Michael Zieve
Reputation
4,223
Next privilege 5,000 Rep.
Approve tag wiki edits
11 31
Impact
~31k people reached

• 18 helpful flags
• 462 votes cast

# 316 Actions

 Sep 3 revised Does bounded-degree base extension yield Zariski-dense Mordell-Weil group? added 9 characters in body Sep 3 asked Does bounded-degree base extension yield Zariski-dense Mordell-Weil group? Aug 29 awarded Yearling Jul 15 awarded Nice Answer May 26 awarded Nice Question Dec 26 comment Seeking conceptual explanation of these nice bijections on roots of unity Thanks, that's exactly the sort of explanation I was looking for. Incidentally, in order that the map $(x:y:z)\mapsto(x:y)$ should give a bijection $E_{ns}(\mathbb{F}_q)\to\mathbb{P}^1(\mathbb{F}_q)$, we need the $a_i$ to be outside $\mathbb{F}_q$, or equivalently we need $A_1^2+4A_2$ to be a nonsquare in $\mathbb{F}_q$. I think this condition is needed in order for your construction to give the desired bijection $\mu_{q+1}\to\mathbb{P}^1(\mathbb{F}_q)$. Also, not surprisingly my bijections of $\mu_{q+1}$ arise by composing one of your bijections with the inverse of another. Dec 26 accepted Seeking conceptual explanation of these nice bijections on roots of unity Oct 26 comment Inverse problem for zeta functions of curves over finite fields Yes, that is the same constraint Felipe mentioned: the Drinfeld-Vladut result follows from the fact that $\#C(\mathbf{F}_q)\le\#C(\mathbf{F}_{q^n})$ for all $n$. An example of non-occurring polynomials is $P_g(T)=(T-\sqrt{q})^{2g}$ when $q$ is a square and $g>(q+1)/(2\sqrt{q})$, as a corresponding curve over $\mathbf{F}_q$ would have $q+1-2g\sqrt{q}$ rational points, which is negative. It is easy to construct many analogous examples when $g$ is large compared to $q$. Sep 24 awarded Autobiographer Aug 29 awarded Yearling Aug 11 comment Inverse problem for zeta functions of curves over finite fields Even for hyperelliptic curves, we aren't anywhere close to having a conjecture of which zeta functions actually occur. But the asymptotic distribution of these zeta functions is known when $q\gg g$ (Katz-Sarnak) and when $g\gg q$ (Kurlberg-Rudnick and subsequent authors). Aug 11 answered Inverse problem for zeta functions of curves over finite fields Aug 6 comment Existence of roots of high order polynomial over finite fields I doubt that efficient algorithms are known -- for instance Bi, Cheng and Rojas recently published a proof that there is an algorithm running in time $p^{m^2+O(m)}$, which of course is very far from $\text{poly}(m)$. Their paper is arxiv.org/abs/1204.1113 Aug 1 comment Which polynomials define extensions of $k(t)$ unramified at the finite places My point is that any answer to this question must take account of the fact that there are several different generators for a given field extension. For instance, evaluating my polynomial $tx^q+x^{q-1}+1$ at $x+1$ gives $tx^q+x^{q-1}+x^{q-2}+...+x+(t+2)$, which is another polynomial defining an extension of $k(t)$ unramified over finite places, and this polynomial has terms of every degree up to $q$. Aug 1 comment Which polynomials define extensions of $k(t)$ unramified at the finite places @JasonStarr: What's the problem? The two polynomials I wrote down define the same extension of $k(t)$, and hence are ramified over the same places of $k(t)$. For example, if $q$ is a power of $p$ then $f(x)=x^q+x+t$ defines an extension of $k(t)$ unramified over finite places, so also $tx^q+x^{q-1}+1$ defines the same extension, and visibly the latter polynomial has a term of degree not $1$ or divisible by $q$ (unless $q=2$). Jul 31 comment Which polynomials define extensions of $k(t)$ unramified at the finite places Note that there must be examples involving other powers of $x$, since the extension gotten by adjoining a root of $f(x)$ is the same as the extension gotten by adjoining a root of $x^{\text{deg}(f)} f(1/x)$. Jul 22 comment Weil's Riemann Hypothesis for dummies? You might like to look into Stichtenoth's book "Algebraic function fields and codes", which gives a completely self-contained exposition of the Riemann hypothesis for curves and all material leading up to it. It is completely algebraic and non-geometric, but it does get to Weil's Riemann hypothesis very quickly starting from nothing. Jul 22 comment Weil's Riemann Hypothesis for dummies? @Dustin: yes, see my answer. Jul 22 answered Weil's Riemann Hypothesis for dummies? Jul 4 comment The equation $f(x)=f(a^n)^k$ always has a solution in $\mathbb{Q}$ The question was copied verbatim from the website Rodgrigo pointed to. It seems strange that the OP did not mention the question's source, or that a solution was given at that source.