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# Michael Zieve

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 Name Michael Zieve Member for 9 months Seen Jun 16 at 23:49 Website Location University of Michigan Age
 May25 awarded ● Popular Question Mar6 comment Affine automorphisms of algebraic function field towersIt depends what you mean by "lots of rational points". For instance, do you require that the tower should be "asymptotically good", in the sense that the number of rational points should grow like a positive constant times the genus as we move out along the steps of the tower? Mar6 revised Affine automorphisms of algebraic function field towersadded 6 characters in body Mar6 answered Affine automorphisms of algebraic function field towers Mar4 awarded ● Enthusiast Feb27 comment Results and conjectures on bounds on degrees of isogenies@Vesselin: do you know a reference where it was suggested that the bound should be polynomial in the degree? Feb21 answered Generators of symmetric group Feb21 comment When is a cyclic cover hyperelliptic?I put in the $g^n$ because $\mu(\nu(x)^2)$ can have poles; multiplying by $g^n$ for a suitable $g$ will convert all the poles into zeroes of multiplicity between $0$ and $n-1$. Feb21 comment When is a cyclic cover hyperelliptic?...irreducibility means that $\gcd(j,n)=1$ and in addition, if $n$ is even then $\mu(x^2)$ is not a square. Feb21 comment When is a cyclic cover hyperelliptic?Nice answer! A minor reformulation: if the curve $y^n=f(x)$ is hyperelliptic, where $f(x)\in\mathbb{C}[x]$ is a monic polynomial in which every root has multiplicity less than $n$, then $f(x)$ is either $(x-a)^i (x-b)^{n-i}\prod (x-c)^{n/2}$ or $(x-a)^i \prod (x-c)^{n/2}$ or $g(x)^n\cdot\mu(\nu(x)^2)^j$ where $\mu(x),\nu(x)\in\mathbb{C}(x)$ have degree one. Conversely, if $f(x)$ has one of these forms and $y^n=f(x)$ is irreducible then $y^n=f(x)$ is hyperelliptic; irreducibility is equivalent to the conditions $a\ne b$ and $\gcd(i,n/2)$ in the first two cases. In the last case, ... Feb21 comment When is a cyclic cover hyperelliptic?Just to clarify the statement of the question: when $C'$ has genus $0$, you might need to allow some of the $\alpha_i$ to equal one another, and also it can happen that $\infty$ is a branch point of your cyclic cover of $\mathbb{P}^1$. Feb14 answered Decomposition of primes in Galois closures of number fields Feb13 awarded ● Critic Feb13 comment Decomposition of primes in Galois closures of number fields@Tommaso, I agree. Although $Q$ has $f=2$ in the extension $M/L$, the two other primes in the $G$-orbit of $Q$ have $f=1$ in the extension $M/L$. The prime of $L$ which lies under these two primes will have $f=2$ in $L/K$. Feb13 comment Decomposition of primes in Galois closures of number fields@Tommaso, yes that's correct, your group theory question has a positive answer in the cyclic case, and therefore the ramification index in $M/K$ is the lcm of the $e_i$'s if every $e_i$ is coprime to $p$, and also the inertial degree is the lcm of the $f_i$'s if every $e_i=1$. Feb13 comment Decomposition of primes in Galois closures of number fieldsOne place to look for counterexamples to your original question is when $G$ is dihedral of order $8$, $H$ is an order-$2$ subgroup of $G$, and $p$ is totally ramified in $L/K$. I think there will be instances of this setup in which $p$ is totally ramified in $M/K$, and also instances in which $p$ is not. Feb13 comment Decomposition of primes in Galois closures of number fieldsAlso, if $p$ is unramified in $L/K$ then $I=1$, so $D$ is cyclic and hence the inertial degree of $p$ in $M/K$ is the least common multiple of the $f_i$'s. For any $L/K$, this works for all but finitely many $p$. Feb13 comment Decomposition of primes in Galois closures of number fieldsThat's correct if $I$ is cyclic, which happens if $p$ doesn't divide any of the $e_i$'s. But if $p$ does divide some $e_i$ then things can be more complicated. Feb13 comment hyperelliptic curves over finite fieldsA peripheral remark: all the known improvements of Weil's bound in the case of hyperelliptic curves over prime fields are done via variants of Stepanov's method, which is based on the construction of an auxiliary rational function of fairly small degree which vanishes to high order at the rational points. I can't find a statement of Stark's result online, but I do see on page 15 of Stohr-Voloch "Weierstrass points..." that Stark's bound can be deduced via their geometric approach to constructing these auxiliary functions. Feb13 revised hyperelliptic curves over finite fieldsadded 2 characters in body; deleted 8 characters in body Feb13 revised hyperelliptic curves over finite fieldsdeleted 137 characters in body Feb13 comment hyperelliptic curves over finite fieldsFelipe, maybe this is what Stark did, but for prime fields you can improve the estimate provided by the Weil bound (for the genus of a pointless hyperelliptic curve) by a factor of $\sqrt{2}$. This was done by Mitkin 1975 "Existence of rational points...", El Baghdadi 1995, and Zannier 1998 "Polynomials modulo $p$...". Feb13 answered hyperelliptic curves over finite fields Feb12 comment hyperelliptic curves over finite fieldsOne fact about hyperelliptic curves (which distinguishes them from arbitrary curves) is that they have quadratic twists, so that a hyperelliptic curve over F_q has at most 2q+2 rational points. There are several constructions of hyperelliptic curves with no points, as well as results improving the `obvious' bound on the genus of a pointless hyperelliptic curve which follows from the Weil bound. Could you make your question more precise, to clarify what type of situation you have in mind? Feb12 comment Mathematicians whose works were criticized by contemporaries but became widely accepted laterThe question says "Ramanujan's work on divergent series was rejected by three leading English mathematicians". Is that true? Or did they just ignore the letters he sent them? Feb9 comment calculate function from its divizorPresumably the second question is asking for an explicit function $H\colon\mathbb{N}\to\mathbb{N}$ such that: for any $q$ and any elliptic curve $C$ over $\mathbb{F}_q$ as in the question, every rational function $f$ on $C$ can be written as (in your version) $f=f_1(x)+y f_2(x)$ where $f_i\in\mathbb{F}_q(x)$ and $\deg(f_i)\le H(\deg(f))$. Feb7 awarded ● Supporter Feb5 comment How does an irreducible polynomial of prime power order split over an extension of prime power degree@Peter: I was just writing a similar comment. Will the original assertion be true for polynomials of prime degree as long as the degree isn't 7, 11, or 31? Feb5 comment How does an irreducible polynomial of prime power order split over an extension of prime power degreeI'm curious how common the group-theoretic counterexamples will be: for instance, do there exist counterexamples for all but finitely many prime powers $p^n$? Feb5 accepted How does an irreducible polynomial of prime power order split over an extension of prime power degree Feb5 answered How does an irreducible polynomial of prime power order split over an extension of prime power degree Feb5 comment How does an irreducible polynomial of prime power order split over an extension of prime power degreeThe answer to your question is no. Let $f(t)\in L[t]$ be any degree-$7$ polynomial whose Galois group $G$ is the simple group of order $168$. Then $G$ contains two conjugacy classes of index-$7$ subgroups. All groups in one of these conjugacy classes have a fixed point, but the groups in the other class have orbits of sizes 3 and 4. So if $K$ is the subfield of the splitting field of $f$ fixed by one of these index-$7$ subgroups of $G$ from the second class, then $f(t)$ factors in $K[t]$ into irreducibles of degrees $3$ and $4$. Feb5 comment How does an irreducible polynomial of prime power order split over an extension of prime power degreeYou're right, I had in mind the case that $K/L$ is Galois. Your question is equivalent to the following: if $G$ is a transitive subgroup of the symmetric group $S_{p^n}$, and $H$ is a subgroup of $G$ of $p$-power index, then must $H$ have an orbit of $p$-power size? There are probably group theorists who can answer this immediately. To see this equivalence, let $G$ be the Galois group of $f(t)$ over $L$, and let $H$ be the Galois group of $f(t)$ over $K$. Writing $M$ for the splitting field of $f(t)$ over $L$, it follows that $H$ is isomorphic to the Galois group of $M/(M\cap K)$. Feb5 comment How does an irreducible polynomial of prime power order split over an extension of prime power degreeYes. Just think about the degrees $[L(c):L]$ and $[K(c):K]$ where $f(c)=0$. You don't need the hypothesis $[L:K]=p^k$. Jan25 comment product of two non-decomposible (closed) polynomialsIf we take the definition of closed to be (2) with $\deg F > 1$, then we get counterexamples whenever $f$ is closed and $g=f$, since then $fg=F(f)$ with $F=t^2$. Jan11 comment The number of elements of order k in PGL(2, q)@Mart: You can find the details in many standard sources, for instance in Suzuki's group theory book, but of course you would gain the most by trying to work them out on your own. Jan11 awarded ● Commentator Jan11 comment The number of elements of order k in PGL(2, q)You can extract the formula you seek from various sources, for instance from the webpage groupprops.subwiki.org/wiki/… Here is a standard way to get the answer. First compute the size of the conjugacy class of an order-k cyclic subgroup C of PGL(2,q), by computing the normalizer N of C in PGL(2,q) and noting that this size is [PGL(2,q):N]. Then determine the number of conjugacy classes of such subgroups C. Jan9 revised Complex curves covered by smooth plane curvesdeleted 567 characters in body; added 246 characters in body Jan8 comment Can every curve be written as $f(x)=g(y)$?Incidentally, in the literature, curves of the form $g(x)=h(y)$ are called "variables separated curves". Jan8 comment Can every curve be written as $f(x)=g(y)$?@jason: your "inclusions" idea works in the hyperelliptic case, namely, the curve $y^2=f(x)$ is isomorphic to $y^2=f(x)h(x)^2$ for any $h$. It seems conceivable that all examples where $\max(d,e)$ is sufficiently large compared to $g$ might behave similarly to the hyperelliptic ones. Jan7 comment Can every curve be written as $f(x)=g(y)$?@jason: I don't understand how there could be finitely many sequences of multiplicities for a fixed genus. Wouldn't this imply that there are only finitely many possibilities for $d$ and $e$ for a fixed genus? Jan7 comment Complex curves covered by smooth plane curvesReally? Then I take it back. I thought it would be straightforward to get that, if there were a morphism from a smooth plane curve of degree $d$ to a generic genus-$g$ curve, then there is a family of morphisms from a Zariski-open subset of degree-$d$ plane curves to genus-$g$ curves. If that's not true, then I was wrong. Jan6 awarded ● Teacher Jan6 awarded ● Editor Jan6 revised Complex curves covered by smooth plane curvesadded 375 characters in body; added 51 characters in body Jan6 answered Complex curves covered by smooth plane curves Jan5 awarded ● Nice Question Jan5 awarded ● Scholar Jan5 comment Can every curve be written as $f(x)=g(y)$?Thanks Jason! That's a very nice argument.