User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:19:58Z http://mathoverflow.net/feeds/user/30180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127316/local-norm-mapping-for-abelian-varieties/128066#128066 Answer by ayanta for Local Norm Mapping for Abelian Varieties ayanta 2013-04-19T11:00:39Z 2013-04-19T11:16:33Z <p>One doesn't need Tate local duality to analyze the good reduction case, and in general the answer is affirmative.</p> <p>First, let's review the general nonsense for norm maps with commutative group functors (to sidestep representability issues). For any finite etale map of schemes $f:S' \rightarrow S$ and any commutative functor $F$ on the category of $S$-schemes there is a natural "norm" map of group functors $N: f_{\ast}(F_{S'}) \rightarrow F$, where $F_{S'}$ denotes the restriction of $F$ to the category of $S$-schemes (represented by base change when $F$ is representable). Indeed, if $T$ is an $S$-scheme then $f_{\ast}(F_{S'})(T) = F(T \times_S S')$ is identified with the group of global sections of the $f_T$-pullback of the restriction of $F_T$ to the small etale site over $T$, so to define $N$ on $T$-points we simply use the natural trace map $(f_T)_{\ast} \circ f_T^{\ast} \rightarrow {\rm{id}}$ on the category of abelian sheaves on the small etale site of $T$. (This trace map is defined more generally using just that $f$ is quasi-finite flat and separated and finitely presented, but later we will need that $f$ is finite etale and the definition of the trace map is much simpler in the case of finite etale $f$ anyway.)</p> <p>For example, taking $F$ to be the functor of points of the Neron model $\mathcal{A}$ of $A$ over $O_K$ and $f$ to correspond to $O_K \rightarrow O_L$, the above norm map on $O_K$-points is precisely the natural norm map $A(L) \rightarrow A(K)$ due to the universal mapping property of Neron models and the compatibility of Neron models with unramified base change. The advantage of this geometric interpretation is that the norm map of smooth finite type $O_K$-groups $$N_{A,L/K}:{\rm{R}}_{O_L/O_K}(\mathcal{A}_{O_L}) \rightarrow \mathcal{A}$$ is an $O_K$-form of the addition map $\mathcal{A}^{[L:K]} \rightarrow \mathcal{A}$. More generally, for any $f$ and $F$ as in the preceding paragraph with $f$ of constant degree $d$, the "norm" map $f_{\ast}(F_{S'}) \rightarrow F$ becomes the addition map $F^{\oplus d} \rightarrow F$ after passing to an etale cover of $S$ that splits $S'$ into a disjoint union of $d$ copies of $S$ (a fiber power of $S'$ over $S$ is such a cover). Indeed, the formation of the norm commutes with base change on $S$, and in the case of a split covering it is clear from the construction that the norm is identified with the componentwise addition map.</p> <p>Now we see that by taking $[L:K]$ divisible by the order of the finite etale component group of the special fiber <code>$\mathcal{A}_0$</code>, $N_{A,L/K}$ lands inside the relative identity component $G := \mathcal{A}^0$ (open complement of the closed union of the finitely many non-identity components in the special fiber) since such an assertion is local for the etale topology on $O_K$ and hence it can be checked by computing with the $O_K$-form of the map given by componentwise addition $\mathcal{A}^{\oplus [L:K]} \rightarrow \mathcal{A}$. Hence, we may replace $\mathcal{A}$ with $G$ and reduce to checking that the norm map $$N_{G,O_L/O_K}:{\rm{R}}_{O_L/O_K}(G_{O_L}) \rightarrow G$$ is surjective on $O_K$-points for <em>any</em> smooth $O_K$-group $G$ of finite type with <em>connected</em> fibers. </p> <p>The kernel $G' = \ker N_{G,O_L/O_K}$ is a form of $G^{[L:K]-1}$ for the etale topology on $O_K$, so it is also smooth of finite type with connected fibers. Hence, for any $O_K$-point $g$ of $G$, the $N_{G,O_L/O_K}$-pullback of $g:{\rm{Spec}}(O_K) \rightarrow G$ is a smooth $O_K$-scheme $X$ of finite type that is a $G'$-torsor for the etale topology, and we just need to show that $X(O_K)$ is non-empty. Since $X$ is <em>smooth</em> and $O_K$ is henselian (e.g., complete), such non-emptiness holds provided that the special fiber $X_0$ has a rational point over the residue field $k$ of $O_K$. </p> <p>We have come down to the task of showing that any torsor for a smooth <em>connected</em> finite type group scheme over $k$ must have a $k$-point. So far there is no number theory in any of this, just general arithmetic geometry over discrete valuation rings. But finally we bring in finiteness of $k$ to apply Lang's theorem which asserts exactly that such torsors always have a rational point when the ground field $k$ is finite.</p> http://mathoverflow.net/questions/127769/applications-of-govorov-lazard-theorem/127786#127786 Answer by ayanta for Applications of Govorov-Lazard Theorem? ayanta 2013-04-17T04:05:16Z 2013-04-17T07:31:48Z <p>An interesting example arises in the consideration of the $n$th symmetric power of a flat scheme morphism (such as for "directly" constructing the Hilbert scheme of $n$ points on a curve and relating it to the Picard scheme, building on one of Weil's original approaches to constructing the Jacobian of a smooth curve). More specifically, if $A$ is a flat $R$-algebra and $S_n$ denotes the $n$th symmetric group then the subalgebra $(A^{\otimes n})^{S_n}$ is $R$-flat and its formation commutes with any base change on $R$. (Thus, more globally, if $X$ is a flat projective scheme over a ring $R$ then the projective $R$-scheme ${\rm{Sym}}^n(X) := (X^n)/S_n$ is $R$-flat and its formation commutes with any base change on $R$.) A key point is that we do not impose any "lazy" hypothesis concerning the size of $S_n$ being a unit in $R$.</p> <p>To see such properties for the subalgebra of symmetric tensors, one forgets about the algebra structure and aims to show more generally that if $M$ is any flat $R$-module then $(M^{\otimes n})^{S_n}$ is $R$-flat and its formation commutes (via the evident map) with any base change on $R$. This module problem is compatible with direct limits in $M$, so by the Lazard theorem we are reduced to the case when $M$ is finite free, which in turn is clear by inspection! See pp. 252-254 in "Neron Models" for a discussion (with references) for the application to Hilbert schemes of curves. In a similar spirit, if $M$ is a flat $A$-module then its symmetric and exterior powers are $A$-flat (as we see by using Lazard's theorem to pass to the direct limit on the elementary case of finite free modules).</p> <p>Another place where the symmetric power of algebras arises is in the construction of the relative Verscheibung morphisms <code>${\rm{Ver}}_{G/S}: G^{(p)} \rightarrow G$</code> for <em>any</em> flat commutative group scheme $G \rightarrow S$ over an <code>${\mathbf{F}}_p$</code>-scheme $S$, compatible with any base change on $S$. For this, one uses the $p$th symmetric power of suitable affine opens in the underlying $S$-scheme. See 4.2-4.3 in Exp. VII$_{\rm{A}}$ in SGA3 for further details (Lazard's theorem arises at the end of 4.2).</p> <p>For yet another application, if $A \rightarrow B$ is a flat map of rings and $I$ is an ideal of $A$ equipped with a divided power structure $\gamma$ then the divided power structure extends (visibly uniquely) to $IB$. The proof involves reducing the existence (or rather, well-definedness) problem to one in which $B$ intervenes only through its underlying $A$-module structure, so one can use Lazard's theorem to reduce the newly formulated problem to the case of finite free modules, where the necessary computations are easy.</p> http://mathoverflow.net/questions/124847/principal-maximal-ideals-in-zx-f/125041#125041 Answer by ayanta for Principal maximal ideals in Z[x]/(F) ayanta 2013-03-20T04:34:41Z 2013-03-20T04:34:41Z <p>Let $A$ be an order in the ring of integers $O_K$ of a number field $K$. We claim that there are infinitely many principal maximal ideals $P$ of $A$. By using localization at rational primes, we have a bijection between the sets of maximal ideals of $A$ and $O_K$ with residue characteristic relatively prime to $N = [O_K:A]$ via $P \mapsto P \cap A$ and $P' \mapsto P'O_K$. </p> <p>In this way, we see that it is harmless to replace $A$ by a sub-order, so we may assume $A = \mathbf{Z} + M O_K$ for an integer $M > 0$. For $x \in 1 + MO_K$, we have $A \cap xO_K = xA$. Indeed, if $y \in O_K$ and $xy = c + Mt$ for $t \in O_K$ then we have to show that $y \in \mathbf{Z} + M O_K$, but this is clear since $xy \equiv c \bmod M O_K$ and $x \equiv 1 \bmod M O_K$. Hence, if $xO_K$ is a prime ideal of $O_K$ then $xA$ is a prime ideal of $A$, so it suffices to construct infinitely many maximal ideals $P$ of $O_K$ admitting a generator congruent to 1 modulo $M O_K$. </p> <p>This latter formulation does not mention the order at all, and is a special case of the more general fact that for any nonzero ideal $J$ of $O_K$ whatsoever, $O_K$ has infinitely many principal maximal ideals $P$ admitting a generator $x \equiv 1 \bmod J$. The existence of infinitely many such $P$ follows from the method of proof of the "abelian" case of the Chebotarev Density Theorem (using generalized ideal class characters in the role of Dirichlet characters in the proof of Dirichlet's theorem on primes in arithmetic progressions). So tacitly here we are using the basic analytic properties of $L$-functions attached to characters of generalized ideal class groups (which can be proved in various ways, such as using $\zeta$-functions and class field theory if one wants to be ahistorical).</p> http://mathoverflow.net/questions/118939/if-m-is-not-a-free-a-module-can-tensoring-with-a-bigger-field-make-it-free/118942#118942 Answer by ayanta for If M is not a free A-module, can tensoring with a bigger field make it free? ayanta 2013-01-15T04:12:24Z 2013-01-15T04:12:24Z <p>Yes, there are such examples even with invertible $M$ and smooth $K$-algebras of dimension 1, with $K$ any field that is not algebraically closed. </p> <p>Choose such a $K$, so we may and do also choose a nontrivial primitive finite extension $F$ of $K$. Consider the projective line over $K$ and remove a closed point $\xi$ such that $F = K(\xi)$. This is an affine open $U = {\rm{Spec}}(A)$ for a Dedekind $A$, and let $M$ be the maximal ideal of $A$ corresponding to an $K$-point. This is invertible as an $A$-module (as for any Dedekind domain) but cannot be principal. Indeed, if $a \in A$ were a generator of $M$ then its divisor on $\mathbf{P}^1_K$ has restriction to $U$ that is a single point of degree 1 yet the condition ${\rm{deg}}_K({\rm{div}}(a)) = 0$ forces the "negative" part of the divisor to have degree $-1$ over $K$, contradicting that this negative part is supported at the closed point $\xi$ with $K$-degree larger than 1.</p> <p>Clearly $A \otimes_K F$ is the coordinate ring of the complement in $\mathbf{P}^1_F$ of ${\rm{Spec}}(F \otimes_K F)$, which contains an $F$-point, so $A \otimes_K F$ is the coordinate ring of a non-empty affine open in the open complement $\mathbf{A}^1_F = {\rm{Spec}}(F[t])$ of an $F$-point in $\mathbf{P}^1_F$. Thus, $A \otimes_K F$ is the localization of the PID $F[t]$ at some nonzero element, so $A \otimes_K F$ is a PID and hence its nonzero ideal $M \otimes_K F$ (corresponding to a single $F$-point) is principal. </p> http://mathoverflow.net/questions/117890/is-rkx-1-ldots-to-kx-1-ldots-a-flat-morphism-what-about-r-to-hatr/117920#117920 Answer by ayanta for Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$? ayanta 2013-01-03T00:59:46Z 2013-01-03T02:12:46Z <p>Let's suppose by $k[[x]]$ we mean the formal power series ring in variables $x_1, x_2, \dots$ which is literally the space of sequences of monomials that individually involve only finitely many variables per monomial (so set-theoretically a direct product of copies of $k$ indexed by such monomials, with a "cofinite" topology). This is of course different from the $(x)$-adic completion of $k[x] := k[x_1,x_2,\dots]$ as noted by Francois, since the latter has as a cofinal system of discrete quotients the rings $k[x]/(x)^m$ of infinite $k$-dimension whereas the former has as a cofinal system of discrete quotients the artinian $k[x_1,\dots,x_r]/(x_1,\dots,x_r)^m$ of finite $k$-dimension.</p> <p>I claim that $k[[x]]$ in the sense I have specified <em>is</em> flat over $k[x]$ (though I also think this is probably completely useless and so I don't claim this is interesting -- maybe just amusing). The key input is buried near the end of volume 1 of SGA3. These methods have no relevance to the $(x)$-adic completion of $k[x]$ (which is an entirely different beast than $k[[x]]$ as defined above).</p> <p>First, some preliminary reductions. We have to show that if $I$ is a finitely generated ideal of $k[x]$ then the injection $I \rightarrow k[x]$ remains injective after tensoring against $k[[x]]$ over $k[x]$. By finite generation, $I$ "comes from" an ideal $I' \subset k[x_1,\dots,x_r]$ for some $r$, and more specifically the natural map $$I' \otimes_{k[x_1,\dots,x_r]} k[x] \rightarrow k[x]$$ is injective since $k[x]$ is certainly flat (even free) over $k[x_1,\dots,x_r]$. So in fact $$I = I' \otimes_{k[x_1,\dots,x_r]} k[x],$$ and hence our problem is to show that the injection $I' \rightarrow k[x_1,\dots,x_r]$ remains injective after tensoring over $k[x_1,\dots,x_r]$ against $k[[x]]$. More specifically, we claim this latter ring map is flat.</p> <p>This final scalar extension process decomposes as a composition of two scalar extensions: $$k[x_1,\dots,x_r] \rightarrow k[[x_1,\dots,x_r]] \rightarrow k[[x]].$$ Since the first step is known to be flat by usual commutative algebra with noetherian ring, we're reduced to proving flatness of the second map. But this is a special case of the Gabriel-Grothendieck theory of pseudo-compact rings in SGA3, in which they systematically develop a good theory of "pseudo-compact modules" and "topological flatness" for "pseudo-compact rings", which are topological rings that are arbitrary inverse limits of artinian rings. This theory includes as a key ingredient a relationship between topological flatness and ordinary flatness when the base ring is <em>noetherian</em> (analogous to completions in the noetherian setting, but logically requiring more work).</p> <p>More specifically, since $A := k[[x_1,\dots,x_r]]$ is noetherian, so any finitely generated $A$-module is finitely presented (and is pseudo-compact for its max-adic topology), for any finitely generated $A$-module $M$ and <em>pseudo-compact</em> $A$-algebra $A'$ the natural map $$M \otimes_A A' \rightarrow M \widehat{\otimes}_A A'$$ is bijective (ultimately because the left side is a cokernel of a map between finite free $A'$-modules and any such map automatically has <em>closed</em> image by a variant of Artin-Rees proved in SGA3). Thus, the preservation of injectivity of the left as a functor in finitely generated $M$ (which is equivalent to $A$-flatness of $A'$) is reduced to topological flatness of $A'$ over $A$. </p> <p>Note that one can "distribute" formal power series over other formal power series when extracting out a <em>finite</em> set of variables into the <em>coefficients</em> over infinitely many variables (think for a minute, using <em>our</em> running definition of "formal power series" for a possibly infinite set of variables). Thus, in our case of interest $A' = k[[x]]$ is a formal power series ring over $A = k[[x_1,\dots,x_r]]$ in infinitely many variables. Thus, we're finally reduced to the question: if $A$ is a pseudo-compact ring (such as $k[[x_1,\dots,x_r]]$) then is $A[[y_1,\dots]]$ topologically flat over $A$? The answer is "yes" because such formal power series rings (in the sense of our running definition) are "topologically free", and a basic fact in the theory is that topological freeness (suitably defined...) implies topological flatness.</p> <p>QED</p> http://mathoverflow.net/questions/117770/does-finite-finite-presented-as-an-algebra-equal-finte-presented-as-a-module/117812#117812 Answer by ayanta for Does 'finite +finite presented as an algebra' equal 'finte presented as a module'? ayanta 2013-01-01T21:28:00Z 2013-01-01T21:28:00Z <p>EGA IV$_1$, 1.4.7.</p> http://mathoverflow.net/questions/117466/henselization-of-valued-field/117467#117467 Answer by ayanta for Henselization of valued field ayanta 2012-12-29T02:11:44Z 2012-12-29T02:11:44Z <p>Same as its importance in commutative algebra. Just to be clear about the definition, for a valued field $K$ with valuation ring $R$, the henselization $K^{\rm{h}}$ is defined to be the valued extension Frac($R^{\rm{h}}$) for the henselization $R^{\rm{h}}$ of $R$ in the sense of commutative algebra (and $R^{\rm{h}}$ is equipped with a preferred valuation extending the one on $R$). </p> <p>This satisfies good properties as if it were a "completion" of $K$ even though it is (separable) <em>algebraic</em> over $K$, and it can be "approximated" using local-etale extensions of $R$; that is really the point. It satisfies Hensel's Lemma and every finite extension $F$ of $K^{\rm{h}}$ admits a unique valuation (necessarily henselian...) extending the one on $K^{\rm{h}}$ (with associated valuation ring that is the integral closure of $R^{\rm{h}}$ in $F$).</p> http://mathoverflow.net/questions/117415/old-books-still-used/117426#117426 Answer by ayanta for Old books still used ayanta 2012-12-28T17:40:09Z 2012-12-28T17:40:09Z <p>The notes of the 1951-2 Artin-Tate seminar on class field theory (published in 1968, and re-issued in LaTeX form a few years ago with a new Introduction by Tate addressing subsequent developments) remains a fundamental reference in algebraic number theory, despite the abundant supply of more recent references on the subject. </p> <p>One reason is that it is the only reference outside the original research literature where one can find a complete treatment (with proofs) of certain key aspects of the theory such as the Grunwald-Wang phenomenon and Weil groups for class formations (especially the case of number fields, which lacks a bare-hands construction as for local fields and global function fields). Come to think of it, the general notion of Weil groups for class formations emerged from that seminar...The style of the proofs remains generally quite fresh.</p> http://mathoverflow.net/questions/117107/independent-generic-general-points-over-some-prime-field/117122#117122 Answer by ayanta for Independent generic/general points over some prime field ayanta 2012-12-24T04:24:56Z 2012-12-24T15:39:35Z <p>You should look in Weil's "Foundations of algebraic geometry"; it isn't as impenetrable as you may fear. He does give all necessary definitions, the key one for your occurring on page 3.</p> <p>Weil's notion of "generic point" was one of his technical innovations, and it is very close to the scheme-theoretic notion, so much more powerful than the weaker concept of "point in general position". A defect is that his (positive-dimensional) varieties have lots of generic points, so in that sense scheme theory has a better notion, but Weil's work is where many important notions in scheme theory were first introduced (in an embryonic form that nonetheless often involved most of the key algebraic difficulties). </p> <p>In Weil's "Foundations", for each characteristic we fix a <em>universal domain</em>: an abstract field $\mathbf{K}$ of infinite transcendence degree over the prime field. A "field" by definition is a <em>subfield</em> $k$ of $\mathbf{K}$ over which $\mathbf{K}$ has infinite transcendence degree. For Weil, if $X$ is a geometrically integral (and separated) $k$-scheme of finite type then a <em>point</em> of $X$ is an element $x \in X(\mathbf{K})$ (a $k$-morphism $x:{\rm{Spec}}(\mathbf{K}) \rightarrow X$). Denoting by $[x]$ the image in $X$ of this $k$-morphism, Weil says $x$ is <em>generic</em> over $X$ if $[x]$ is its generic point. </p> <p>The field $k(x)$ is defined to be the residue field of $X$ at $[x]$ (so it is a finitely generated extension of $k$) and because $x$ is a $k$-morphism from ${\rm{Spec}}(\mathbf{K})$ we see that there is a canonically associated $k$-embedding $k(x) \hookrightarrow \mathbf{K}$ (so $k(x)$ is a "field"). Note that <em>distinct</em> points $x, y \in X(\mathbf{K})$ can satisfy $[x] = [y]$ in $X$ and even $k(x) = k(y)$ inside $\mathbf{K}$, namely when the canonical $k$-embeddings of $k(x)$ and $k(y)$ into $\mathbf{K}$ do not respect the equality of images inside $\mathbf{K}$. For example: taking $\mathbf{C}$ to be the universal domain in characteristic 0, generic points of the affine line over $k = \mathbf{Q}$ are identified with elements of $\mathbf{C}$ that are transcendental over $\mathbf{Q}$, so $\pi$ and $\pi+1$ are distinct generic points of the affine line over $\mathbf{Q}$ even though $\mathbf{Q}(\pi) = \mathbf{Q}(\pi+1)$ as subfields of $\mathbf{C}$ (such equality is stronger than merely saying that the fields $\mathbf{Q}(\pi)$ and $\mathbf{Q}(\pi+1)$ are $\mathbf{Q}$-isomorphic). </p> <p>Weil says that points $x, y \in X(\mathbf{K})$ are <em>independent over $k$</em> if the image domain of the canonical $k$-algebra map $k(x) \otimes_k k(y) \rightarrow \mathbf{K}$ defined using multiplication via the canonically associated $k$-embeddings into $\mathbf{K}$ (!) has fraction field whose transcendence degree over $k$ is equal to the sum of those of $k(x)$ and $k(y)$ over $k$. (See page 3 of the Foundations, especially the third paragraph.) For example, if $k(x) \otimes_k \overline{k}$ is a domain (equivalently, $[x]$ has Zariski closure in $X$ that is geometrically integral over $k$, or as Weil would say "$k(x)$ is regular over $k$") then $k(x) \otimes_k k(y)$ is a domain and hence "independence over $k$" in such cases is precisely the condition that $k(x) \otimes_k k(y) \rightarrow \mathbf{K}$ is injective. For example: taking $\mathbf{C}$ to be the universal domain, the generic points $\pi$ and $e$ of the affine line over $\mathbf{Q}$ are independent over $\mathbf{Q}$ if and only if $\pi$ and $e$ are algebraically independent over $\mathbf{Q}$ (the latter property being an unsolved problem), whereas the pair of generic points $\pi^5 - 2$ and $\pi^3 + 7 \pi - 4$ are certainly not independent over $\mathbf{Q}$. The notion of "independence" (over $k$) for a finite set of points is defined similarly using the tensor product over $k$ of several subfields of $\mathbf{K}$.</p> <p>So in modern terms, the lemma of Nagata is saying that if $C$ is a degree-$d$ curve inside the projective plane over $\mathbf{K}$ (I really mean $\mathbf{K}$ and not the prime field) and if $x_1, \dots, x_{16}$ are $\mathbf{K}$-points of $C$ that lie in the usual affine plane with coordinates that are collectively algebraically independent over the prime field then the multiplicities of $C$ at the points $x_i$ have sum at most $4d$. Of course, one has to look at Nagata's stuff to decide if his curves are meant to be geometrically reduced or geometrically irreducible, etc. </p> http://mathoverflow.net/questions/128872/reference-for-rationality-in-algebraic-groups-in-the-language-of-schemes Comment by 2013-04-27T16:25:02Z 2013-04-27T16:25:02Z @Jason: The existence of the split form (usually called the Existence Theorem) is proved in those Luminy notes over $\mathbf{Z}$. The existence and uniqueness of a quasi-split inner form is also treated there, in the cohomological exercises. http://mathoverflow.net/questions/128872/reference-for-rationality-in-algebraic-groups-in-the-language-of-schemes Comment by 2013-04-27T13:37:19Z 2013-04-27T13:37:19Z What do you mean by &quot;(such as existence of....Borel subgroups defined over the base field)&quot;? There are plenty of examples that have no Borel subgroup over the ground field (i.e., are not quasi-split). Do you mean rational conjugacy of minimal parabolics, that every connected reductive group has a unique quasi-split inner form, or something else? http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p/128196#128196 Comment by 2013-04-20T22:17:56Z 2013-04-20T22:17:56Z @Geoff: In the final sentence, did you mean &quot;finite index&quot; instead of &quot;finite&quot;? Yet the kernel can have nontrivial torsion when $R$ is ramified or $p = 2$. But via $p$-adic logarithm, the kernel of reduction mod $pR$ (not just mod the maximal ideal of $R$, in case of ramification) is a torsion-free subgroup for odd $p$, and likewise mod $4R$ when $p = 2$, so the finite subgroup is a subgroup of ${\rm{GL}}_n(R/pR)$ for odd $p$, and ${\rm{GL}}_n(R/4R)$ for $p = 2$. So that yields a crude bound $C(n,q,e)$ where $q$ is residue field size and $e = e(R|\mathbf{Z}_p)$ is ramification degree. http://mathoverflow.net/questions/127907/extensions-of-affine-group-schemes Comment by 2013-04-20T04:03:18Z 2013-04-20T04:03:18Z If a $k$-group homomorphism $f:G' \rightarrow G$ admits a $k$-scheme section $s$ then we get a $k$-scheme isomorphism between $G'$ and $G \times \ker(f)$. So if $G'$ is connected and smooth then this would force its direct factor scheme $\ker(f)$ to be connected and smooth. But $\mu_n = \ker({\rm{SL}}_n \twoheadrightarrow {\rm{PGL}}_n)$ is never connected and smooth when $n &gt; 1$, so there is no section to ${\rm{SL}}_n \rightarrow {\rm{PGL}}_n$ when $n &gt; 1$. http://mathoverflow.net/questions/127365/inertia-subgroup-in-the-ordinary-reduction-case-when-p2/127628#127628 Comment by 2013-04-19T16:58:20Z 2013-04-19T16:58:20Z I agree that an essential input is the analytic inverse function theorem (applied to the finite etale cover of trivializations of the $n$-torsion). I agree that Krasner's Lemma provides initial intuition for believing the result, and one could certainly use the analytic inverse function theorem to give a proof of Krasner's Lemma that is a bit more advanced than the usual proof, but I don't think the proof of the actual result in question here (&quot;local constancy&quot; for the Galois representation in the rational-point fibers of a finite etale group over an analytic space) uses Krasner's Lemma. :) http://mathoverflow.net/questions/127365/inertia-subgroup-in-the-ordinary-reduction-case-when-p2/127628#127628 Comment by 2013-04-19T12:58:12Z 2013-04-19T12:58:12Z One needs to do more than Krasner's Lemma in order to control the structure of the Galois representation: Krasner's Lemma only controls very coarse information, the splitting field. One has to use the Henselian property of analytic local rings or other input from non-archimedean analytic geometry to rigorously justify the local constancy assertion being made here. http://mathoverflow.net/questions/127907/extensions-of-affine-group-schemes Comment by 2013-04-19T11:36:38Z 2013-04-19T11:36:38Z Vanishing of such cohomology does not suffice to make a section. For $n &gt; 1$, consider ${\rm{GL}}_n \rightarrow {\rm{PGL}}_n$. This has sections Zariski-locally on the base (even as schemes over $\mathbf{Z}$), but there is no section over the entire base, over any field. Indeed, if $s$ were a global section then WLOG $s(1)=1$ and composing $s$ with det would define a global unit on ${\rm{PGL}}_n$ with value $1$ at $1$. The only such unit is $1$ (since ${\rm{PGL}}_n$ is semisimple), so $s$ would be a section to ${\rm{SL}}_n \rightarrow {\rm{PGL}}_n$, contradiction via connectedness. http://mathoverflow.net/questions/127769/applications-of-govorov-lazard-theorem/127806#127806 Comment by 2013-04-19T10:09:26Z 2013-04-19T10:09:26Z The special case of Lazard's theorem over $\mathbf{Z}/n\mathbf{Z}$ can be proved much more easily than the general case (we can assume $n$ is a prime power to be over an artin local ring, and then lift a basis over the residue field and use the vanishing of a power of the maximal ideal, etc. -- this appears very early on in Matsumura's book &quot;Commutative Ring Theory&quot;, for example), so it could be debatable to say that Lazard's theorem plays a &quot;crucial role&quot; in the proof of the proper base change theorem. But the final sentence is not debatable. :) http://mathoverflow.net/questions/127907/extensions-of-affine-group-schemes Comment by 2013-04-18T03:34:57Z 2013-04-18T03:34:57Z If $G_1$ isn't finite type then there cannot be $G'_2$ of finite type in which $G_1$ is a closed subscheme, so what are you trying to do in the first question? And even with algebraicity throughout, if $G_1$ is not $k$-smooth then Galois cohomology is absolutely the wrong notion to consider, so some explanation of the motivation for the 2nd question would be helpful. (There's always fpqc topology, etc., but saying what is useful to consider is much informed by knowing the context to which you plan to apply an answer.) http://mathoverflow.net/questions/127807/exponential-map-for-finite-group-schemes Comment by 2013-04-18T03:14:28Z 2013-04-18T03:14:28Z Does &quot;finite free&quot; group scheme mean &quot;structure algebra is locally free of finite rank&quot; as a module over the base ring? If so then over a $\mathbf{Q}$-algebra any such group scheme is finite etale, so the Lie algebra vanishes. What are you trying to do, and why? http://mathoverflow.net/questions/127706/a-question-about-r-points-of-an-complex-reductive-group Comment by 2013-04-17T04:44:46Z 2013-04-17T04:44:46Z @Jim Humphreys: it is not generally true that the center of a connected reductive (non-ss) group is connected, so it is unclear what you meant in the first part of the second sentence of your comment. For example, if $G = {\rm{SL}}_n$ and $d|n$ with $1 \le d &lt; n$ then the central pushout of $G$ along $\mu_d \hookrightarrow {\rm{GL}}_1$ is non-ss connected reductive with disconnected center. (Direct product if $d = 1$.) Likewise, $G := {\rm{Spin}}_{2n}$ for $n &gt; 2$ has center $\mu_2 \times \mu_2$ or $\mu_4$ and we can form a non-ss connected reductive central pushout along a $\mu_2$. http://mathoverflow.net/questions/124847/principal-maximal-ideals-in-zx-f/125041#125041 Comment by 2013-03-24T04:12:33Z 2013-03-24T04:12:33Z Dear Martin: The advantage of being at a department as strong as Muenster is that it has a lot of people who know number theory quite well both at this level and way beyond, so you can also ask any of them in person. Proceeding in that way you may give more understanding of the reasoning than by waiting for someone on MO to declare &quot;Yes, I agree with this argument&quot; (assuming I have not made a blunder, which I do not believe I have). http://mathoverflow.net/questions/124847/principal-maximal-ideals-in-zx-f/125041#125041 Comment by 2013-03-23T01:16:50Z 2013-03-23T01:16:50Z Dear Aakumadula: Here is the argument I had in mind. Since &quot;ideal generated by&quot; is inverse to the operation of &quot;intersection&quot; when considering maximal ideals relatively prime to the index (so to speak), we just use a bit of transitivity: if $A'' \subset A' \subset A$ is a containment of orders and $P$ is a maximal ideal of $A$ whose residue characteristic doesn't divide $[A:A'']$ then the prime ideals $P' = A' \cap P$ and $P'' = P' \cap A'' = P \cap A''$ satisfy $P'' A' = P'$ (!) so if $P \cap A''$ is principal then such a generator in $A''$ also generates $P'$ in $A'$. http://mathoverflow.net/questions/124847/principal-maximal-ideals-in-zx-f Comment by 2013-03-19T00:49:21Z 2013-03-19T00:49:21Z @Aakumadula: I misremembered the bijection between the sets of maximal ideals $P$ of $O_K$ with residue char. not dividing $[O_K:A]$ and maximal ideals $P'$ of $A$ with residue char. not dividing $[O_K:A]$ ($P \mapsto P \cap A$, $P' \mapsto P'O_K$), as we see by localizing at rational primes. Better idea: for $N := [O_K:A]$ we have $1 + NO_K \subset A$, so it suffices to find infinitely many maximal ideals of $O_K$ that are principal and admit a generator congruent to 1 modulo $N$. Now we apply &quot;Dirichlet's proof&quot; of Chebotarev, with characters of generalized ideal class groups. http://mathoverflow.net/questions/124847/principal-maximal-ideals-in-zx-f Comment by 2013-03-18T13:02:58Z 2013-03-18T13:02:58Z @Aakumadula: To complete your thought, for any number field $K$ and order $A$ in the ring of integers $O_K$, the finiteness of the index of $A$ in $O_K$ as an abelian group implies that maximal ideals of $O_K$ whose residue characteristic does not divide $[O_K:A]$ lie entirely inside $A$. Thus, principal maximal ideals of $O_K$ whose residue characteristic does not divide $[O_K:A]$ do the job (i.e., they are also principal maximal ideals of $A$).