bio | website | math.lsa.umich.edu/~zieve |
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location | University of Michigan | |
age | ||
visits | member for | 2 years, 11 months |
seen | yesterday | |
stats | profile views | 1,378 |
I am a professor at the University of Michigan.
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 |
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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. |
Jul 2 |
awarded | Curious |
Jun 29 |
comment |
Relations among the height of algebraic numbers
Related question: mathoverflow.net/questions/64643/… |
Jun 29 |
comment |
Relations among the height of algebraic numbers
This is not possible unless you bound the degrees of a and b as well. The trace of $a+1$ is $\text{tr}(a)+\text{deg}(a)$. |