User sebastian petersen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:28:58Z http://mathoverflow.net/feeds/user/8680 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96932/reed-muller-codes Reed-Muller-Codes Sebastian Petersen 2012-05-14T18:06:14Z 2012-07-20T03:06:06Z <p>Let $F$ be the field with two elements, $V_m=F^{2^m}$.Let $R(r, m)\subset V_m$ be the binary Reed-Muller Code. Define $R_m:=R(1, m)$. Then the dimension of $R_m$ is $1+m$ and its minimal distance is $d(R_m)=2^{m-1}$. (Cf. for example the book of Luetkebohmert). Hence the information rate is $I(R_m)=\frac{1+m}{2^m}$ while the relative minimal distance of $R_m$ is $rd(R_m)=\frac{1}{2}$.</p> <p>Now $R_5$ has the same relative minimal distance as $R_4$ (and thus should have approximately the same error correction abilities), while the information rate of $R_4$ is much better ($I(R_4)\approx 0.31$ while $I(R_5)\approx 0.19$).</p> <p>I furthermore read in Luetkebohmert (and elsewhere) that $R_5$ was used in practice, e.g. during a Mariner space shuttle mission.</p> <p>Question: Given the values above, why would anybody use $R_5$ and not $R_4$?</p> <p>(I hope this question is not too easy for MO, but I do not see the answer at the moment.)</p> <p>Edit: It seems that the code $R_5$ might have been used for Mariner, because then one has really one code word per pixel of the image. (Compare the comments)</p> <p>Still let me note that the $R_m$ all have the same relative minimal distance, that there exists an efficient (poly time?) algorithm to decode them, and that $I(R_m)$ tends rapidly to zero when $n\to\infty$. And I find this a bit strange: For an "interesting" or even "celebrated" family of codes $(C_m)_m$, which all have the same relative minimal distance, I would naively have expected something like $(I(C_m))_m$ should be a decreasing (but bounded or even converging) sequence. </p> http://mathoverflow.net/questions/59555/mathematics-seminar-for-non-mathematicians Mathematics seminar for "non-mathematicians" Sebastian Petersen 2011-03-25T13:10:54Z 2012-05-15T21:20:55Z <p>Next term I am leading a seminar for students, who will become teachers for elementary school i.e. for kids of age 6-10. The students in the seminar will have no mathematical background beyond the "Abitur". They are supposed to give a 45 minutes talk on a (not too difficult) mathematical topic, and they have to write an exposition of a few pages. My first ideas cover geometry of triangles, basics around Fibonacci sequences ect. Did somebody on MO teach a similar class already?</p> <p>In brief: I would appreciate further suggestions for suitable topics very much!</p> http://mathoverflow.net/questions/58732/torus-based-cryptography Torus based cryptography Sebastian Petersen 2011-03-17T08:56:21Z 2012-03-20T15:10:24Z <p>In cryptography one needs finite groups $G$ in which the discrete logarithm problem is infeasible. Often they use the multiplicative group $\mathbb{G}_m(\mathbb{F}_p)$ where $p$ is a prime number of bit length $500$, say.</p> <p>Rubin and Silverberg suggested (cf. [1]) to use certain tori instead, if the goal is to minimize the key size. In the easiest case, this comes down to using the group $$T_2(p)=ker(Norm: \mathbb{F}_{p^2}^\times\to \mathbb{F}_p^\times).$$</p> <p>If I understood correctly, then the underlying philosopy seems to be: *The group $T_2(p)$ should be as secure as $\mathbb{F}_{p^2}^\times$, but its size is only $p+1$.* (So, if you use groups of type $T_2(p)$ instead of groups of type $\mathbb{G}_m(\mathbb{F}_p)$, then you can achive the same security with half the key size.)</p> <p><strong>Question.</strong> What are the reasons, be they heurisical or strictly provable, to believe in this philosopy?</p> <p>Denote by $\mathbb{G}'_m$ the quadratic twist of the algebraic group $\mathbb{G}_m$. It is easy to see that $T_2(p)$ is isomorphic to $\mathbb{G}'_m(\mathbb{F}_p)$. (This isomorphism is easy to compute). The philosophy predicts: The quadratic twist of the multiplicative group should be better than the multiplicative group itself.</p> <p>(Compare with elliptic curves: If $E/\mathbb{F}_p$ is an elliptic curve, then I would certainly not expect its quadratic twist to be better than $E$ itself.)</p> <p><strong>Remark:</strong> I concentrated on the simplest case above. One also considers certain groups $T_n(p)$ which are expected to be as secure as $\mathbb{F}_{p^n}^\times$, while their size is only $\approx\varphi(n)p$. Lemma 7 in [1] is meant to explain this. However, I would be keen on a more detailed explanation. </p> <p>[1] Lect. Notes in Comp. Sci. 2729 (2003) 349-365. (available at <a href="http://math.stanford.edu/~rubin/" rel="nofollow">http://math.stanford.edu/~rubin/</a>)</p> http://mathoverflow.net/questions/81870/a-good-reduction-property A good reduction property Sebastian Petersen 2011-11-25T11:45:11Z 2011-11-25T11:45:11Z <p>Let $R$ be a normal noetherian domain and $K$ the quotient field of $R$. Let $X$ be a smooth algebraic $K$-scheme and $x\in X(K)$ a $K$-rational point of $X$. Does there exist a tripel $(U, s, i)$ consisting of a smooth algebraic $R$-scheme $U$, a section $s:Spec(R)\to U$ and an open immersion $i: U\times_R Spec(K)\to X$ such that $s$ corresponds to $x$, i.e. such that $i\circ (s\times_R Spec(K)): Spec(K)\to X$ agrees with $x$?</p> <p>My main interest is the case where $R$ is local and $dim(X)=1$.</p> <p>(Maybe it is easy. I read something in a paper which more or less comes down to that, but I do not get it at the moment.)</p> http://mathoverflow.net/questions/77444/semistable-reduction-theorem-over-higher-dimensional-schemes Semistable reduction theorem over higher dimensional schemes Sebastian Petersen 2011-10-07T11:17:48Z 2011-10-07T11:17:48Z <p>Let $k$ be a field, $S/k$ a smooth variety with function field $K$ and $U$ a nonempty open subscheme of $S$. For every finite separable extension $E/K$ we denote by $S^E$ (resp. $U^E$) the normalization of $S$ (resp $U$) in $E$. Let $A/U$ be an abelian scheme with generic fibre $A_\eta$. For every prime number $\ell$ different from $char(k)$ and every finite separable extension $E/K$ we consider the representation $\rho_{\ell, E}$ of $\pi_1(U^E)$ on the $\ell$-torsion part $A_\eta[\ell]$. Let $H(E)$ be the kernel of the epimorphism $\pi_1(U^E)\to \pi_1(S^E)$. </p> <p>Question: Does there exist a finite separable extension $E/K$ such that for every prime number $\ell\neq char(k)$ the group $\rho_{\ell, E}(H(E))$ is generated by its $\ell$-Sylow subgroups?</p> <p>Remark: It is known that the answer is yes in the special case $dim(S)=1$. This is a consequence of the semistable reduction theorem of Grothendieck.</p> http://mathoverflow.net/questions/56187/fundamental-group-of-a-semiabelian-variety Fundamental group of a semiabelian variety Sebastian Petersen 2011-02-21T17:08:11Z 2011-05-09T11:22:13Z <p>Let $K$ be an algebraically closed field. Assume for simplicity that $K$ has characteristic zero. Let $A/K$ be a semiabelian variety. Let $n$ be an integer coprime to $char(K)$. Denote by $\pi_1(A)$ the etale fundamental group of $A$. Is it true that $Hom(\pi_1(A), {\mathbb{Z}}/n)$ is isomorphic to $A[n]^\vee$ (edit)? What happens, if I replace $A$ by an arbitrary connected algebraic group over $K$?</p> <p>Remark: The answer is yes'' in the case where $A/K$ is proper. (The key ingredient to the proof of this fact is the theorem of Serre-Lang, that every finite etale cover $X/A$ carries the structure of an abelian variety, provided $A/K$ itself is an abelian variety.) </p> http://mathoverflow.net/questions/53899/finiteness-property-of-fundamental-groups Finiteness property of fundamental groups Sebastian Petersen 2011-01-31T16:11:05Z 2011-03-14T12:55:30Z <p>Let $K$ be an algebraically closed field of characteristic zero. Let $X/K$ be a smooth variety. Is it true that the \'etale fundamental group $\pi_1(X)$ is topologically finitely generated. </p> <p>I know that the answer is yes'' in the following two cases:</p> <p>1) $X/K$ is proper.</p> <p>2) $\dim(X)=1$. (In this case we can write down a presentation of $\pi_1(X)$ very explicitly.)</p> <p>Can there anything go wrong in the general case? (It would be wonderful to have a reference which one can simply quote.)</p> http://mathoverflow.net/questions/53899/finiteness-property-of-fundamental-groups/58423#58423 Answer by Sebastian Petersen for Finiteness property of fundamental groups Sebastian Petersen 2011-03-14T12:55:30Z 2011-03-14T12:55:30Z <p>Thanks, Dono Arapura, for explaining how one proves this statement, using the Riemann existance theorem in SGA1.</p> <p>I mention for the sake of completeness: Today in the morning, I found that there is a SGA-reference for my question as it stands above: SGA 7.1, II.2.3.1. (I was surprised, that it is SGA 7.1, not SGA 1.) </p> http://mathoverflow.net/questions/57802/what-properties-define-open-loci-in-families/57831#57831 Answer by Sebastian Petersen for What properties define open loci in families? Sebastian Petersen 2011-03-08T14:12:20Z 2011-03-11T15:13:26Z <p>You find plenty of theorems of this type in EGA IV/Part 3, see especially EGA IV.9. </p> <p>Let $f: X\to S$ be a morphism of finite presentation. (We do not have to make a flatness or properness assumption in what follows.) For example let $P$ be one of the following properties:</p> <ul> <li>being geometrically irreducible</li> <li>being geometrically connected</li> <li>being geometrically regular</li> <li>being geometrically normal</li> <li>being geometrically reduced</li> <li>having property R_k geometrically</li> </ul> <p>Let $U$ be the set of points $s\in S$ such that $X_s/k(s)$ has property P.</p> <p>Then $U$ is at least locally constructible. (cf. IV.9.7.7 and IV.9.9.4)</p> <p>Hence: If $S$ is irreducible and noetherian and $U$ contains the generic point of $S$, then $U$ contains a nonempty open set. </p> <p>(But in EGA IV / Part 3 there are much more results of that flavour...)</p> http://mathoverflow.net/questions/57969/stalks-of-etale-sheaves/58068#58068 Answer by Sebastian Petersen for Stalks of etale sheaves Sebastian Petersen 2011-03-10T13:14:51Z 2011-03-10T13:29:11Z <p>It is enough to prove this in the case where $Y$ is $Spec$ of a strictly Henselian ring. I think, one sees the main point in the argument already in the following special case:</p> <p>Let $Y=Spec(K)$ and $X=Spec(E)$ where $E/K$ is a finite separable extension of fields. Denote $\pi: X\to Y$ the canonical map. Let $\overline{E}$ be an algebraic closure of $E$ and $\overline{x}$ (resp. $\overline{y}$) be the corresponding geometric point of $X$ (resp $Y$). Let $L/K$ be a finite separable extension contained in $\overline{E}$ and containing the Galois closure of $E$. If $F$ is a sheaf on $X$, then $\pi_*F(Spec(L))=F(Spec(L\otimes_K E))$ and $Spec(L\otimes_K E)= \coprod_{i=1}^d Spec(L)$, where $d=[E:K]$. Hence $\pi_* F(Spec(L))=F(Spec(L))^d$. Now take an inductive limit over such $L$, in order to obtain $\pi_*F_{\overline{y}}=F_{\overline{x}}^d$.</p> <p>Hope this is of some use...</p> http://mathoverflow.net/questions/57830/picard-group-fundamental-group-and-deformation/57837#57837 Answer by Sebastian Petersen for Picard group, Fundamental group, and deformation Sebastian Petersen 2011-03-08T14:45:03Z 2011-03-10T13:24:24Z <p>I am not sure whether this is exactly what you want. But still, I see the following relation between $Pic$ and $\pi_1$:</p> <p>Let $k$ be an algebraically closed field and $X/k$ a proper variety. Then there is an isomorphism $H^1(X, \mathbb{Z}/n)=Hom(\pi_1(X), \mathbb{Z}/n)$. Furthermore, from the long exact cohomology sequence associated to $$0\to \mathbb{Z}/n\to \mathbb{G}_m\to \mathbb{G}_m\to 0$$ we obtain an isomorphism $$H^1(X, \mathbb{Z}/n)\cong H^1(X, \mathbb{G}_m)[n].$$ Here we used that $k$ is algebraically closed and that $X$ is proper over $k$. By a generalization of Hilbert 90 we have $H^1(X, \mathbb{G}_m)=Pic(X)$. (Cf. Milne's book, Chapter III, Proposition 4.9.) Hence, after all, we see that there is an isomorphism $$Hom(\pi_1(X), \mathbb{Z}/n)\cong Pic(X)[n].$$ I hope, this is of some use...</p> <p>(The cohomology groups used above are the etale ones.)</p> http://mathoverflow.net/questions/56984/alterations-factor-as-modification-finite-map/56998#56998 Answer by Sebastian Petersen for Alterations factor as modification + finite map Sebastian Petersen 2011-03-01T13:27:03Z 2011-03-01T13:49:49Z <p>As Sandor points out, this is Stein factorization. Let $X$, $Y$ be varieties over a field $K$. Let $h: X\to Y$ be a proper morphism. Then $h_*(O_X)$ is coherent and $Y':=Spec(h_*(O_X))$ (cf. EGA II.1.3 for the definition of $Spec$ of a sheaf of quasicoherent algebras) is finite over $Y$. Consider the Stein factorization $$h:X\buildrel h' \over\longrightarrow Y'\buildrel g\over \longrightarrow Y$$ of $h$. (It is true by Zariski's connectedness theorem (cf. EGA III.4.3.1 - III.4.3.4) that $h'$ has non-empty geometrically connected fibres and $g$ is finite, but this will not be needed here). </p> <p>Now assume that $h$ is an alteration, i.e. that $h$ is surjective and that there exists a nonempty open subscheme $U\subset Y$ such that $h^{-1}(U)\to U$ is finite. Replacing $U$ by a smaller set we may assume that $U=Spec(A)$ is affine. Define $U'=g^{-1}(U)$ and $V=h^{-1}(U)$. Then $V=Spec(B)$ for some $A$-algebra $B$, because $h$ is affine as a finite morphism. We thus have $h_*(O_X)(U)=B$. Hence it follows from the definition of the sheaf Spec that $U'=Spec(B)$ and the restriction $h': V\to U'$ is an isomorphism. Hence $h': X\to Y$ is a modification. </p> <p>Note that this argumentation does not use the full strength of Zariski's connectedness theorem in an essential way here. Most of the things are in a sense "a tautology". The only somewhat deeper fact entering is the fact that $h_*(O_X)$ is coherent, because of the properness assumption. </p> <p>Concerning your comment, the following is clearly true: If $f: X\to Y$ is a finite morphism and $f_*(O_X)=O_Y$, then $f$ is an isomorphism.</p> http://mathoverflow.net/questions/56378/semisimplicity-of-etale-cohomology-representations Semisimplicity of étale cohomology representations Sebastian Petersen 2011-02-23T10:30:02Z 2011-02-23T15:21:28Z <p>Let $K$ be a number field and $G=Gal(\overline{K}/K)$ the absolute Galois group of $K$. Let $\ell$ be a prime number.</p> <p>Let $A/K$ be an abelian variety. Then the representation of $G$ on $V_\ell(A)$ is semisimple. This is the famous theorem of Faltings (Invent. Math. 73).</p> <p>Now let $X/K$ be a smooth projective variety and $0\le q\le 2\dim(X)$, and define $\overline{X}=X_{\overline{K}}$.</p> <p><strong>Question.</strong> Is it known that the representation of $G$ on $H^q(\overline{X}, \mathbb{Q}_\ell)$is semisimple?</p> <p><strong>Remark.</strong> The answer is yes for $q=1$, because $H^1(\overline{X}, Q_\ell)$ is dual to $V_\ell(A)$ where $A$ is the Albanese variety of $X$.</p> <p>I would also be interested in the case where the number field $K$ is replaced by a global function field (say), and $\ell$ is assumed to be coprime to the characteristic. </p> http://mathoverflow.net/questions/56187/fundamental-group-of-a-semiabelian-variety/56299#56299 Answer by Sebastian Petersen for Fundamental group of a semiabelian variety Sebastian Petersen 2011-02-22T17:08:29Z 2011-02-22T17:27:52Z <p>I can give a partial answer, or at least a strategy towards my question, myself. I hope that this will also put my question into the right context, and increase its chances of being answered completely. (I admit that it was not so well-formulated at the beginning.)</p> <p>Let $K$ be algebraically closed of characteristic zero (for simplicity). Let $X/K$ be a separated connected algebraic $K$-scheme. Then we have $H^1(X, \mathbb{Z}/n)=Hom(\pi_1(X), \mathbb{Z}/n)$. On the other hand we have an exact sequence $$1\to \Gamma(X, {\cal{O}}_X)^\times/n\to H^1(X, \mathbb{Z}/n)\to H^1(X, {\mathbb{G}}_m)[n]\to 1\ (*),$$ associated to the short exact sequence of etale sheaves $$1\to \mu_n\to \mathbb{G}_m\to \mathbb{G}_m\to 1$$ (and noting that $\mu_n=\mathbb{Z}/n$ because $K$ is algebraically closed).</p> <p>Furthermore $H^1(X, \mathbb{G}_m)=Pic(X)$, the group of invertible sheaves on $X$. If $X$ is proper in addition, then $\Gamma(X, {\cal{O}}_X)^\times=K^\times$. Hence there is an isomorphism $$Hom(\pi_1(X), \mathbb{Z}/n)=H^1(X, \mathbb{Z}/n)\cong Pic(X)[n]$$ for every connected proper $K$-scheme.</p> <p>Now consider the special case where $A/K$ is an abelian variety. Then $$Pic(X)[n]=A^\vee[n]=A[n]^\vee=Hom(A[n], \mathbb{Z}/n);$$ hence we obtain in fact a canonical isomorphism $$Hom(\pi_1(A), \mathbb{Z}/n)=Hom(A[n], \mathbb{Z}/n).$$ This is where the "well-known" isomorphism $\pi_1(A)\cong \prod_\ell T_\ell(A)$ comes from. (The fact that $\pi_1(A)$ is abelian has to be shown in addition.) </p> <p>One also sees that $H^1(A, Z_{\ell})$ is dual to the Tate module $T_\ell(A)$ ($\ell$ a prime number).</p> <p>Now let $B$ be a semiabelian variety. I asked the above question, because I wanted to know, whether the situation is similar in the case of a semiabelian variety. For example, I wanted to know:</p> <p>i) Is $Hom(\pi_1(B), \mathbb{Z}/n)$ (canonically) isomorphic to $Hom(B[n], \mathbb{Z}/n)$?</p> <p>ii) Is there a useful relation between $\pi_1(B)$ and $\prod_\ell T_\ell(B)$, where $T_\ell(B)=lim_i B[\ell^i]$ is the Tate module (defined in a naive way analoguos to the proper case). Are these groups canonically isomorphic?</p> <p>iii) Is there a useful relation between $H^1(B, Z_\ell)$ and $T_\ell(B)$? Is the first $\mathbb{Z}_\ell$-module canonically isomorphic to the dual of the second?</p> <p>I think, this case of semiabelian varieties is somewhat different, because $\Gamma(B, {\cal{O}}_B)^\times/n$ does not vanish any more, unless $B$ is proper. </p> <p>But nevertheless these questions still make sense to me. Further comments / answers are appreciated very much. </p> http://mathoverflow.net/questions/51114/composition-and-intersection-of-residue-fields Composition and intersection of residue fields Sebastian Petersen 2011-01-04T12:02:02Z 2011-01-04T13:28:35Z <p>Let $A$ be a normal ring with quotient field $K$. Let $L/K$ be a finite separable extension. Let $E_1/K$ and $E_2/K$ be extensions of $K$ contained in $L$. Let $B_1$ (resp. $B_2$) be the normalization of $A$ in $E_1$ (resp. $E_2$) and let $C$ be the normalization of $A$ in $L$. Then $$K\subset E_i\subset L\ \mbox{and}\ A\subset B_i\subset C\ \mbox{for}\ i=1, 2.$$ Let $\mathfrak P$ be a prime ideal of $C$. Define $\mathfrak p_1:=\mathfrak P\cap B_1$, $\mathfrak p_2:=\mathfrak P\cap B_2$ and $\mathfrak p:=\mathfrak P\cap A$. Then, on the level of residue fields, we have $$k(\mathfrak p)\subset k(\mathfrak p_i)\subset k(\mathfrak P)\ \mbox{for}\ i=1, 2.$$</p> <p>*Assume that $B_1$ and $B_2$ are finite etale $A$-algebras.*</p> <p><strong>Questions:</strong></p> <p>a) Does $L=E_1E_2$ (composite field) imply $k(\mathfrak P)=k(\mathfrak p_1) k(\mathfrak p_2)$?</p> <p>b) Does $K=E_1\cap E_2$ imply $k(\mathfrak p)=k(\mathfrak p_1)\cap k(\mathfrak p_2)$?</p> <p><strong>Comments:</strong></p> <p>a) For my application it would be enough to know the answers in the special case where $E_1/K$ and $E_2/K$ are Galois extensions.</p> <p>b) Somebody told me that the answer to a) would be no'', if we omitted the hypothesis that the algebras $B_i/A$ are etale, even in the case where $K$ is a number field.</p> http://mathoverflow.net/questions/50160/etale-cohomology-analogue-for-the-semistable-reduction-theorem Etale cohomology analogue for the semistable reduction theorem Sebastian Petersen 2010-12-22T14:46:41Z 2010-12-22T14:46:41Z <p>Let $K$ be a field, $X/K$ a smooth projective variety, $l\neq char(K)$ a prime number and $q\ge 0$. Then we define $\overline{X}:=X_{K_{sep}}$ and denote by $\rho_{X, l}^{(q)}$ the representation of $Gal_K$ on $H^q(\overline{X}, \mathbb{Z}_l)$. </p> <p>Note that if $X$ is an abelian variety, then $\rho_{X, l}^{(1)}$ is just the representation of $Gal_K$ on the dual of the Tate module $T_l X$ (and $\rho_{X, l}^{(q)}$ is the representation of $Gal_K$ on $\bigwedge^q T_l(X)^\vee$). </p> <p>Now assume that $S$ is a noetherian integral scheme with function field $K$.</p> <p><strong>Question 1:</strong> Is it true that there is a non-empty open subscheme $U\subset S$ such that $\rho_{X, l}^{(q)}$ factors through $\pi_1(U[1/l])$ for every prime number $l\neq char(K)$?</p> <p><strong>Remark 1:</strong> The answer is yes'', provided $X$ is an abelian variety.</p> <p>From now on assume that $S=Spec(R)$ with a henselian discrete valuation ring $R$. Then $K$ is the quotient field of $R$. Let $k$ be the residue field of $R$ and $p$ the characteristic of $k$. ($p$ is zero or a prime number.) If $L/K$ is a finite separable extension, then we denote by $I_L\subset Gal_L$ the corresponding inertia group. Let $l\neq p$ be a prime number. Denote by $I_{L, l}$ the maximal pro-$l$ quotient of $I_L$. Then $I_{L, l}$ is procyclic and we choose a generator $g_{L, l}$ of $I_{L, l}$. We say that a representation $\rho: Gal_K\to Aut(T)$ of $Gal_K$ on a finitely generated free $\mathbb{Z}_l$-module $T$ is <em>semistable over $L$</em>, if $\rho$ factors through $I_{L, l}$ and $(\rho(g_{L, l})-Id)^2=0$. Still $X/K$ is a smooth projective variety.</p> <p><strong>Question 2:</strong> Is it true that there is a finite Galois extension $L/K$ such that $\rho_{X, l}^{(q)}$ is semistable over $L$ for every prime number $l\neq p$ and every $q\ge 0$? Is it at least known to be true that there is a finite Galois extension $L/K$ such that $\rho_{X, l}^{(q)}(I_L)$ is a pro-$l$ group for every prime number $l\neq p$ and every $q\ge 0$?</p> <p><strong>Remark 2:</strong> Again the answer is yes'' in the special case where $X$ is an abelian variety. (Cf. SGA 7). </p> <p>Any answer or hint on the literature would be very helpful; my own search did not yield much.</p> http://mathoverflow.net/questions/49642/division-fields-of-abelian-varieties-over-function-fields Division fields of abelian varieties over function fields Sebastian Petersen 2010-12-16T13:07:00Z 2010-12-17T09:35:02Z <p>Let $k$ be a finitely generated field (for example a finite field or a number field) and $K/k$ a finitely generated regular extension with $trdeg(K/k)=1$. Let $A/K$ be a principally polarized abelian variety. For every prime number $l\neq char(K)$ let $K_l:=K(A[l])$ be the field obtained by adjoining the coordinates of the $l$-torsion points of $A$ to $K$. For such a function field extension $K_l/K$ it is natural to consider the algebraic closure $k_l$ of $k$ in $K_l$, i.e. the field of elements of $K_l$ which are algebraic over $k$. In brief: I am wondering what we can say about $k_l$.</p> <p>Of course, the existence of the Weil pairing afforded by the principal polarization forces $k(\mu_l)\subset k_l$ for every prime $l\neq char(K)$. This inclusion needs not be an equality. (Consider the case where $A$ has abelian subvarieties defined over $k$.)</p> <p>Let us call an abelian variety $B/K$ <em>weakly isotrivial</em>, if there is a non-zero abelian variety $B_0/\overline{k}$ and a $\overline{K}$-homomorphism $B_{0, \overline{K}}\to A_{\overline{K}}$ with finite kernel.</p> <p><strong>Question:</strong> Suppose that $A$ is not weakly isotrivial. Is it true that there is a constant $l_0(A/K)$ such that $k_l=k(\mu_l)$ for every prime number $l\ge l_0(A/K)$? (If no, is this true after replacing $K$ by a finite extension and $k$ by its algebraic closure in this extension''?) </p> <p><strong>Remarks:</strong> </p> <p>a) For example if $char(K)=0$, $End(A)=\mathbb{Z}$ and $\dim(A)=2, 6$ or odd, then the answer to this question is yes''. The proof we have for that case is somewhat special, however, because then we have exceptionally good control over the associated monodromy groups. (We use that $Gal(K_l/K)=GSp(2\dim(A), \mathbb{F}_l)$ for almost all primes $l$, some group theory and the Mordell-Lang theorem.) </p> <p>b) The special case $\dim(A)=1$ is contained in the book on modular forms by S. Lang.</p> <p>c) It is clear in the situation of the question that there is a constant $l(A/K)$ such that $[K_l:K]>[k_l:k]$ for all primes $l\ge l(A/K)$. Otherwise we would have $K_l=k_l K$ and hence $K_l\subset \overline{k} K$ for infinitely many primes $l$. This would imply $|A(\overline{k}K)[l]|=l^{2\dim(A)}$ for infinitely many primes $l$, which is not the case by the Mordell-Lang theorem. </p> http://mathoverflow.net/questions/46541/how-to-introduce-notions-of-flat-projective-and-free-modules/46627#46627 Answer by Sebastian Petersen for How to introduce notions of flat, projective and free modules? Sebastian Petersen 2010-11-19T13:35:33Z 2010-11-19T13:41:24Z <p>The question reminds me of the time when I was studying mathematics. I had attended a course on algebra (some basic theory of group, rings, categories, loads of Galois theory and some valuation theory, which was the main area of research of the lecturer). Commutative algebra was the first subject I studied on my own, because I wanted to attend a course on algebraic curves, and because I needed some commutative algebra for my diploma thesis. This is not a direct answer to your question, but maybe some thoughts from that time are of use...</p> <p>Literature: As a student I found Bourbaki, Nagata (local rings) and Matsumura (Comm. rings) too difficult as a starting point. I liked the book of Atiyah-MacDonald and I loved the book "Kommutative Algebra" of Brüske, Ischebeck, Vogel. The Brüske-Ischebeck-Vogel book is out of print and available online <a href="http://wwwmath.uni-muenster.de/u/ischebeck/SkriptBrskeIschebeckVogel.pdf" rel="nofollow">http://wwwmath.uni-muenster.de/u/ischebeck/SkriptBrskeIschebeckVogel.pdf</a>. If I had to teach a course on commutative algebra, then I would surely have this book on my desk again. Unfortunately it is written in German, but nevertheless it might be worth to have a look...</p> <p>Free, projective and flat modules: I remember that I needed them for my thesis project and my thesis contained a section summarizing these things. I think at that time I was fine with the account in the BIV book. They define M projective iff $Hom(M, -)$ is exact, flat iff $M\otimes -$ is exact and injective if $Hom(-, M)$ is exact. I found this natural at a first reading, and later on the step to $Ext$ and $Tor$ was quite natural as well. (Maybe I was influenced a bit by the fact that categories and functors were always in the air in Munich at that time and were mentioned in my algebra course.) Having these definitions at hand, one can directly go into the proof of theorems comparing these classes of modules...</p> http://mathoverflow.net/questions/46405/global-spec-and-vector-bundles/46467#46467 Answer by Sebastian Petersen for Global Spec and Vector Bundles Sebastian Petersen 2010-11-18T09:51:38Z 2010-11-18T09:51:38Z <p>Let $L$ be a locally free ${\cal O}_X$-module of finite rank. Define $V=Spec(Sym(L^\vee))$. Then $$Mor_X(X, V)={\cal O}_X-Alg(Sym(L^\vee), {\cal O}_X)=Hom(L^\vee, {\cal O}_X)=L(X).$$ The universal mapping property of the (global) Spec is in EGA II.1.</p> http://mathoverflow.net/questions/44831/flatness-of-the-canonical-projections/44836#44836 Answer by Sebastian Petersen for Flatness of the canonical projections Sebastian Petersen 2010-11-04T15:46:22Z 2010-11-04T15:46:22Z <p>Yes. If $f: X\to S$ is a flat morphism and $S'\to S$ is an arbitrary morphism, then the projection $f_{S'}: X\times_S S'\to S'$ is flat. A reference is EGA IV.2.1.4.</p> http://mathoverflow.net/questions/44082/elliptic-curves-over-finite-fields/44092#44092 Answer by Sebastian Petersen for Elliptic curves over finite fields Sebastian Petersen 2010-10-29T09:19:52Z 2010-10-29T09:19:52Z <p>1) Silverman's book mentioned above is surely a very good reference. For the first steps you can also consider Silverman-Tate "Rational points on elliptic curves". As you already use Hartshorne, I assume that you are familiar with basic notations in algebraic geometry and scheme theory. So you may wish to switch to references on general abelian varieties at some point. (Elliptic curves are abelian varieties of dimension $1$.) For abelian varieties I suggest Milne's article in the Storrs book "Arithmetic geometry" and the wonderful "prebook" on ablian varieties written by Moonen and van der Geer. To my knowledge this prebook is not published yet, but it can be downloaded on the homepage of Ben Moonen.</p> <p>2) This is true, as mentioned above. An alternative definition: An elliptic curve is an abelian variety of dimension $1$. Here an abelian variety over $K$ is a (geometrically) integral proper group scheme over $K$. </p> <p>3) I can give a few basic informations on Mordell-Weil groups: Let $K$ be a field and $E/K$ an elliptic curve. For $n$ coprime to $char(K)$ you have $E(\overline{K})[n]\cong ({\mathbb Z}/n)^2$, hence you know that $E(K)[n]$ is always isomorphic to a subgroup of $({\mathbb Z}/n)^2$. If $char(K)=p>0$, then there is an integer $f\in{0, 1}$ such that $E(\overline{K})[p^i]\cong ({\mathbb Z}/p^i)^f$ (and consequently $E(K)[p^i]$ is isomorphic to a subgroup of $({\mathbb Z}/p^i)^f$) for all $i\ge 1$.</p> <p>If $K$ is finitely generated (over its prime field), then it is known that $E(K)$ is a finitely generated ${\mathbb Z}$-module by the so called Mordell-Weil-Lang-Neron theorem. (Cf. The article of Brian Conrad "Chows $K/k$-trace and $K/k$-image, and the Lang-Neron theorem (via schemes)".)</p> <p>If $K$ is finite with $|K|=q$, then clearly $E(K)$ is finite. In addition to the information above, you then have the Hasse-Weil bound on the size of $E(K)$: $$||E(K)|-q-1|\le 2\sqrt{q}.$$</p> <p>4) I am not sure whether I interpret this question in the right way. But you can of course take your favorite elliptic curve $E$ over ${\mathbb F}_5$, given by an explicit Weierstrass equation, and and use a computer to make a list of the points in $E({\mathbb F}_5)$. (Just check which of the $31$ points in ${\mathbb P}_2({\mathbb F}_5)$ lie on $E$.)</p> http://mathoverflow.net/questions/43778/logarithmic-differentials Logarithmic differentials Sebastian Petersen 2010-10-27T09:27:59Z 2010-10-27T11:42:15Z <p>I have a general und a more special question. I begin with the general one: If $X/k$ is an algebraic scheme over a field $k$ and $D$ is a divisor with normal crossings on it, then there is the so-called sheaf of log-differentials $\Omega_{X/k}(\log(D))$ on it. Is there a good reference for the very basic properties of this kind of differetials? In particuar I am interested in behavior under pullback and analogues to the two exact sequences one has for usual differentials.</p> <p>Now I come to the special question. Let $k=\mathbb F_q$ and $B/k$ a smooth projective curve. Let $F: B\to B$ be the Frobenius morphism. ($F$ is the idendity on the underlying topological space of $B$ and the map ${\cal O}_B(U)\to {\cal O}_B(U)$ induced by $F$ is $x\mapsto x^q$ for every open $U\subset B$.) Let $S=\sum a_P P$ be an effective divisor on $B$. Is it true that $F^* \Omega_{X/k}(\log(S))=\Omega_{X/k}(\log(F^* S))$ (rather then $F^* \Omega_{X/k}(\log(S))=\Omega_{X/k}(\log(S))$)? (Here $F^* S=\sum a_P^q P$.)</p> <p>My interest in these matters arise from an attempt to understand the paper "Purely inseparable points on curves of higher genus" <a href="http://www.mathjournals.org/mrl/1997-004-005/1997-004-005-004.pdf" rel="nofollow">http://www.mathjournals.org/mrl/1997-004-005/1997-004-005-004.pdf</a> I find the results in this paper very interesting. For example they have been applied towards full Mordell-Lang in positive characteristic, and I have other applications in mind. </p> <p>But I am very puzzled. In the proof of the Theorem in this paper, I do not understand why $\Omega_B((F^n)^{-1} S)=\Omega_B(S)$ and why $\deg(P^*\omega)$ should be bounded. Also I do not see a reason why the separable map $g$ occuring later in the proof should be non-constant. </p> <p><strong>Remark:</strong> In this paper the ground field $k$ seems to be an issue that should be discussed a bit. The paper starts with "Let $k$ be a field of characteristic $p>0$". I think the Corollary is false as it stands, at least in the case where $k$ is algebraically closed with $trdeg(k/{\mathbb F}_p)\ge 1$. (In that case, if $K=k(X)$ and $C/K$ is a smooth projective curve of genus $\ge 2$ which is defined over $k$ will have $|C(k)|=\infty$, $|C(K)|=\infty$ and $|C(K^{\frac{1}{p^\infty}})|=\infty$, but of course $C$ needs not be birational to a curve defined over a <em>finite</em> field.) Also there seem to be counterexamples to the Theorem itself, if $k$ is a general field of positive characteristic. These counterexamples disappear, if we assume $k$ finitely generated over its finite prime field, and I think the whole paper was meant to adress that case. So at the moment I try to understand everything in the case where $k$ is finitely generated of positive characteristic. But until now I failed even in the easiest case where $k$ is finite ... </p> <p>EDIT: Maybe I should state here the Claim, whose proof I finally want to understand.</p> <p><strong>Claim:</strong> Let $k$ be a finitely generated field of positive characteristic. Let $K/k$ be a finitely generated field extension with $trdeg(K/k)=1$. Let $C/K$ be a smooth projective curve over $k$ of genus $\ge 2$. Assume that $C$ is not birational over $\overline{K}$ to a curve which is defined over a finite field. Then $C(K^{\frac{1}{p^\infty}})$ is finite. </p> http://mathoverflow.net/questions/43529/decomposition-of-an-algebraic-group-in-an-affine-and-a-proper-part Decomposition of an algebraic group in an affine and a proper part Sebastian Petersen 2010-10-25T15:16:16Z 2010-10-26T05:06:23Z <p>Let $K$ be a perfect field. In what follows, an algebraic group $G/K$ is by definition a group scheme of finite type over $K$. </p> <p>The following seems to be well-known:</p> <p><strong>Theorem:</strong> Let $G/K$ be a connected smooth algebraic group. Then there is a connected smooth <em>affine</em> normal closed subgroup $N$ of $G$, an abelian variety $A/K$ and a homomorphism $G\to A$ with kernel $N$ such that the sequence $$0\to N\to G\to A\to 0$$ is exact for the fppf-topology (say). </p> <p>Can someone give me a proper reference or a hint why this is true? (Checking the literature I find on the one hand plenty of references treating affine algebraic groups, and on the other hand references containing the theory of abelian varieties, but I was surprised not to find a reference containing a proof of this Theorem about the "mixed case".)</p> http://mathoverflow.net/questions/31756/monodromy-groups-of-families-of-abelian-varieties-a-reference-request/43156#43156 Answer by Sebastian Petersen for Monodromy groups of families of abelian varieties: a reference request Sebastian Petersen 2010-10-22T10:00:37Z 2010-10-22T10:00:37Z <p>Part of what you are looking for seems to have been done recently. See Appendix B and Section 4 (especially Thm. 14) of the recent preprint "Expander graphs, gonality and variation of Galois representations" of Ellenberg, Hall and Kowalski. <a href="http://arxiv.org/abs/1008.3675" rel="nofollow">http://arxiv.org/abs/1008.3675</a></p> http://mathoverflow.net/questions/43054/basic-properties-of-neron-models Basic properties of Neron models Sebastian Petersen 2010-10-21T14:51:36Z 2010-10-21T15:04:58Z <p>Let $R$ be a dvr with residue field $k$ and quotient field $K$. Define $S=Spec(R)$. Let $A/K$ be an abelian variety. To my knowledge the Neron model of $A$ is a group scheme ${\cal N}/S$ with generic fibre $A$, which represents the functor $$Y\mapsto Mor_K(Y\times_S Spec(K), A)$$ on the category of smooth $S$-schemes. The morphism ${\cal N}\to S$ is smooth, in particular it is flat and locally of finite type.</p> <p><strong>Question 1.</strong> Is it true that ${\cal N}\to S$ is of finite type?</p> <p>I know that the special fibre ${\cal N}\times_S Spec(k)$ is in general not connected and that the component group of the special fibre is an important invariant.</p> <p>But what about the scheme ${\cal N}$ itself?</p> <p><strong>Question 2.</strong> Is it true that ${\cal N}$ is connected?</p> <p>I strongly assumed that the answer to question 2 is "yes". (My reason to believe this: Assume ${\cal N}$ is not connected. Then ${\cal N}=U\cup V$ for nonempty disjoint open subsets $U$ and $V$ of ${\cal N}$. Then $A=U_K\cup V_K$ and $U_K$, $V_K$ are disjoint open subsets of $A$. Furthermore $U\to S$, $V\to S$ are flat morphisms, hence $U_K$ and $V_K$ are non-empty. This is a contradiction, because $A$ is connected.)</p> <p>However I saw in the book of Bosch-Lutkebohmert-Raynaud examples of non-connected Neron models. And I saw in Deligne's articles on Hodge theory the expression "connected Neron model of $A$" (as opposed to "Neron model of $A$"). Hence I am very puzzled...</p> <p>(I have to admit that I did not yet go through the construction of a representing object of the functor above. That is probably the reason why I cannot help myself at the moment.)</p> http://mathoverflow.net/questions/42959/symmetric-products-of-projective-varieties/43043#43043 Answer by Sebastian Petersen for Symmetric products of projective varieties Sebastian Petersen 2010-10-21T12:24:49Z 2010-10-21T12:24:49Z <p>To fix ideas, let $K$ be a field and $X/K$ be a seperated $K$-scheme of finite type. Let $G$ be a finite group operating on $X$ via $K$-morphisms. The operation is said to be admissible provided every orbit of $G$ is contained in an open affine subset of $X$. If the operation is admissible, then there is a pair $(Y, p)$ consisting of a seperated $K$-scheme $Y$ of finite type and a finite, surjective morphism $p\in Hom(X, Y)^G$, such that the map $$Hom(Y, Z)\to Hom(X, Z)^G,\ f\mapsto f\circ p$$ is bijective for all schemes $Z/K$. Then $(Y, p)$ is said to be the quotient of $X$ mod $G$. (Cf. SGA I, V.1 for such constructions.)</p> <p>From now on assume that $X/K$ is projective. Then $X$ is a closed subscheme of $P_n$ and every finite subset of $X$ is contained in an open affine subset of $X$. We see: If a finite group $G$ acts on $X/K$ by $K$-morphisms, then the operation is automatically admissible.</p> <p>Now take a projective $K$-scheme $V/K$. Then $V^n=V\times\cdots\times V$ is projective over $K$ (Segre-embedding) and hence the natural operation of the symmetric group $S_n$ on $V^n$ will be admissible. Consequently the quotient exists in that case. </p> <p>Aside: It can happen that $V^n/S_n$ is non-smooth, even if $V^n$ is smooth over $K$.</p> http://mathoverflow.net/questions/42660/multiplication-by-n-on-commutative-algebraic-groups Multiplication by $n$ on commutative algebraic groups Sebastian Petersen 2010-10-18T15:47:54Z 2010-10-18T18:43:55Z <p>Let $K$ be a field of characteristic zero. Let $G/K$ be a group scheme of finite type. Assume that $G$ is commutative and connected. For a natural number $n$ denote by $n_G: G\to G$ the multiplication by $n$ morphism. Is it true that $n_G$ is surjective with finite kernel?</p> <p>(I know that the answer is yes provided $G$ is an abelian variety. But the proof of this fact makes use of a very ample invertible sheaf on $G$, so it does not carry over to the general case directly.) </p> http://mathoverflow.net/questions/40989/why-is-an-absolute-value-generated-by-a-simple-subvariety-of-a-variety-v-well-beh/41134#41134 Answer by Sebastian Petersen for Why is an absolute value generated by a simple subvariety of a variety V well-behaved? Sebastian Petersen 2010-10-05T12:07:26Z 2010-10-05T12:07:26Z <p>Another approach might be the following:</p> <p>Let $k$ be a field and $V/k$ a variety, i.e. a seperated $k$-scheme<br> of finite type which is geometrically integral. Let $F$ be its function field. Let $x\in V$ be a point whose local ring ${\cal O}_x$ is a d.v.r. Denote by $v$ the discrete valuation of $F$ corresponding to this d.v.r. Let $F_v$ be the completion of $F$ at $v$.</p> <p>For your question the following fact seems to be important. $V$ is an excellent scheme, because it is of finite type over a field. (c.f. EGA IV.2.7.8). This implies that $F_v/F$ is a separable field extension.</p> <p>Let $E/F$ be a finite (not necessarily separable) field extension. Then the canonical map $E\otimes_F F_v\to \prod_{w/v} E_w$ is bijective. (Note that $E\otimes_F F_v$ is reduced, because $F_v/F$ is separable.)</p> <p>This shows that $v$ is well-behaved (if I understood correctly what well-behaved means).</p> http://mathoverflow.net/questions/40997/fast-computation-of-multiplicative-inverse-modulo-q/41002#41002 Answer by Sebastian Petersen for Fast computation of multiplicative inverse modulo q Sebastian Petersen 2010-10-04T09:41:42Z 2010-10-04T09:41:42Z <p>For arbitrary $q$ (not necessarily prime) the Euclidean algorithm is pretty fast in solving the problem "decide whether the residue class of $a$ is a unit and compute the inverse, if it exists". </p> http://mathoverflow.net/questions/40768/reduced-varieties-with-no-regular-points/40779#40779 Answer by Sebastian Petersen for Reduced varieties with no regular points? Sebastian Petersen 2010-10-01T17:44:45Z 2010-10-01T17:44:45Z <p>I can just add a reference to Angelo's comment. If $X$ is a scheme locally of finite type over a field $k$ (or more generally over a complete local noetherian ring $k$), then $Reg(X)$ is open. Cf. EGA IV.6.12.5 and IV.6.12.8. The results seem to go back to Nagata and Zariski.</p> <p>If $X$ is of finite type over a field $k$ and reduced, then Angelo's argument shows that $Reg(X)$ is non-empty as well. Hence $Reg(X)$ will also contain closed points in that case.</p> http://mathoverflow.net/questions/96932/reed-muller-codes/97646#97646 Comment by Sebastian Petersen Sebastian Petersen 2012-06-04T09:05:25Z 2012-06-04T09:05:25Z Thanks a lot for the detailed answer! http://mathoverflow.net/questions/96932/reed-muller-codes Comment by Sebastian Petersen Sebastian Petersen 2012-05-15T11:42:58Z 2012-05-15T11:42:58Z @ Marc Wildon: This is indeed very interesting for me, especially the information about meeting the Plotkin bound. I will go figure. @ quid, Chris Wuthrich: Thanks for the comments. I agree: It seems plausible to me that simplicity of the coding scheme and availability of an efficient decoding algorithm might have been important reasons for using $R_5$ in Mariner. http://mathoverflow.net/questions/96932/reed-muller-codes Comment by Sebastian Petersen Sebastian Petersen 2012-05-14T19:42:44Z 2012-05-14T19:42:44Z @ Mark Wildon: OK, maybe this was the reason. What seems strange to me is that the codes $R_m$ are said to be especially important because of their good error correcting properties. Their relative minimal distance is the same for all m, while the inf-rate tends rapidly to zero for large m. Is it right to say that they are only important for quite small values of $m$, like $m=3, 4$, while they are bad for large $m$. Still somewhat puzzeled :-) http://mathoverflow.net/questions/96932/reed-muller-codes Comment by Sebastian Petersen Sebastian Petersen 2012-05-14T19:38:44Z 2012-05-14T19:38:44Z @Noam Elkies: In the book they write it was $R_5$, which fits well with the information, also given there, that it could correct 7 errors per code word and was a $[32, 6, 16]$-code. The dual to $R_5$ is equivalent to $R(3, 5)$ which is a $[32, 26, 4]$-code if I compute it right. Thus with much better inf-rate, but poor minimal distance. http://mathoverflow.net/questions/96932/reed-muller-codes Comment by Sebastian Petersen Sebastian Petersen 2012-05-14T19:31:10Z 2012-05-14T19:31:10Z @quid: As far as I understand, every pixel was a number in 0 ... 63, and thus required six bit to store, which fits well with the fact that the dimension of $R_5$ is $6$. So it seems to be one code word per pixel. I am now wondering whether this is the only (simple) reason. On the other hand, if use some other binary code, can always subdivide the (long) bit stream appropriately, so it still sounds strange to me... http://mathoverflow.net/questions/81870/a-good-reduction-property Comment by Sebastian Petersen Sebastian Petersen 2011-12-02T14:19:13Z 2011-12-02T14:19:13Z Dear Ariyan, thanks for the helpful references! There is also the paper of Pop &quot;Henselian implies large&quot; which contains useful information in the first section after the intro. My impression is that the answer is yes. Maybe I write a brief note on this for internal use (i.e. of course not to be published). I this sees the day, I send you a copy. Keep me informed if you have any news. Best, Sebastian http://mathoverflow.net/questions/80669/affine-morphism Comment by Sebastian Petersen Sebastian Petersen 2011-11-25T11:36:21Z 2011-11-25T11:36:21Z This is not well-posed in my opinion. Do you mean that $G$ is to be an affine group scheme and that furthermore $G$ acts on $X$ and that $Y$ is the quotient of $X$ modulo $G$? Maybe you should first edit the question in order to make it precise. http://mathoverflow.net/questions/59555/mathematics-seminar-for-non-mathematicians Comment by Sebastian Petersen Sebastian Petersen 2011-03-26T15:43:16Z 2011-03-26T15:43:16Z @Ottem: To be honest, I never understood what Community Wiki is. But feel free to make it such a Communiki Wiki, if this fits better, or let me know how to do that. http://mathoverflow.net/questions/59555/mathematics-seminar-for-non-mathematicians Comment by Sebastian Petersen Sebastian Petersen 2011-03-26T15:42:02Z 2011-03-26T15:42:02Z @JNR: I have not much to add to the answer of unknown(google). In fact those attending the seminar will have 13 years of school behind them. (But we will soon switch to a new system where one only has to go to school for 12 years before the so-called Abitur.) http://mathoverflow.net/questions/58732/torus-based-cryptography/58792#58792 Comment by Sebastian Petersen Sebastian Petersen 2011-03-19T07:42:01Z 2011-03-19T07:42:01Z Thanks a lot for the precise answer! Best regards, Basti http://mathoverflow.net/questions/58732/torus-based-cryptography Comment by Sebastian Petersen Sebastian Petersen 2011-03-17T12:36:32Z 2011-03-17T12:36:32Z At Franz: One argument is favour of the security of T_2(p) seems to be that it does not lie in a proper subfield of $mathbb{F}_{p^2}$. But still, I do not see why one should expect that it is in fact as secure as $mathbb{F}_{p^2}$. http://mathoverflow.net/questions/58732/torus-based-cryptography Comment by Sebastian Petersen Sebastian Petersen 2011-03-17T12:34:17Z 2011-03-17T12:34:17Z @Gerry: I added a reference. http://mathoverflow.net/questions/57830/picard-group-fundamental-group-and-deformation/57837#57837 Comment by Sebastian Petersen Sebastian Petersen 2011-03-10T13:26:07Z 2011-03-10T13:26:07Z @Theo Johnson-Freyd: I still think, my claim is correct. I added a reference. http://mathoverflow.net/questions/57830/picard-group-fundamental-group-and-deformation/57837#57837 Comment by Sebastian Petersen Sebastian Petersen 2011-03-09T15:18:35Z 2011-03-09T15:18:35Z To be sure: For me $Pic(X)$ is the group of isom. classes of invertible sheaves of $O_X$-modules, with tensor product as multiplication. Why should this be torsion? Take for example $X$ a smooth proper curve over an algebraically closed field $k$. Then $Pic(X)$ can be identified with the divisor class group, and the degree man gives a surjection to $\mathbb{Z}$ with kernel $J_C(k)$. Thus $Pic(X)=\mathbb{Z}\oplus J_C(k)$. This shows it is non-torsion. And, I think, even $J_C(k)$ will have infinite rank as a $\mathbb{Z}$-module, unless $k$ is algebraic over a finite field... What did I miss? http://mathoverflow.net/questions/56984/alterations-factor-as-modification-finite-map/57000#57000 Comment by Sebastian Petersen Sebastian Petersen 2011-03-01T13:52:15Z 2011-03-01T13:52:15Z Ah - it seems we wrote almost in parallel :-)