User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:52:01Z http://mathoverflow.net/feeds/user/30379 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120016/galois-action-on-special-fiber-of-a-stable-model/120031#120031 Answer by pranavk for Galois action on special fiber of a stable model pranavk 2013-01-27T15:35:09Z 2013-01-28T05:29:01Z <p>The answer is affirmative when $X_K$ is smooth and geometrically connected with genus $\ge 2$ (which I am guessing are implicit hypotheses), without restriction on the residual characteristic nor on the generic characteristic. The key ingredients are theorems of Grothendieck concerning semistable reduction and the link between stable reduction of curves and semistable reduction of Jacobians (proved by Deligne and Mumford).</p> <p>By increasing $K$ if necessary, the "minimality" hypothesis on $L$ reduces the problem to showing that if the ${\rm{Gal}}(L/K)$-action on the special fiber $X_k$ of $X_{R_L}$ is trivial then $X$ has stable reduction over $R$. By Deligne-Mumford, it is equivalent to deduce that the Jacobian $J$ of $X_K$ has semistable reduction. By Grothendieck's inertial criterion for semistable reduction, this in turn is equivalent to deducing that the action of $G_K = {\rm{Gal}}(K_s/K)$ on $V_{\ell}(J)$ is unipotent for some (equivalently, any) prime $\ell \ne {\rm{char}}(k)$. </p> <p>By a theorem of Raynaud, $P := {\rm{Pic}}^0_{X_{R_L}/R_L}$ is a scheme, and more specifically is semi-abelian with generic fiber $J_L$. The subgroup <code>$G_L = {\rm{Gal}}(K_s/L)$</code> of $G_K$ acts unipotently on $V_{\ell}(J)$, and Grothendieck's orthogonality theorem (applied over $R_L$) gives much more: $V_{\ell}(J)^{G_L} = V_{\ell}(P_k)$ with $V_{\ell}(J)/V_{\ell}(J)^{G_L}$ canonically dual to $V_{\ell}(T)$ where $T$ is the maximal torus of $P_k$ (here using the auto-duality of $J$ over $K$). This "canonicity" includes $G_K$-equivariance, so to prove the unipotence of the $G_K$-action on $V_{\ell}(J)$ it suffices to prove the triviality of the ${\rm{Gal}}(L/K)$-action on $V_{\ell}(P_k)$. But this latter Galois action is just the composition of functoriality of $V_{\ell}$ applied to the ${\rm{Gal}}(L/K)$-action on $P_k = {\rm{Pic}}^0_{X_k/k}$ arising from the ${\rm{Gal}}(L/K)$-action on $X_k$. (Various implicit compatibility verifications are left as an exercise.) This latter action is assumed to be trivial, so we are done. </p> http://mathoverflow.net/questions/118141/does-every-polynomial-diophantine-equation-have-solutions-modulo-p/118150#118150 Answer by pranavk for Does every polynomial diophantine equation have solutions modulo p? pranavk 2013-01-05T20:11:56Z 2013-01-05T21:02:57Z <p>I assume that you're asking if there exists such a solution in $\mathbf{F}_p$ for infinitely many $p$, or for all sufficiently large $p$ under some condition (since for the single equation $x^2 + 1 = 0$ you can't hope for all large $p$ in general). </p> <p>This can be solved using the Chebotarev Density Theorem and some estimates of Deligne from Weil II (as a modern replacement for the uniform Lang-Weil estimates for "equi-characteristic families", which I think is what Voloch and Arapura have in mind). The details below may look long, but ultimately the basic ideas are simple and the arguments are entirely standard in the subject.</p> <p>Let's rephrase the question more intrinsically. Let $A$ be a finitely generated $\mathbf{Z}$-algebra admitting a ring map $f:A \rightarrow \mathbf{C}$. That encodes your initial hypotheses (in a more general form), upon writing $A$ as a quotient of a polynomial ring over $\mathbf{Z}$ modulo some (finitely generated) ideal. By the Nullstellensatz over $\mathbf{C}$, the existence of $f$ just says that $A_{\mathbf{C}} := A \otimes_{\mathbf{Z}} \mathbf{C}$ is nonzero, and that in turn is equivalent to $A_{\mathbf{Q}}$ being nonzero. Your question seems to be whether for infinitely many $p$ or perhaps all sufficiently large $p$ there is a ring homomorphism $A \rightarrow \mathbf{F}_p$. </p> <p>In terms of the finite type affine $\mathbf{Z}$-scheme $X = {\rm{Spec}}(A)$, you're asking if non-emptiness of the generic fiber $X_{\eta}$ over $\mathbf{Q}$ forces the fiber $X_p := X \bmod p$ to have an <code>$\mathbf{F}_p$</code>-rational point for infinitely many $p$ or (under some hypothesis on $X_{\eta}$) all sufficiently large $p$. More generally, for <em>any</em> finite type $\mathbf{Z}$-scheme $X$ with non-empty generic fiber $X_{\eta}$ over $\mathbf{Q}$, $X_p$ has an <code>$\mathbf{F}_p$</code>-rational point for infinitely many $p$, and even for all sufficiently large $p$ provided that <code>$X_{\eta}$</code> is geometrically irreducible over $\mathbf{Q}$ (equivalently, the $\mathbf{C}$-fiber $X_{\mathbf{C}}$ is irreducible, or more concretely $X_{\mathbf{Q}}$ is irreducible and $\mathbf{Q}$ is algebraically closed in the function field of $(X_{\mathbf{Q}})_{\rm{red}}$). </p> <p>The existence of an <code>$\mathbf{F}_p$</code>-point for infinitely many $p$ is a straightforward consequence of the Chebotarev Density Theorem, as follows. Choose a closed point $x \in X_{\eta}$, so its residue field $K$ is a number field. By general "spreading out" principles (ultimately just denominator-chasing in the affine case), the map $x:{\rm{Spec}}(K) \rightarrow X_{\eta}$ between respective $\mathbf{Q}$-fibers of the finite type schemes ${\rm{Spec}}(O_K)$ and $X$ over ${\rm{Spec}}(\mathbf{Z})$ "spreads out" to a map ${\rm{Spec}}(O_K[1/N]) \rightarrow X_{\mathbf{Z}[1/N]}$ over $\mathbf{Z}[1/N]$ for a suitable dense open ${\rm{Spec}}(\mathbf{Z}[1/N]) \subset {\rm{Spec}}(\mathbf{Z})$. Thus, to make an $\mathbf{F}_p$-point of $X_p$ for $p \nmid N$ it suffices to do the same for ${\rm{Spec}}(O_K[1/N])$, which is to say that we seek a prime ideal $\mathfrak{p}$ of $O_K$ over $p \nmid N$ at which the residual degree over $\mathbf{F}_p$ is 1. For example, it suffices to find infinitely many rational primes $p > N$ that are totally split in $K$, and the Chebotarev Density Theorem applied to the Galois closure of $K$ over $\mathbf{Q}$ provides a healthy supply of such $p$.</p> <p>The deeper case is to show that $X_p(\mathbf{F}_p)$ is non-empty for <em>all</em> large $p$ when $X_{\eta}$ is geometrically irreducible. Note that such a hypothesis rules out cases like $X = {\rm{Spec}}(O_K)$ for a number field $K \ne \mathbf{Q}$, for which we know that the desired assertion is false. As Arapura and Voloch observe, the key input is the so-called Lang-Weil estimate. To make this precise (since Lang and Weil did not use the framework of "families" across varying characteristics, as their work pre-dated the advent of schemes), we need some kind of "uniformity" in our understanding of <code>$\#X(\mathbf{F}_p)$</code> as $p$ varies; this is sort of "orthogonal" to the more traditional question Weil would have asked concerning the behavior of <code>$\#X_p(\mathbf{F}_{p^n})$</code> as $n$ grows with $p$ <em>fixed</em>. </p> <p>No doubt Lang and Weil would have been able to adapt their method for "equi-characteristic families" so it applies to the "family" $f:X \rightarrow {\rm{Spec}}(\mathbf{Z})$ across varying residue characteristics, but nowadays it seems that the most efficient and elegant way to proceed is to apply Deligne's Weil II estimates, as follows. Let $d$ be the dimension of $X_{\eta}$. By general "spreading out" principles, the fibers of $f$ over some dense open ${\rm{Spec}}(\mathbf{Z}[1/M])$ are all of dimension $d$. Fix a prime $\ell$ and work over $\mathbf{Z}[1/\ell M]$ now. By the general principles in etale cohomology, the higher direct images with proper supports <code>$R^if_{!}(\mathbf{Q}_{\ell})$</code> are constructible $\mathbf{Q}_{\ell}$-sheaves on the etale site of ${\rm{Spec}}(\mathbf{Z}[1/\ell M])$ which vanish for $i > 2d$, and by choose a sufficiently divisible nonzero $N$ divisible by $\ell M$ we can arrange that each <code>$R^i f_{!}(\mathbf{Q}_{\ell})$</code> has lisse restriction over $S = {\rm{Spec}}(\mathbf{Z}[1/N])$, say with rank $r_i$. In particular, $r_{2d} = 1$ due to the <em>geometric irreducibility</em> of the generic fiber, and more specifically by excision and Poincare duality we know from the geometric irreducibility (and the lisse condition over $S$) that <code>$R^{2d} f_{!}(\mathbf{Q}_{\ell}) = \mathbf{Q}_{\ell}(-d)$</code> as etale sheaves over $S$. Moreover, for $p \nmid N$, the Grothendieck-Lefschetz trace formula gives <code>$$\#X(\mathbf{F}_p) = \sum_{i=0}^{2d} (-1)^i {\rm{Tr}}(\phi_p| R^i f_{!}(\mathbf{Q}_{\ell})_{\overline{\mathbf{F}}_p})$$</code> where $\phi_p$ denotes "geometric Frobenius" on the mod-$p$ geometric stalk of $R^i f_{!}(\mathbf{Q}_{\ell})$.</p> <p>In Galois-theoretic terms, this is saying <code>$$\#X(\mathbf{F}_p) = \sum_{i=0}^{2d} (-1)^i {\rm{Tr}}({\rm{Frob}}_p^{-1}|{\rm{H}}^i_c(X_{\overline{\mathbf{Q}}}, \mathbf{Q}_{\ell}))$$</code> using the natural action of ${\rm{Gal}}(\overline{\mathbf{Q}}/\mathbf{Q})$ on these cohomologies of the geometric generic fiber (which are all unramified away from $N$) with ${\rm{Frob}}_p$ an "arithmetic Frobenius" at $p$. The term for $i = 2d$ is the action of <code>${\rm{Frob}}_p^{-1}$</code> on <code>$\mathbf{Q}_{\ell}(-d)$</code>, which is $p^d$. </p> <p>Now for the big input: Deligne's Weil II ensures that the $\phi_p$-action on the $i$th compactly supported $\ell$-adic cohomology of the geometric mod-$p$ fiber has <em>all</em> eigenvalues (inside <code>$\overline{\mathbf{Q}}_{\ell}$</code>) inside the subfield <code>$\overline{\mathbf{Q}} \subset \overline{\mathbf{Q}}_{\ell}$</code> and <em>every</em> archimedean absolute value of each of these is bounded above by $p^{i/2}$. Hence, by computing such traces as a sum of eigenvalues inside this subfield <code>$\overline{\mathbf{Q}}$</code> and fixing an embedding of <code>$\overline{\mathbf{Q}}$</code> into $\mathbf{C}$ (to make sense of absolute values of all of these eigenvalues at the same time) we get the estimate <code>$$\#X(\mathbf{F}_p) \ge p^d -\sum_{i=0}^{2d-1} r_i p^{i/2}$$</code> where $r_i$ is the dimension of <code>${\rm{H}}^i_c(X_{\overline{\mathbf{Q}}},\mathbf{Q}_{\ell})$</code> (i.e., the constant rank of the lisse <code>$R^i f_{!}(\mathbf{Q}_{\ell})$</code> over $S$).</p> <p>This final estimate illuminates the crux of the "uniformity" as we vary $p$: the controlling factors are the ranks $r_i$ of the fibral cohomologies, the "glue" for varying $p$ being that the $i$th compactly supported $\ell$-adic cohomologies of the geometric mod-$p$ fibers that are relevant to the Lefschetz trace formula have a <em>common</em> dimension $r_i$ as we vary across the primes $p \nmid N$ (precisely because these cohomologies are the stalks of a single lisse $\ell$-adic sheaf on $S$). It also illuminates the significance of the geometric irreducibility hypothesis via controlling the form of the "main term" $p^d$. (If we hadn't imposed geometric irreducibility of the generic fiber then the contribution from top-degree cohomology would have been an Artin representation controlled by the algebraic closure of $\mathbf{Q}$ in the residue fields at the generic points of $X_{\eta}$, and in this way the Chebotarev Density Theorem would intervene to account for counterexamples in the absence of the geometric irreducibility hypothesis.)</p> <p>Anyway, the right side is $p^d$ minus a polynomial in $\sqrt{p}$ of degree at most $2d-1$, so it is $p^d - O(p^{d-1/2})$ as $p$ grows ("Lang-Weil"!), and hence is nonzero for $p$ sufficiently large (depending just on the $d$ and the ranks $r_i$ of the cohomologies of the geometric generic fiber).</p> http://mathoverflow.net/questions/117959/laurent-polynomials/117967#117967 Answer by pranavk for Laurent Polynomials pranavk 2013-01-03T14:53:58Z 2013-01-03T14:53:58Z <p>Thinking geometrically in terms of the map ${\rm{Spec}}(R[x,1/x]) \rightarrow {\rm{Spec}}(R)$ and noting that being a unit amounts to being nonzero in the residue field at every prime, an element $f = \sum a_i x^i \in R[x,1/x]$ is a unit if and only if it has unit restriction to every fiber, which is to say that for every prime ideal $P$ of $R$ (with residue field $k(P)$) the image $f(P) := \sum a_i(P) x^i$ in $k(P)[x,1/x]$ is a unit. But since $k(P)$ is a field, this latter condition is exactly that $f(P)$ is a $k(P)^{\times}$-multiple of a power of $x$. </p> <p>That is, there is exactly one $i$ (depending perhaps on $P$) such that $a_i(P) \ne 0$, which can be equivalently expressed as the condition that $a_i(P)a_j(P) = 0$ in $k(P)$ for all $i \ne j$ and $\sum a_i(P) \ne 0$ in $k(P)$. Varying over all $P$, this necessary and sufficient condition says exactly that (1) $a_i a_j$ is nilpotent in $R$ when $i \ne j$ and (2) $\sum a_i \in R^{\times}$. </p> <p>In the presence of (1), squaring the sum in (2) (which has no effect on whether or not it is a unit) and noting that adding a nilpotent element has no effect on being a unit shows that (2) can be replaced with (2') $\sum a_i^2 \in R^{\times}$ (thereby recovering the formulation in shatich's answer).</p> http://mathoverflow.net/questions/117942/the-fibres-of-smooth-projective-families-over-all-geometric-points-have-isomorphi/117954#117954 Answer by pranavk for The fibres of smooth projective families over all geometric points have isomorphic cohomology; are these isomorphisms 'functorial'? pranavk 2013-01-03T13:33:52Z 2013-01-03T13:33:52Z <p>The affirmative answer involves just going back to how the comparison isomorphism between cohomologies of different geometric fibers are defined by means of specialization maps for higher direct image sheaves on the base space, making sure we are using the "same" strict henselizations to make the specialization maps on the two sides (for $p$ and $p'$), and noting that such specialization maps for a fixed strict henselization are functorial in the sheaf on the base space.</p> <p>To be precise, let $p:P \rightarrow S$ and $p':P' \rightarrow S$ be scheme morphisms and $f:P' \rightarrow P$ an $S$-morphism. Choose an etale abelian sheaf $F$ on $P$. For its pullback $F' = f^{\ast}F$ on $P'$ there are naturally induced "$f$-pullback" maps <code>$\theta^i_F: R^i p_{\ast}(F) \rightarrow R^i p'_{\ast}(F')$</code> (in the style of "base change" morphisms, applied to the typically non-cartesian but commutative square with bottom side the identity map of $S$, top side $f$, and left and right sides $p'$ and $p$ respectively). Having set up this notation, put it to the side for a moment. </p> <p>For any etale abelian sheaf $G$ on $S$ and points $s, \eta \in S$ with $s$ a specialization of $\eta$, upon choosing a geometric point $\overline{s}$ over $s$ and a geometric point $\overline{\eta}$ over $\eta$ on Spec(<code>$O_{S,\overline{s}}^{\rm{sh}}$)</code> we have a natural map <code>${\rm{sp}}_{\overline{s},\overline{\eta},G}:G_{\overline{s}} \rightarrow G_{\overline{\eta}}$</code>. In the above setting, if $p$ is a smooth proper surjection and $F$ is lcc with torsion orders invertible on $P$ then $R^i p_{\ast}F$ is an lcc sheaf (smooth and proper base change theorems) and this specialization map for $G = R^i p_{\ast}(F)$ on $S$ is an isomorphism. That isomorphism is <em>by definition</em> the comparison isomorphism between cohomologies of $F$ on the geometric fibers at $\overline{s}$ and $\overline{\eta}$ (isomorphism depending on the "choice" of strict henselization). </p> <p>So ultimately the affirmative answer to your question (implicitly assuming we are using a single such choice of strict henselization underlying the fibral comparison isomorphisms for both $p$ and $p'$!) amounts to the observations that (i) the formation of the specialization map ${\rm{sp}}_{\overline{s},\overline{\eta},G}$ for general $G$ is functorial in $G$ with respect to any morphism between abelian etale sheaves on $S$ (such as the maps $\theta^i_F$), (ii) in the special case that $S$ is a geometric point, $\theta^i_F$ is the natural pullback map in degree-$i$ cohomologies associated to $F$ and $F'$.</p> <p>As this argument makes plain, we have to use the same strict henselization on both sides or else it all breaks down. In terms of ACL's comment above in the topological analogue over the complex numbers (using constant coefficients), this corresponds to identifying the cohomologies on fibers over the various points of the contractible $U$ via their pullback identifications with the cohomologies on the total spaces $P_U$ and $P'_U$ over $U$ (i.e., the choice of strict henselization in the algebraic theory plays the role of $U$ in the topological case). That is, if we use a different contractible open $V$ in $S$ containing the same pair of points $s_1, s_2 \in S$ then we generally get a different isomorphism between cohomologies on the $s_i$-fibers (for both $P$ and $P'$), and we certainly wouldn't claim any compatibility for such comparison isomorphisms if we use $U$ for the definition of fibral comparison for $P$ and use $V$ for the definition of fibral comparison for $P'$.</p> http://mathoverflow.net/questions/127706/a-question-about-r-points-of-an-complex-reductive-group Comment by 2013-04-16T14:16:14Z 2013-04-16T14:16:14Z Since $\mathbf{C}$ is algebraically closed, $Z_e$ is a <i>split</i> group of multiplicative type, so $Z_e$ is a <i>split</i> torus. Thus, for any $\mathbf{C}$-algebra $R$, the obstruction to $G(R((t)))/Z_e(R((t))) \rightarrow (G/Z_e)(R((t)))$ being bijective is a class in the etale cohomology set $H^1(R((t)),Z_e)$, which is a power of ${\rm{Pic}}(R((t)))$. For $R$ a field or even artinian local ring, this Pic is trivial and so bijectivity holds. Thus, you have bijectivity on infinitesimal points over $\mathbf{C}$, which probably implies an isomorphism as smooth ind-schemes, yes? http://mathoverflow.net/questions/127643/algebraic-machinery-for-algebraic-geometry/127662#127662 Comment by 2013-04-16T03:34:18Z 2013-04-16T03:34:18Z Fulton's &quot;Algebraic Curves&quot; does an excellent job of introducing commutative algebra in a geometric context, and its selection of exercises does an amazing job at conveying the rich interaction of geometry and algebra beyond what is done in the text. I recommend trying Fulton's book alongside Reid's, and then you can decide for yourself which you prefer. http://mathoverflow.net/questions/127581/describe-the-set-of-rational-points-on-the-curve Comment by 2013-04-15T04:55:47Z 2013-04-15T04:55:47Z Why is this labeled &quot;cryptography&quot;? Is it homework for a cryptography class (as warm-up to discussing elliptic curves)? http://mathoverflow.net/questions/127524/isomorphic-maximal-commutative-semi-simple-sub-algebras-of-m-nc Comment by 2013-04-14T14:47:52Z 2013-04-14T14:47:52Z @ofir: Any such $A$ is a product of copies of $\mathbf{C}$, so a faithful representation $A \hookrightarrow {\rm{M}}_n(\mathbf{C})$ on $\mathbf{C}^n$ from a commutative semisimple $\mathbf{C}$-algebra $A$ sends the primitive idempotents to {\em distinct} pairwise orthogonal commuting nonzero idempotent linear operators on $\mathbf{C}^n$ whose sum is the identity operator. This is exactly a decomposition of $\mathbf{C}^n$ as a direct sum of nonzero subspaces. The &quot;maximal&quot; way to do this is with an ordered $n$-tuple of independent lines, and in a suitable basis all $n$-tuples look the same... http://mathoverflow.net/questions/126103/parabolic-type-subgroups-of-glv/126212#126212 Comment by 2013-04-14T05:01:02Z 2013-04-14T05:01:02Z Dear Hung Nguyen: Thanks for the explanation, though I'm a bit puzzled as to why your friend thinks that the class of subgroups as in your question is a &quot;natural&quot; one. Of course, the notion of a &quot;natural&quot; subgroup of ${\rm{GL}}(V)$ is a matter of taste, but parabolic subgroups and their unipotent radicals and Levi factors seem to be rather more &quot;natural&quot; than the things in your question. But maybe your friend has some good reason to regard the subgroups in your question as &quot;natural&quot;? http://mathoverflow.net/questions/127290/n-canonical-embedding Comment by 2013-04-12T04:29:15Z 2013-04-12T04:29:15Z @Artin009: Just because those MIT notes say it that way doesn't mean it is the right way to think about it (and those notes don't explain the exercise or the precise meaning of the terminology; curious that the notes mask all evidence of who wrote them, as far as I can tell). I recommend looking at my above suggested reference in Deligne-Mumford, where you'll see them carry out the actual cohomological computations using duality on the semistable curve, and I hope that should clarify the situation for you. http://mathoverflow.net/questions/127290/n-canonical-embedding Comment by 2013-04-12T00:37:28Z 2013-04-12T00:37:28Z @Artin009: In the absence of smoothness, is &quot;separating points and tangent vectors&quot; the right way to think about things? Anyway, have you read the proof of very ampleness for $n &gt; 2$ in Theorem 1.2 of the Deligne-Mumford paper? (Please also consider to use a name other than &quot;Artin009&quot;.) @Will: Since $\omega$ is not the pushforward of a line bundle on the normalization, why is RR on the normalization relevant? Isn't duality on the nodal curve a more appropriate argument? For example, that is what Deligne and Mumford do. http://mathoverflow.net/questions/126881/elliptic-curve-over-a-scheme-is-a-group-scheme/126910#126910 Comment by 2013-04-10T13:08:52Z 2013-04-10T13:08:52Z @anon: Did you mean to write that there is a down-to-earth proof that there is <i>at most one</i> group law with the given zero when the base is reduced (in which case the proof is immediate from consideration of the generic fibers over the base, due to flatness and separatedness considerations over the base)? To say &quot;there is a unique&quot; (which I read as including an existence assertion) over any base or to say &quot;unique&quot; over a non-reduced base both seem to lie beyond the reach of &quot;more down-to-earth&quot; arguments. http://mathoverflow.net/questions/126881/elliptic-curve-over-a-scheme-is-a-group-scheme Comment by 2013-04-10T05:12:25Z 2013-04-10T05:12:25Z Strictly speaking, the argument in GIT has a projectivity hypothesis on the abelian scheme (due to the role of projectivity for the existence of Hilbert schemes), and there are non-projective abelian schemes over non-normal noetherian domains. If one uses algebraic spaces, which didn't exist at the time that GIT was written, then Hilbert functors are representable without projectivity hypotheses and so the proof of Grothendieck's theorem works without projectivity hypotheses. http://mathoverflow.net/questions/126881/elliptic-curve-over-a-scheme-is-a-group-scheme/126910#126910 Comment by 2013-04-09T01:10:02Z 2013-04-09T01:10:02Z There is a general theorem of Grothendieck to the effect that a smooth proper morphism equipped with a section and geometrically connected fibers is an abelian scheme if it is so on a single geometric fiber, but the proof isn't in any sense down-to-earth. I am amazed to hear that there could be a &quot;down-to-earth&quot; proof of the <i>existence</i> of the group scheme structure even just when the base is an artin local ring (for which &quot;generic fiber&quot; is the special fiber). Anon, what method do you have in mind which isn't the deformation-theoretic proof of Grothendieck's version? http://mathoverflow.net/questions/126669/base-change-of-affine-group-schemes-with-respect-to-frobenius-map Comment by 2013-04-06T16:06:58Z 2013-04-06T16:06:58Z I recommend ignoring that exercise, since such an assertion is rather misleading. (The point is that the Frobenius is an automorphism of $k$, from which the isomorphism claim in his weak sense is obvious.) http://mathoverflow.net/questions/126669/base-change-of-affine-group-schemes-with-respect-to-frobenius-map Comment by 2013-04-06T09:32:05Z 2013-04-06T09:32:05Z [I assume you intend for the isomorphism to be of group schemes over $k$.] http://mathoverflow.net/questions/126669/base-change-of-affine-group-schemes-with-respect-to-frobenius-map Comment by 2013-04-06T09:30:04Z 2013-04-06T09:30:04Z No, since they're not generally isomorphic at all. http://mathoverflow.net/questions/126454/etale-fundamental-group-of-punctured-formal-neighborhood Comment by 2013-04-05T04:34:59Z 2013-04-05T04:34:59Z Whoops, the argument I had in mind only shows (via Abyhankar's Lemma) that the kernel has no nontrivial prime-to-$p$ quotients, which is of course much weaker than being pro-$p$, so I have nothing useful to say about your question. More specifically, assuming regularity and excellence but not &quot;henselian local&quot;, the maximal prime-to-$p$ quotient of the kernel is the quotient $\prod_{\ell} {\mathbf{Z}}_{\ell}(1)$ that you know. Please talk to experts near you in Princeton to learn about Abhykanar's Lemma beyond the 1-dimensional case (as higher dimensions is where the real strength of it lies). http://mathoverflow.net/questions/126560/finite-type-artin-stack-over-mathbb-c Comment by 2013-04-04T21:56:47Z 2013-04-04T21:56:47Z Any map $f:X \rightarrow Y$ between quasi-separated algebraic spaces locally of finite type over an affine scheme $B$ with $X$ of finite type over $B$ is itself of finite type. Indeed, $Y$ is covered by q-c opens $U_i$, preimages of which are open in $X$, so finitely many preimages cover $X$. Hence, $f$ lands inside a quasi-compact open $U \subset Y$, and $U \rightarrow B$ is finite type (being q-c and lft), so $X \rightarrow U$ is finite type. Thus, we want $U \hookrightarrow Y$ to be finite type. For $V \rightarrow Y$ with $V$ affine, $U \times_Y V$ is q-c since $Y$ is quasi-separated.