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stats | profile views | 836 |
Sep 18 |
comment |
Solving z^n=a+ib using only radicals of positive real numbers
@Hugo: If you unravel the proof of my comment, you'll find that for the application to cyclotomic polynomials it is no different from the usual Galois theory argument for analyzing constructibility of polygons. So that particular example isn't as compelling an illustration as it may seem to be at first glance. |
Sep 18 |
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Solving z^n=a+ib using only radicals of positive real numbers
The restriction to degree 3 can be relaxed (and Cardano bypassed): if $f$ is an irreducible polynomial of any odd degree $> 1$ over a field $F$ of characteristic 0 then its splitting field over $F$ contains a root of unity of odd prime order and so cannot be embedded into $\mathbf{R}$. In particular, when $F$ is given as a subfield of $\mathbf{R}$ and the Galois group of $f$ is abelian over $F$ (e.g., cyclic), so the splitting field is generated over $F$ by a single root of $f$, then it is impossible to find any root of $f$ inside a radical tower over $F$ inside $\mathbf{R}$. |
Sep 17 |
awarded | Good Answer |
Sep 17 |
awarded | Commentator |
Sep 17 |
awarded | Nice Answer |
Sep 16 |
revised |
Mostow's theorem on algebraic groups
added 4 characters in body |
Sep 16 |
comment |
Can we control the size of the intersection of two abelian subfactors of an abelian variety ?
Take $B$ and $B'$ to be non-CM non-isogenous elliptic curves, pick a common $K$-subgroup $G$, and define $A = (B \times B')/G$. The only elliptic curves that are abelian subvarieties of $A$ are the evident copies of $B$ and $B'$, so there's nothing you can do to "fix" the problem in such cases. One way to avoid such a situation is to assume that $A$ admits a polarization of degree prime to $p$, but that's a rather severe assumption. |
Sep 16 |
answered | Mostow's theorem on algebraic groups |
Sep 16 |
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Two definitions of smoothness?
@LMN: Yes. When I was writing my comment about this argument I was planning to say "(the Krull-Akizuki trick)" but then I couldn't fit it all into one box, so I removed that phrase and hoped you would know the (standard) argument I had in mind for that step. |
Sep 15 |
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Two definitions of smoothness?
@LMN: The main content in relating the two definitions is as follows. First, by flatness every irreducible component of $X$ dominates one of $Y$ and so under Def. 2 each generic fiber is of pure dimension $n$. To prove the same for all non-empty fibers, by base change to dvr's over $Y$ and the purity hypothesis for the generic fibers we can pass to irreducible components (after the dvr base change, using the easy nature of flatness over a dvr!) to reduce to the case of an irreducible scheme fppf over a dvr. Now apply results about flatness and dimension theory (or EGA IV$_3$, 14.3.10). |
Sep 15 |
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Two definitions of smoothness?
Hypothesis $(\ast)$ cannot be removed, as you can see via the special case when $Y={\rm{Spec}}(k)$ and $X={\rm{Spec}}(K)$ for a finite extension of fields. The $K$-vector space $\Omega^1_{X/Y}$ has some dimension $n≥0$, and $n>0$ if and only if $K/k$ is not separable (equivalently, not etale). So we need some condition to relate n to fiber dimensions. Hartshorne's $(\ast)$ is focusing exclusively on dimensions of generic fibers, so some work is needed to show that Definition 2 forces all fibers to have pure dimension $n$. Definition 1 is conceptually superior. |
Sep 15 |
comment |
Mostow's theorem on algebraic groups
(cont'd) Of course, char. 0 is being used again at the very end, to see that the formation of $G'$-invariants is (right-)exact on the category of algebraic linear representations of the possibly disconnected reductive $G'$. Indeed, by the analogue of Hochschild-Serre (or bare hands) we reduce the cohomological vanishing (or equivalently, the exactness) to 3 cases: (i) connected semisimple, (ii) torus, and (ii) finite etale. Lie algebras over $K$ settle (i), and scalar extension to $\overline{K}$ reduces (ii) and (iii) to the easy cases of split tori and finite constant groups. |
Sep 15 |
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Mostow's theorem on algebraic groups
Let $U = R_u(G)$. To prove existence of a $K$-rational splitting and uniqueness up to $U(K)$-conjugacy, WLOG $U = V$ is a vector group. Conjugation of $G$ on $V$ defines an action of $G' = G/V$ on $V$ that is linear since char($K$)=0 (!). Also, $G \rightarrow G'$ is an etale $V$-torsor, so it admits a section by vanishing of degree-1 quasi-coherent etale cohomology of $G'$. Thus, the problem is vanishing of Hochschild cohomology H$^i(G',V)$ (algebraic cochains) for $i=1,2$. On the category of algebraic linear representations, Hochschild cohomology is the derived functor of "invariants". QED |
Sep 12 |
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Artin/Popescu approximation for (some) big rings
It may be helpful if you indicate why you pose the question (e.g., idle curiosity or something more substantial). For example, some non-archimedean geometry (most naturally, Berkovich spaces) ensures that for a finite system of polynomial equations over $A$ (or even something more general), any solution in $B$ can be approximated arbitrarily well by a solution in $A$. So if that is your goal then the Artin-Popescu (and "smoothening") viewpoint in such generality is unnecessary. |
Sep 12 |
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A question on an intuitive way to look at stacks
@QcH: The zero map is another illustration of the problem. Johan indeed meant to say that the square he drew is cartesian (which is just another way to express the isomorphism to the pullback, as you have said it; it is also the way KM85 express things). Another bonus of writing to Johan is that he may still have some t-shirts lying around and so you may get a free one in this way. |
Sep 10 |
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Restricting representations to lattices
The Zariski-density of $G(\mathbf{Z})$ in $G(\mathbf{R})$ for any "split semisimple" group $G$ over $\mathbf{Z}$ (e.g., ${\rm{SL}}_n$) admits a direct proof more elementary than the proof of Borel's theorem. Indeed, by considering pairs of opposite root groups and the open cell structure relative to a positive system of roots for a split maximal $\mathbf{Z}$-torus, we get $\mathbf{Z}$-subgroups $U \simeq \mathbf{G}_a$ such that the subgroups $U_{\mathbf{Q}}$ generate $G$ over $\mathbf{Q}$. Thus, we just need to note that $\mathbf{Z}$ is Zariski-dense in $\mathbf{G}_a$ over $\mathbf{Q}$! |
Sep 10 |
comment |
Restricting representations to lattices
@Misha: A more basic kind of counterexample is that if the connected semisimple $G$ isn't simply connected then $G(\mathbf{R})$ may be disconnected (e.g., ${\rm{PGL}}_{2n}$ and various other $G$ with "fundamental group" of even order). In such cases the projection from $G(\mathbf{R})$ onto its nontrivial finite component group leads to rather non-algebraic representations. This highlights the significance of the connectedness of $G(\mathbf{R})$ when $G$ is simply connected (see my comments on the answer below). |
Sep 10 |
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a question on TITS' note “Reductive groups over local fields”
@Igor Rivin: My belief was based on my expertise in the relevant math, but I concede it was just an educated guess (which seems to have been correct, but that's beside the point). More importantly, the mixture of English errors in the question and the "name" of the OP suggest that English fluency (as opposed to English proficiency) may be missing here, hence my Jeopardy joke. Math is universal, yet MO practice leaves little choice but to post questions in English, and I think in cases lacking English fluency one should cut the OP some slack if an expert can surmise the likely intent. |
Sep 10 |
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Restricting representations to lattices
@Igor (cont'd): At the final step above I am hauling out the big guns: if $G'$ is a connected semisimple $\mathbf{R}$-group that is simply connected then $G'(\mathbf{R})$ is connected. The only proof I know is to deduce it from considerations with maximal compact subgroups, Cartan involutions, and a theorem of Steinberg (or topological precursor by Cartan) on Zariski-connectedness of centralizers of semisimple automorphisms of simply connected semisimple groups, way too much to explain here. I would love to know a literature reference (beyond the "classical" anisotropic case). |
Sep 10 |
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Restricting representations to lattices
@Igor: Let's show for linear algebraic $\mathbf{R}$-groups $G'$ and $G$ with $G'$ connected semisimple and simply connected, analytic $f:G'(\mathbf{R}) \rightarrow G(\mathbf{R})$ are algebraic. The Lie algebra of $\Gamma_f$ is a perfect subalgebra of $\mathbf{g}' \times \mathfrak{g}$, so by 7.9 in Borel's LAG it arises from a connected closed $\mathbf{R}$-subgroup $H$ of $G' \times G$. Clearly $H$ is semisimple, and $H \rightarrow G'$ is an isogeny, so an isomorphism as $G'$ is simply connected. This gives $G' \rightarrow G$ recovering $f$ on $G'(\mathbf{R})^0 = G'(\mathbf{R})$ (Steinberg!) |