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Answer by kreck for Finite generation and Henselization

2013-06-14T23:52:52Z

You probably meant to assume $R$ is noetherian. And Hironaka seems to have suppressed his appeal to Zariski's Main Theorem, as we'll see below (or maybe someone else sees a more elementary procedure, which is certainly possible).

Consider the schematic support ${\rm{Spec}}(S/I)$ of the $S$-finite $A$ in ${\rm{Spec}}(S)$ (here, $I$ is the annihilator ideal of $A$ in $S$). Its special fiber over ${\rm{Spec}}(R)$ has underlying reduced scheme that coincides with that of the schematic support of $A/mA$ over $S/mS$, and this latter schematic support is $k$-finite (with $k := R/m$) since $A/mA$ is $k$-finite. Thus, $S/I$ has $k$-finite special fiber over $R$ since such finiteness is insensitive to killing nilpotents.

Since $R$ is noetherian, so $I$ is finitely generated, there is a residually trivial etale neighborhood $({\rm{Spec}}(R'),\xi)$ of $(m,z) \in {\rm{Spec}}(R[z])$ such that $I$ has a finite set of generators coming from $R'$ (via the unique map $R'_{\xi} \rightarrow S$ over $R[z]_{(m,z)}$). Letting $I' \subset R'$ be the ideal generated by those elements of $R'$, we see that $R'/I'$ has henselization $S/I$ at $\xi$ with $k$-finite special fiber over $R$. But henselization is compatible with quotients, so the special fiber of $R'_{\xi}/I'$ over $R$ is an essentially finite type local $k$-algebra with $k$-finite henselization, so $R'_{\xi}/I'$ has $k$-finite special fiber over $R$. In other words, ${\rm{Spec}}(R'/I') \rightarrow {\rm{Spec}}(R)$ is quasi-finite at $\xi$.

By openness of the quasi-finite locus of a finite type map between noetherian schemes, we can localize $R'$ around $\xi$ so that $R'/I'$ is a quasi-finite $R$-algebra. Hence, by Zariski's Main Theorem, ${\rm{Spec}}(R'/I')$ is Zariski-open in a finite $R$-algebra. But $R$ is henselian, so by Hensel's Lemma (in the EGA form) applied to lifting idempotents from the special fiber we know that every finite $R$-algebra is a direct product of finite local $R$-algebras. Consequently, by shrinking some more around the point $\xi$ in the special fiber we can arrange that ${\rm{Spec}}(R'/I')$ is Zariski-open in a finite local $R$-scheme with closed point $\xi$, so this Zariski-open locus is the entire space. In other words, we get to the case that $R'/I'$ is $R$-finite and local. Thus, it is equal to its own henselization, which is $S/I$ by design.

We have proved that $S/I$ is $R$-finite, yet $A$ is an $S/I$-module, so $A$ s $R$-finite. QED

Answer by kreck for Are extensions of profinite groups profinite?

2013-01-02T17:26:59Z

[EDIT: I assume $X$ is given the subspace topology and $G$ the quotient topology.]

The answer to (iii) (and (i) and (ii)) is "yes". Is this not treated in the book "Profinite groups" (which I've never looked at)? I'm less sure about the "explicit" request, since (even for finite $X$) it seems a bit hard to "see" an open subgroup of $E$ that omits a given nontrivial $x \in X$ (then we could shrink it a bit to also be normal and intersect with preimages in $E$ of open normal subgroups of $G$ to get what you want).

Here is the proof for (iii) in the affirmative.

Step 1 (Hausdorff nonsense): Firstly, since $G$ is (I presume) given the quotient topology of $E$ and that is Hausdorff, its identity is closed and hence $X$ is closed in $E$. I assume $X$ is meant to have the subspace topology and so since its identity point is closed (as $X$ is Hausdorff) it follows that the identity of $E$ is a closed point. Thus, since $E$ is a topological group, it follows that $E$ is Hausdorff. That was pretty boring, and perhaps you were assuming $E$ to be Hausdorff at the outset. Quomodocumque.

Step 2 (reformulation): Now we recall that among Hausdorff topological groups, the profinite ones are precisely those that are compact and totally disconnected (i.e., only non-empty connected subsets are points). This is proved in Montgomery-Zippin and elsewhere I presume. So we just have to check that $E$ inherits each such property separately from $X$ and $G$, and these are elementary as follows.

For the total disconnectedness it is equivalent to say that the only connected closed subgroup is the trivial one (since the connected component of the identity point is closed and visibly a subgroup). But such a subgroup of $E$ has trivial image in the totally disconnected $G$ and so lies in the totally disconnected $X$ and thus is trivial, so $E$ is totally disconnected.

Step 3 (properness): Next, we verify the compactness. There may be a clever way to see it using open covers or nets, but I don't see such an argument offhand (since I don't recall in what generality one knows that quotient maps between topological groups admit local continuous cross-sections), so here is a direct argument using topological properness (for Hausdorff spaces) in the sense of Bourbaki. Maybe the argument can be done more efficiently; I just give what comes to mind at the moment.

Recall that a separated continuous map between topological spaces defined to be proper when it is universally closed (in the category of all topological spaces), and that this is equivalent to the map being closed with quasi-compact fibers. In particular, since properness is preserved under composition and $G$ is proper over a point, to prove the Hausdorff $E$ is compact it suffices to show that $f$ is proper.

More specifically, since $f$ has compact fibers (translates of $X$), we just have to show that $f$ is closed. That is, if $C$ is a closed set in $E$ then we want to show that $f(C)$ is closed in $G$. Since $G$ has the quotient topology from $E$, this means that $f^{-1}(f(C))$ is closed in $E$. To prove this closedness, we'll use compactness of $X$ in another way.

The map $X \times E \rightarrow E \times_G E$ defined by $(x,e) \mapsto (xe, e)$ is a topological isomorphism (respecting 2nd projections), so if $C$ is a closed set in $E$ then $X \times C$ goes over to a closed set in $E \times_G E$, and this closed set is $S := f^{-1}(f(C)) \times_{f(C)} C$. Note that its image under the first projection to $E$ is $f^{-1}(f(C))$. But $E \times_G E$ is also identified with $E \times X$ respecting first projections (via $(e,x) \mapsto (e, ex)$, say), and this first projection is proper since $X$ is compact Hausdorff. In particular, this first projection is a closed map, so $f^{-1}(f(C))$ is closed in $E$ because of the closedness of $S$ in the fiber product. That completes the proof of compactness of $E$.

QED

Answer by kreck for If a formal power series over the complex numbers satisfies a polynomial identity, does it imply that the power series has a radius of convergence?

2012-12-14T12:59:17Z

The result holds allowing several variables $z_1,\dots,z_n$, by using Artin approximation. (The method of proof below applies verbatim over non-archimedean fields of any characteristic, where "analytification" below may be taken in the naive sense over such fields or in the sense of rigid-analytic geometry. A variant on the argument, again using Artin approximation -- or rather its generalization proved by Popescu -- shows that if $R$ is any excellent normal local noetherian domain then its henselization $R^{\rm{h}}$ is the subring of elements of $\widehat{R}$ that satisfy a 1-variable polynomial equation over $R$ of positive degree; recall that for any local noetherian ring $R$, $R^{\rm{h}}$ is local noetherian and the map $R \rightarrow R^{\rm{h}}$ induces an isomorphism between completions.)

To make a precise statement about convergent power series, let $\Phi \in \mathbf{C}[w,z_1,\dots,z_n]$ involve $w$, and let $P \in \mathbf{C}[\![z_1,\dots,z_n]\!]$ be a formal power series such that $P(0,\dots,0) = 0$ and $\Phi(P,z_1,\dots,z_n) = 0$. We claim that $P$ converges on a ball around $(0,\dots,0)$ with positive radius. Moreover, we claim that $P$ lies in the subring of $\mathbf{C}[\![z_1,\dots,z_n]\!]$ given by the henselization $R^{\rm{h}}$ of the algebraic local ring $R = \mathbf{C}[z_1,\dots,z_n]_{(z_1,\dots,z_n)}$.

Since $\widehat{R}$ is a domain and $\Phi \in R[w]$ has positive $w$-degree, the equation $\Phi = 0$ has at most finitely many solutions in $\widehat{R}$. Thus, there is an exponent $e > 0$ such that distinct solutions in $\widehat{R}$ are distinct modulo the $e$th power of the maximal ideal $\mathfrak{m}$ of $\widehat{R}$. By the Artin approximation theorem, for any $f \in \widehat{R}$ satisfying $\Phi(f,z_1,\dots,z_n)=0$ and any $m > 0$ there exists $f_m$ in the henselization $R^{\rm{h}}$ such that $\Phi(f_m,z_1,\dots,z_n)=0$ and $f_m \equiv f \bmod \mathfrak{m}^m$. Taking $m = e$, the solutions $f, f_e \in \widehat{R}$ to $\Phi=0$ must coincide! In other words, all solutions to $\Phi=0$ in $\widehat{R}$ lie in $R^{\rm{h}}$.

By construction, $R^{\rm{h}}$ is a direct limit of local-etale $R$-algebras, so there exists a local-etale map $R \rightarrow R'$ such that all solutions to $\Phi=0$ in $\widehat{R}$ lie in $R'$ (via the canonical isomorphism $\widehat{R} \rightarrow \widehat{R'}$ and the inclusion of $R'$ into its own completion). By definition of "local-etale", there is an etale map $h:V \rightarrow \mathbf{A}^n_{\mathbf{C}}$ and a point $v \in h^{-1}(0)$ such that $O_{V,v} = R'$ as $R$-algebras. (In particular, $V$ is smooth.) Since $h$ is etale, it follows from the Zariski local structure theorem for etale morphisms and the analytic inverse function theorem in several complex variables that the analytification $h^{\rm{an}}$ is a local isomorphism. In particular, $O_{V^{\rm{an}},v}$ is identified via $h^{\rm{an}}$-pullback with the local ring $O_{(\mathbf{A}^n_{\mathbf{C}})^{\rm{an}},0}$ of convergent power series in $z_1,\dots,z_n$ at the origin.

Passing to completions on this identification of analytic local rings, we recover the identification of $O_{V,v}^{\wedge} = \widehat{R'}$ with $\widehat{R}$ induced by $h$, so it follows that under the inclusion $$R' = O_{V,v} \subset O_{V^{\rm{an}},v} = O_{(\mathbf{A}^n_{\mathbf{C}})^{\rm{an}},0}$$ the element of $R'$ that "is" $P$ (provided by Artin approximation) maps to a convergent power series near the origin that has Taylor expansion at the origin equal to $P$. Hence, $P$ has positive radius of convergence. QED

Answer by kreck for two non-degenerate quadratic forms on $GF(2)^2r$

2012-12-06T06:44:13Z

The short answer: Robert Wilson's nice book "The finite simple groups" has a huge amount of information about quadratic spaces (and other linear algebra structures) over finite fields of all characteristics. (This has nothing to do with "Einstein on the Beach" by a different Robert Wilson.)

The long answer: over any finite field $k$ of any characteristic there are exactly two isomorphism classes of non-degenerate quadratic forms of even rank $n \ge 2$, generalizing what you have listed for even rank $2r$ in characteristic 2, and to see this conceptually it seems best to use the formalism of algebraic groups.

The "split" form $Q_n^+$ is defined to be $$x_1 x_2 + \dots + x_{n-1}x_n,$$ and to define another one we first define $Q^{-}_2$ to be the norm form for the quadratic extension $k'$ of $k$ (given by homogenizing an irreducible quadratic polynomial over $k$). Note that the equation $Q^{-}_2(x,y) = 1$ has $k$-rational solution set of size $(q^2-1)/(q-1) = q+1$ for $q := |k|$, whereas $Q^{+}_2=1$ has solution set of size $q-1$, so indeed $Q^{-}_2 \not\simeq Q^{+}_2$. We define $Q^{-}_n(x_1,\dots,x_n) = Q^{-}_2(x_1,x_2) + Q^{+}_{n-2}(x_3,\dots,x_n)$ for even $n \ge 4$. Witt cancellation is valid in even rank without restriction on the characteristic, so it shows that $Q^{-}_n \not\simeq Q^{+}_n$ for all even $n \ge 2$.

For even $n \ge 2$, to show that there are no more isomorphism classes than these we first recall that the set of isomorphism classes of rank-$n$ non-degenerate quadratic spaces $(V,Q)$ over a field $K$ are in bijection with the pointed Galois cohomology set ${\rm{H}}^1(K,{\rm{O}}_n)$. Thus, the exact sequence (for any even $n$) $$1 \rightarrow {\rm{SO}}_n \rightarrow {\rm{O}}_n \rightarrow \mathbf{Z}/2\mathbf{Z} \rightarrow 1$$ of $K$-groups defines a map of pointed sets ${\rm{H}}^1(K,{\rm{O}}_n) \rightarrow {\rm{H}}^1(K,\mathbf{Z}/2\mathbf{Z})$. Since $\mathbf{Z}/2\mathbf{Z}$ is commutative, by the "twisting method" we see that the fiber through the class of $(V,Q)$ is identified with the image of ${\rm{H}}^1(K,{\rm{SO}}(Q))$ in ${\rm{H}}^1(K,{\rm{O}}(Q))$. For finite $K$ (and still assuming $n$ is even), the pointed set ${\rm{H}}^1(K,{\rm{SO}}(Q))$ vanishes, due to Lang's theorem (since ${\rm{SO}}(Q)$ is smooth and connected). Hence, for finite $k$ and even $n$ the map of sets
$${\rm{H}}^1(k,{\rm{O}}_n) \rightarrow {\rm{H}}^1(k,\mathbf{Z}/2\mathbf{Z})$$ is injective. The target has size 2, so we're done for even $n$.

The case of odd $n$ is easier to understand since over any field $K$ we have ${\rm{O}}_n = {\rm{SO}}_n \times \mu_2$ as $K$-group schemes (where projection to $\mu_2$ is the determinant) with ${\rm{SO}}_n$ smooth and connected. Lang's theorem implies that for finite $k$ the natural map $${\rm{H}}^1(k,{\rm{O}}_n) \rightarrow {\rm{H}}^1(k,\mu_2) = k^{\times}/(k^{\times})^2$$ is bijective. Thus, again the size is 2 when the characteristic is odd, and in addition to the "split" form $Q^+_n = x_0^2 + Q^+_{n-1}$ another non-degenerate one of rank $n$ is $Q^{-}_n = x_0^2 + Q^{-}_{n-1}$. (We have $Q^{-}_n \not\simeq Q^{+}_n$ due to Witt's cancellation theorem once again, with the caveat that in odd rank Witt cancellation is only valid away from characteristic 2.)

The cohomology set is trivial when $k$ has characteristic 2 (so $Q^+_n := x_0^2 + Q^+_{n-1}$ is the only example up to isomorphism of odd rank $n$ over finite fields of characteristic 2; if we make the definition $Q^{-}_n := x_0^2 + Q^{-}_{n-1}$ then it happens to be isomorphic to $Q^{+}_n$ over $k$).

Comment by

2013-06-14T22:29:17Z

Don't you have a mentor for the summer project who can give you advice?

Comment by

2013-06-14T13:14:40Z

Provided you know what whatever definition you use agrees with inverting $p$ upon the inverse-limit definition applied to a Galois lattice, you want to identify ${\rm{H}}^2(K,\mathbf{Z}_p(1))= \invlim {\rm{H}}^2(K,\mu_{p^n})$ with $\mathbf{Z}_p$, or more specifically to show ${\rm{H}}^2(K,\mu_{p^n})\simeq \mathbf{Z}/(p^n)$ compatibly with change in $n$ ("$p$-mult." on left, reduction on right). Since ${\rm{Br}}(K)=\mathbf{Q}/\mathbf{Z}$, pass to $p$-power torsion and look at $p$-multiplication from $p^{n+1}$-torsion into $p^n$-torsion, and relate Kummer sequences for $p^n$ and $p^{n+1}$, etc.

Comment by

2013-06-14T06:38:46Z

@Jacob: I don't know what a "coherator" is (and suspect it doesn't matter), and the final question in my initial comment was not meant as a joke. If you know the trick to prove in the scheme case that the kernel of $O_X\rightarrow \mathcal{End}(F)$ is quasi-coherent (even though the target is generally not) then you should see it works the same for algebraic spaces, thereby answering your own question. So please tell us what approach you have tried (first for schemes) and where you got stuck, so one can know what advice would be helpful to you.

Comment by

2013-06-13T10:23:21Z

Do you know how to prove that this definition "works" for schemes, using the Zariski topology? (Even on an affine scheme the sheaf-Hom among quasi-coherent sheaves need not be quasi-coherent, let alone the sheaf associated to the "expected" Hom-module, unless one imposes a finite presentation hypothesis; finite type is insufficient.) If "no" then figure that out before contemplating the case of algebraic spaces. And if "yes" then please explain why you can't apply the same argument to answer your question for algebraic spaces.

Comment by

2013-06-13T03:20:15Z

@Alberto: Referring to that part of SGA3 for this is puzzling, since Theorem 3.2 in VI$_{\rm{A}}$ is easier to parse. But the usual textbooks construct $G \rightarrow Q$ with kernel $N$ in down-to-earth ways, so "all" you need to identify $Q$ with a sheaf quotient $G/N$ is that a surjective homomorphism $f$ between smooth affine groups is flat (and even smooth in char. 0). As in my previous comment: use translations to get flatness from generic flatness, and then fppf descent to get smoothness of $f$ from that of $\ker f$. And as Angelo notes, smooth surjections have etale-local sections.QED

Comment by

2013-06-12T17:09:50Z

The "unconventional" aspect includes not getting the piece of paper from high school as well, but the educational aspect is indeed entirely conventional (and even exceptional): taking classes as a registered student for a few years at each of Bronx High School of Science and MIT.

Comment by

2013-06-12T13:25:18Z

The answer is affirmative for any excellent normal noetherian local domain (no need to bring in $\mathbf{C}$). Excellence is inherited by henselization, so we may also assume $R$ is henselian. Let $K={\rm{Frac}}(R)$, so $\widehat{K}:={\rm{Frac}}(\widehat{R})$ is separable over $K$. Hence, if $K'$ is nontrivial finite over $K$ inside $\widehat{K}$ and $R'$ is the $R$-finite integral closure of $R$ in $K'$ then $R'$ is local and $K'/K$ is separable, so $\widehat{K}\otimes_K K'$ contains the non-domain $K'\otimes_K K'$. But it is a localization of the ring $\widehat{R}'$ that is a domain.QED

Comment by

2013-06-12T05:48:46Z

The phrase "According to SGA3" could use some more precision.

Comment by

2013-06-12T05:43:53Z

@ag: Please formulate a more precise question. Do you mean to ask about the role of DM stacks in the consideration of moduli for certain kinds of coherent sheaves (vector bundles?) on certain kinds of schemes (curves, surfaces...)? If you put no effort into writing a clear and precise question then you can't expect to get a useful answer.

Comment by

2013-02-28T12:33:10Z

@anon: Ah, so I was misreading; thanks for the correction.

Comment by

2013-02-28T12:02:22Z

@Dmitry: My memory was a bit faulty, sorry. Looking back at that paper (which is indeed the one I had in mind), the 2nd paragraph of section 2 indicates that one can establish the analogues of what Artin proved in his earlier paper(s), but not something stronger.

Comment by

2013-02-28T01:13:04Z

Even though it appears to be hopeless to say anything in general about commutative pseudo-reductive groups, Totaro's paper contains quite a bit of interesting information about commutative pseudo-reductive groups (in that it makes several different kinds of constructions). So it definitely worth a close look (even if it may have the main effect of just convincing you that the commutative case is even more unwieldy than you had hoped).

Comment by

2013-02-27T23:42:34Z

By Lemma 9.3 of Totaro's paper "Pseudo-abelian varieties" (see the arxiv version), for any field $k=k_s$ and 1-dimensional $k$-wound smooth connected unipotent $k$-group $U$ there is a commutative pseudo-reductive extension of $U$ by $\mathbf{G}_m$. So the minimal possible dimension of 2 is realized over any separably closed field $k$ that isn't algebraically closed (as the "Rosenlicht construction" yields such $U$ over any such $k$). Given $k \ne k_s$, Totaro's construction over $k_s$ descends to some finite separable extension of $k$. Making "$k$-anisotropic" examples seems harder.

Comment by

2013-02-27T22:36:50Z

For (1), it still seems reasonable only to consider finitely many analytic equations (since the local ring is noetherian, after all), and at least in the non-archimedean case I believe there is a paper of Siegfried Bosch on this generalization of Artin's result (i.e., considering analytic equations over $R$, not just polynomial equations over $R$). I don't remember the exact title, but if you search for papers of Bosch with "Artin" or "approximation" in the title then you should find it.

Comment by

2013-02-27T06:10:14Z

The topological space of the formal scheme $\widehat{X}$ consists of the single point $P$ (so $\widehat{X} - P$ is empty). The functor of global sections on $\widehat{X}$ with supports at $P$ is the same as the one without mention of the supports, so it is an exact functor (even coincides with the identity functor, so to speak). Hence, $H^j_P(\widehat{X},\cdot)$ vanishes for $j > 0$ and is the identity functor for $j = 0$. Is there some motivation to guide the way to an interesting reformulation, or is this idle curiosity?