$. Now if $A$ itself is of finite injective dimension, it is obviously a dualizing complex.) By the way, I learned this local duality by the torsion-complete equivalence.”

Xu: That’s fine. Can you prove that a Gorenstein ring is Cohen–Macaulay?

“If $A$ is Gorenstein then$$R\mathrm{Hom}(k,A[d])=R\mathrm{Hom}(k,E(k))$$is on degree $0$, so ${\mathrm{Ext}}^{<d}(k,A)=0$.”

Xu: How can you characterize Cohen–Macaulay using the dualizing complex?

“$A$ is Cohen–Macaulay iff $\omega$ is concentrated on degree $d$.”

Xu: Give me examples of these kinds of rings, and a ring that is Cohen–Macaulay but not Gorenstein.

“Regular rings like $k[x_1,\dots ,x_n]$ are always CM and Gorenstein. For a CM but not Gorenstein example, one-dimensional integral domains are always CM ... OK, let me give a zero-dimensional one. $k[x,y]/(x^2,xy,y^2)$ is CM but not Gorenstein.”

Xu: Talk about the completion. Is the completion of a UFD still a UFD?

“No, but I don’t know exactly the example.”

Xu: What properties are preserved?

“Say $(R,\mathfrak{m})$ is Noetherian local with completion $\hat{R}$. The preserved ones are Cohen–Macaulay, Gorenstein, and of course regular.”

Xu: How can you prove this?

“I can’t find a unified way ... For CM we only need that this is faithfully flat and they are of the same dimension ... I think for others this works too.”

Bhatt: Wait. For regularity ...

“Oh I need the maximal ideal of $\hat{R}$ is $\mathfrak{m}\hat{R}$.”

Xu: How about normality?

“I don’t think completion preserves normality.”

Xu: How about $({\text{S}}_n)$?

“$({\text{S}}_n)$ will be fine ... but $({\text{R}}_n)$ behaves badly. It will behave well if $R$ is a G-ring.” (I was wrong, and Xu didn’t realize it either. Completion is very bad and doesn’t preserve $({\text{S}}_n)$. See stacks 0AL7.)

Xu: Good. Let’s say a finite group $G$ acts on ${\mathbb{C}}^n$. What can you say about the singularity of the quotient ${\mathbb{C}}^n/G$?

“I don’t know.”

Xu: Then make a guess.

“I have no idea. I haven’t seen this before.”

Xu: What is the dualizing complex of ${\mathbb{C}}^n$?

“It’s ${\Omega }^n$.”

Xu: What is an explicit generator and how does $G$ act?

“It’s $\mathrm{d}{x}_1\cdots \mathrm{d}{x}_n$ and $G$ acts by determinant.”

Xu: Then when will it descend to ${\mathbb{C}}^n/G$?

“Ah, when $G$ maps into ${\mathrm{SL}}_n$. So it’s Gorenstein iff $G$ maps into ${\mathrm{SL}}_n$ and ... it’s always CM since it’s a direct summand of something regular!”

Xu: Correct. I think you can ask him some analysis.

Zhang: First, complex analysis. What can you say about a holomorphic function on a punctured disk? When does it extend to the origin?

“When it’s bounded.”

Zhang: Can you relax that condition?

“I think we only need that it grows slower than $\frac{1}{z}$.”

Zhang: Can you prove it?

“I can’t handle this.” (I only needed to multiply it by $z$, but I forgot.)

Zhang: OK. Classify the singularities of a holomorphic function.

“Removable singularities, poles, and essential singularities.”

Zhang: Give me an example of an essential singularity.

“$e^{1/z}$ at $0$.”

Zhang: Is there a smooth but non-analytic function?

“$\{\begin{array}{ll}{e}^{-1/x},& x>0.\\ 0,& x\le 0.\end{array}$”

Zhang: What’s an entire function?

“A function that is holomorphic on the entire plane.”

Zhang: Give me one with zeroes exactly the integers.

“$\sin (\pi z)$.”

Zhang: How about one with zeroes only non-positive integers? (I almost started to write the Weierstrass product.) Do you know what’s the Gamma function?

“$\Gamma (z)={\int }_0^{+\infty }{t}^{z-1}{e}^{-t}dt$.”

Zhang: What are its poles?

“Non-positive integers, all of order one.”

Zhang: Simple poles. Then $\frac{1}{\Gamma (z)}$ is the desired function. If I have a sequence of discrete points on the plane, is there an entire function with zeroes on these points?

“Yes. The Weierstrass product. Say your sequence is $a_1,a_2,\dots$ and they are all nonzero, then$$\prod _{n=1}^{\infty }\left(1-\frac{z}{a_n}\right)e^{\text{truncation of the Taylor series of }-\log (1-z/a_n)}$$will have the precribed zeroes, and it will converge if the numbers of terms in the truncations grow fast enough.”

Zhang: What’s Picard’s little theorem?

“It says that if an entire function avoids two points in the plane, then it must be constant. This is essentially because the universal cover of a plane punctured two points is the unit disk.”

Zhang: Can you prove it?

“We first draw $|z|=1$ and $\Re (z)=\pm 1$ in the upper half plane (I drew them on the board); then use Riemann mapping theorem to map the area here to the whole upper half plane; then use Schwarz reflection, and this (the source) will fill the upper half plane and this (the target) will avoid two points and cover the plane infinitely many times. Now we have the covering map, and the upper half plane is the same as the disk.”

Zhang: Tell me the definition of a harmonic function.

“A function with Laplacian $0$.”

Zhang: Say there is a harmonic function on the punctured disk. Can you always realize it as the real part of a holomorphic function?

“No. $\log |z|$.”

Zhang: Is a positive harmonic function on the whole plane necessarily constant?

“Yes. The plane is simply connected and I can realize it as the immaginary part of a holomorphic function, which will be constant.” (In fact there is a better argument that works for ${\mathbb{R}}^n$: if $u$ is such a function then for $x,y\in {\mathbb{R}}^n$, let $r=|x-y|$, and for all $R>0$ we have$$u(x)=\frac{1}{m(B_R)}{\int }_{B_R(x)}u\le \frac{1}{m(B_R)}{\int }_{B_{R+r}(y)}u=\frac{m(B_{R+r})}{m(B_R)}u(y)$$by the mean value property. Now let $R\to \infty$ we get $u(x)\le u(y)$ for all $x,y\in {\mathbb{R}}^n$, so $u$ is constant.)

Zhang: Now real analysis. What is $L^p$-convergence?

“The $L^p$-norm is $\Vert f{\Vert }_p={\left(\int |f{|}^p\right)}^{1/p}$ and we say $f_1,f_2,\dots$ converge to $f$ in $L^p$ if $\Vert f_n-f{\Vert }_p\to 0$ as $n\to \infty$.”

Zhang: Does this imply $f_n$ converge almost everywhere?

“No. It only implies convergence in measure.”

Zhang: An example?

“Let the space be $[0,1]$. Let $f_1(x)=\{\begin{array}{ll}1,& x\in [0,1/2];\\ 0,& \text{otherwise};\end{array}$ $f_2(x)=\{\begin{array}{ll}1,& x\in [1/2,1];\\ 0,& \text{otherwise};\end{array}$ $f_3(x)=\{\begin{array}{ll}1,& x\in [0,1/4];\\ 0,& \text{otherwise};\end{array}$ $f_4(x)=\{\begin{array}{ll}1,& x\in [1/4,1/2];\\ 0,& \text{otherwise};\end{array}$ ...”

Zhang: OK. If a sequence of continuous functions converges to a function, is it necessarily continuous?

“No. $x^n$ on $[0,1]$.”

Zhang: If it decreasingly converges to $0$, does it necessarily converge uniformly? (I was stuck.) Do you know Dini’s convergence theorem?

“I think I used to know it but I forgot ... Yeah, this seems correct.”

Zhang: Fine. Can you say about $L^p$ when $p<1$? Is it a Banach space?

“No. It doesn’t satisfy the triangle inequality. In fact I once learned that for some space for $p<1$ we have $(L^p{)}^{\ast }=0$, when I was studying functional analysis.”

Zhang: Good.

Bhatt: A little more algebra. Say $G$ is a profinite group and we have a representation $G\to {\mathrm{GL}}_2({\mathbb{F}}_p)$. Can you lift it to ${\mathrm{GL}}_2({\mathbb{Z}}_p)$?

“I don’t know. I haven’t seen this before.”

Bhatt: How about lift it to ${\mathrm{GL}}_2(\mathbb{Z}/p^2)$?

“I have no idea ... Wait. A profinite group can be finite, so if I set $G={\mathrm{GL}}_2({\mathbb{F}}_p)$ and the representation to be $\mathrm{id}$, then it cannot be lifted.”

Bhatt: Right. What if $G$ is the Galois group of a field $K$? (I still have no idea.) Can you quantify whether or not it can be lifted?

“We can take the short exact sequence$$1\to 1+p{M}_2(\mathbb{Z}/p^2)\to {\mathrm{GL}}_2(\mathbb{Z}/p^2)\to {\mathrm{GL}}_2({\mathbb{F}}_p)\to 1$$and take $H^{\ast }(K_{\text{eˊt}},-)$ ... so it’s a map $H^0(K_{\text{eˊt}},{\mathrm{GL}}_2({\mathbb{F}}_p))\to H^1(K_{\text{eˊt}},1+p{M}_2(\mathbb{Z}/p^2))$ ...”

Bhatt: It’s a homomorphism. It’s $H^1$.

“Ah, right. So it is $H^1(K_{\text{eˊt}},{\mathrm{GL}}_2({\mathbb{F}}_p))\to H^2(K_{\text{eˊt}},1+p{M}_2(\mathbb{Z}/p^2))=H^2(K_{\text{eˊt}},{\mathbb{F}}_p^4)$ ...”

Bhatt: No, it’s not ${\mathbb{F}}_p^4$; there is an action.

“OK, then it’s $M_2({\mathbb{F}}_p)$ with ${\mathrm{GL}}_2({\mathbb{F}}_p)$ acting by conjugation.”

Bhatt: This is a $p$-torsion group so you’re allowed to pass to prime-to-$p$ extensions.

“Right. So we can assume $G_K$ is pro-$p$. Then what can I do?”

Bhatt: What can you say about a pro-$p$ group mapping to ${\mathrm{GL}}_2({\mathbb{F}}_p)$?

“It should factor through $\left[\begin{array}{cc}1& {\mathbb{F}}_p\\ 0& 1\end{array}\right]$.”

Bhatt: And what field extension does it correspond to? How do you lift it to $\mathbb{Z}/p^2$?

“It corresponds to a degree-$p$ cyclic extension ... and there must be a degree-$p^2$ cyclic extension above by Kummer, and in characteristic $p$ by Artin–Schreier; so we can lift the representation.” (Except in the real closed case, which I forgot.) (Obviously one can lift it all the way up to ${\mathbb{Z}}_p$ by Kummer and Artin–Schreier.)

Bhatt: Great. This was Serre’s argument.

Xu: I will ask you some complex analysis. Can you find a function holomorphic only on the puntured ${\mathbb{A}}^2$?

“No. Hartogs.”

Xu: What does it correspond to, in commutative algebra?

“If you have a line bundle, then a rational section ... we need some regularity here ... OK, let $A$ be a normal Noetherian integral domain, then $A={\bigcap }_{\mathrm{ht}(\mathfrak{p})\le 1}{A}_{\mathfrak{p}}$ in the fraction field.”

Xu: Good.

Bhatt: We haven’t covered the ‘real’ algebraic number theory yet since there were no global fields. So tell me about the finiteness of the class number.

“First let me define the class group. Let $F$ be a global field; then its class group $C_F$ is $\mathrm{Pic}({\mathcal{O}}_F)$, or more concretely, given by the short exact sequence$$1\to F^{\times }\to I_F\to C_F\to 1$$where $I_F$ is the group of fractional ideals.”

Bhatt: What about the function field case?

“Well, for a function field $F=K(X)$, this should be $\mathrm{Pic}(X)$ or ${\mathrm{Pic}}^0(X)$ ... it should be ${\mathrm{Pic}}^0$ for otherwise it would be infinite.”

Bhatt: Then why is it finite?

“We have the Minkowski bound. First recall the Minkowski theorem: for a lattice $L\subseteq V={\mathbb{R}}^n$ and a convex symmetric set $A\subseteq V$, if $m(A)>2^n\mathrm{vol}(L)$, then there exists nonzero $a\in A\cap L$. Then since ${\mathcal{O}}_F\subseteq {\mathcal{O}}_F{\otimes }_{\mathbb{Z}}\mathbb{R}=F{\otimes }_{\mathbb{Q}}\mathbb{R}$ is a lattice, and for an ideal $\mathfrak{a}$ we have $\mathrm{vol}(\mathfrak{a})=N(\mathfrak{a})\mathrm{vol}({\mathcal{O}}_F)$ ... let’s take $\rho >0$ and consider$$A=\{x\in F{\otimes }_{\mathbb{Q}}\mathbb{R}\mid |x{|}_v<\rho ,\forall v\mid \infty \}$$... then for $2^r{\pi }^s{\rho }^n>2^{n}N(\mathfrak{a})\mathrm{vol}({\mathcal{O}}_F)$ we have nonzero $x\in A\cap \mathfrak{a}$. And $x{\mathfrak{a}}^{-1}\subseteq {\mathcal{O}}_F$ will be an ideal of norm less than a constant so there are only finitely many classes.”

Bhatt: How about function fields?

“${\mathrm{Pic}}^0(X)$ is a proper variety, and we’re taking points over ${\mathbb{F}}_q$, a finite field, so it must be finite.”

Bhatt: OK. I think that’s enough. You may leave the room.

“Can I go home?”

Xu: No, you should wait outside until we tell you the result.

(Then I went to the restroom, and when I came back they went out of the room and told me that I passed.)

(The general was a lot of fun. I had hardly spent time preparing it and I learned many interesting new things during the exam. Don’t be afraid of being stuck since they’re supposed to ask until you get stuck.)