bio | website | math.lsa.umich.edu/~zieve |
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location | University of Michigan | |
age | ||
visits | member for | 2 years, 7 months |
seen | yesterday | |
stats | profile views | 1,318 |
I am a professor at the University of Michigan. In 2013-2014 I am on leave from Michigan, and I am at Shing-Tung Yau's Mathematical Sciences Center attached to Tsinghua University.
Dec 26 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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Relations among the height of algebraic numbers
Related question: mathoverflow.net/questions/64643/… |
Jun 29 |
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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 $\text{tr}(a)+\text{deg}(a)$. |
Jun 18 |
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Quadratic twist of curve defined over finite field
An elementary exposition of twists in the case of curves over finite fields is Proposition 5.2.8 of Stichtenoth's book "Algebraic Function Fields and Codes". The proof there is essentially Chebotarev's field-crossing argument, and makes the proof of the function field analogue of Chebotarev's density theorem seem quite natural, which might be helpful for thinking about the proof in the number field case. |
Jun 5 |
revised |
Lagrange Interpolation and integer polynomials
added 196 characters in body |