User syed - MathOverflow

Reference for notation $H^0(C, mK)$

2013-02-11T11:06:53Z

I am reading the draft of "Equations of Riemann Surfaces of Genus 4, 5 and 6 wih Large Automorphism groups" and the author starts using the notation $H^0(C, mK)$ on page 3, without explaining it. As the author is studying the action of the automorphism group of $C$ on $H^0(C, mK)$ and refers to Farkas who studies the action of the automorphism group of $C$ on holomorphic q-differentials, I suspect that $H^0(C, mK)$ is the first cohomology group computed using q-differentials, but I'm not sure.

My question is that if my guess is correct, then how does one even define that cohomology (do we still have $d^m d z^m = 0$ so we have our exact sequence to define the cohomology?). And if I'm totally wrong then please tell me what that notation means.

In any case I would appreciate if someone guides me to a reference so I can read more on this object.

Answer by Syed for Computing the fixed field of an automorphism of a function field

2012-01-20T10:10:21Z

Colin Weir, suggested the following algorithm to solve the problem in non-rational case, I thought for the sake of others who probably have the same question, I'll post it, here:

Suppose that $\sigma$ is an automorphism of $k(x,y)$. Using above theorem we can find a $x^\sigma$ such that $k(x^\sigma) = k(x)^{\sigma}$. Now, we can re-compute $k(x,y)$ as $k(x^\sigma, y)$. Using degree argument, now one can easily prove that $k(x^\sigma)[\textrm{All elementary symmetric polynomials in } \{y, \sigma(y), \sigma^2(y),...,\sigma^{d-1}(y)\}]$ is equal to $k(x^\sigma, y)^\sigma$. In practice, you add these symmetric polynomials one by one, till your tower reaches the desirable degree.

Computing the fixed field of an automorphism of a function field

2012-01-05T19:18:24Z

Let say we have a function field $k(x,y)$ defined by $f(x,y)$ over $k$, with $\sigma \in Aut(k(x,y)/k)$ and. Suppose, I'm not that out of luck, so that either of $\prod \sigma^i(x)$ or $\sum \sigma^i(x)$ (and the same for $y$) don't fall in $k$, let call them $x^\sigma$ and $y^\sigma$. Then I can compute $k(x^\sigma, y^\sigma)$ and it is not hard to see that if $(\deg(x),\deg(y)) = 1$, that would be my fixed field and I can compute it by finding the min poly of one over another.

However, for example in the case of hyperelliptic involution, we have $\sum \sigma^i(y)=0$. Or there are situations that finding such an $x,y$ is not an easy question. For example, in Algebraic Function Fields and Codes of Stichtenoth, Question 6.9, he asks for such an element $t$ in $F_q(x)$ such that $t^{Aut(F_q(x)/F_q)} = t^{PGL(2,q)}$ is not in $F_q$, and I couldn't solve it (So, it's hard for me at least. I can of course use computer algebra for a particular $q$ but this not what the question asks).

So, I was wondering what is a fail-free way of choosing these generators, such that the fixed field algorithm always works (to prevent them from falling into the constant field and have relatively prime degree). If I use all symmetric polynomials of $Order(\sigma)$ variables, is there a guarantee that at least one of them won't let me down?

Or, if is there better, fixed field computation algorithm there, please tell me (the fixed field algorithm for number field doesn't work straight forward because it could be that $k(x) \not \subseteq k(x^\sigma)$ but one can fixed this if they change the underlying rational function field to the latter, under condition that $x^\sigma$ doesn't fall into $k$, which was my problem to begin with).

Long story short, please tell me what is the fixed field algorithm for automorphisms of (global) function field, that normal people use?

Thanks a lot

post scriptum: I ran into this theorem stated in link text with no proof or reference (beside that Dr. Peter Muller suggested it to the authers (whoever he is)),

[knowing that one can embed a group of automorphisms of rational function field into the field] Let $G = {g_1, . . . , g_m} \subseteq K(x)$ be a finite group. Let $P(t) = \prod^{m}_1 (t−g_i) ∈ K(x)[t]$. Then any non–constant coefficient of $P(t)$ generates $F^G$.

Beside the fact that without having the proof it's hard to generalize it to the nonrational case, it also doesn't guarantee that it doesn't happen that all coeffients of $P(t)$ are constant. In any case, I thought it might help the person who's going to help me ;)

postquam post scriptum: I pasted the "fixed field" functions (for number fields) from both Magma and Pari, here:link text. I see that Magma basically is doing the same thing as I guessed, computing lots of symmetric polynomials and adding them to the base field till the relative degree is the size of the subgroup. For PARI, I don't understand what's the significance of "fixedfieldorbits" and "vandermondeinversemod". I thought It might be helpful. They both lack the function to compute the fixed field of a function field.

Follow-up on @paul garrett's proposed solution to Stichtenoth's problem.

If I understood the proposed method to generate the generator of $F_q(x)^{Aut(F_q(x))}$ correctly, following (sage) code should be able to generate it:

kx.<x> = FunctionField(FiniteField(q))
w = GL2q(Matrix([[0,1],[1,0]]))
Ns = [GL2q(Matrix([[1,n],[0,1]])) for n in range(0,q)]

invElm = (x^q - x)^(q-1)
t = invElm;
for n in Ns:
    ninvElm = PGLAction(PGL2GL, GL2PGL.Image(n*w), invElm);
    t += ninvElm

print t

Unfortunately, the result is always $(x^q-x)^{(q-1)}$ because summing up over n*w is always zero. Unless, I chose the wrong set of automorphism to apply (this is image of identity plus sum of images of n*w for n =0,..,q-1).

There is another part to that question (that wasn't hard to solved) before asking for finding t. It's to find the ramification locus of $F_q(x)^{Aut(F_q(x))}$ and to prove that all places of deg 2 are conjugates. It probably helps.

But, anyway, my question is not the Stichtenoth's question, I just brought-up it as an example that my problem isn't trivial.

Answer by Syed for The image of generator under an automorphism of a cyclic function field

2011-12-27T08:21:15Z

OK, finally, I think I got it, but it is not that trivial to simply be omitted from the proof (If I complicated it and there's is a straight forward way to see it please tell me):

We have $\sigma(y)^q \in K(X)$. Expanding $\sigma(y)^q$, we see that every term in the expansion has the form $y^{\sum s_i i}B_i^{s_i}$ such that $\sum s_i = q$ but also we needs that $\sum s_i i \equiv 0 \mod q$ for all terms. These two equations, generate a non-singular homogeneous system in $\mathbb{Z}/q\mathbb{Z}$ and hence the only solution to the system is $s_i \equiv 0$, for all $i$'s. Which means $s_i = k_iq$ but they are non-negative and sum-up to $q$, so the only legitimate term in the expansion is when for one $i = k, s_i = q$ and for $ i \neq k, s_i = 0$.

The image of generator under an automorphism of a cyclic function field

2011-12-27T07:37:03Z

I'm reading the proof of Lemma 4.1 [1] which says:

"Let $F = K(x,y), y^q = f(x)$, where $q$ is a prime different from characteristic of $K$. Let $Z := Gal(F/K(x))$ and we have $Z < G < Aut(F/K)$ Then:

$Z < Z(G) \iff$ $Z$ is normal in $G$ and for all $\sigma \in G$ there exists $0 \neq B_{\sigma} \in K(x)$ with $\sigma(y) = y.B_{\sigma}$."

My problem is just on the second line, so I'm writing everything up to that point:

"Proof: "$\Rightarrow$" Let $\phi: y \mapsto \xi y, x \mapsto x$ with $\xi$ primitive $q$-th root of unity. Then $Z = <\phi>$. Now let $\sigma \in G$, then $\sigma(y) = y^k.B_{\sigma}$, with $k \in \mathbb{Z}, (k,q) =1$ and $0 \neq B_{\sigma} \in K(x)$..." (the rest is routine to show $k=1$)

My problem is that why only one power of $y$ appears in the image of $\sigma(y)$ while all powers of $y$ makes a basis for $F/K(X)$ why not $\sigma(y) = \sum_{i = 0}^{q-1}y^i.B_{i}$.

Thank you very much indeed for helping me.

[1] Rolf Brandt, Über die Automorphismengruppen von algebraischen Funktionenkörpern,Universität-Gesamthochschule Essen, 1988

Which rational subfields are corresponding to the two dimensional subspaces of holomorphic differentials

2011-10-04T17:28:04Z

I implemented the algorithm that Felipe Voloch's suggested in his reply to the question:

http://mathoverflow.net/questions/56693/subfields-of-a-function-field

the algorithm is here:

http://mathoverflow.net/questions/56693/subfields-of-a-function-field/56774#56774

I considered the function field $F/k$ ($k$ of positive characteristic) defined by following genus 4 curve:

Y^4 +(2*x^7+ 4*x + 4)*Y^2+ x^14

When I ask Magma to list all subfields between $k(x)$ and $k(x,Y)$, Magma gives me:

F1: Y^2 + (2*x^7 + 4*x + 4)*Y + x^14 Genus 2

F2: Y^2 + 4*x^15 + 4*x^14  Genus 0

F3: Y^2 + 4*x^21 + 4*x^15 + 4*x^14 Gesus 2

I used Felipe's algorithm in this way: I generated all two dimensional subspaces of the homomorphic differentials of $F$, for each of these subspaces let $[v_1, v_2]$ be a basis, I looked at $k(v_2/v_1)$ (according to the algorithm). What I observed was that all rational subfields of the form $k(v_2/v_1)$ are subfields of either of $F1, F2$ or $F3$.

So my question: while there are infinite rational subfields of $F$ which are not contained in $F1, F2$ or $F3$ why all of $k(v_2/v_1)$ are subfield of these subfields?

So basically I'm asking two questions:

What is the characteristics of the subfields of a function field $F$ which contains all $k(v_2/v_1)$ rational subfields such that $v_2, v_1$ are linearly independent homomorphic differentials?
A mathematical proof that explain this phenomenon.

I checked the above observation for few different $k = \mathbb{F}_{5}, \mathbb{F}_7$ and 11 and I got the same result.

Why do subspaces of the space of Global holomorphic differentials of a function field correspond to its subfields

2011-08-10T22:10:40Z

I'm asking this question as a follow up to the Felipe Voloch's answer to this question:

http://mathoverflow.net/questions/56693/subfields-of-a-function-field

which you can read it here:

http://mathoverflow.net/questions/56693/subfields-of-a-function-field/56774#56774

(I just didn't have much faith in attracting attention to an already-answered question, so I started a new one).

So, my question is that I have no idea why the mentioned algorithm works? Basically how one proves that every subfield of $F$ corresponds to a subspace of the space of Global homomorphic differentials of $F$. Unfortunately, the "(proof left to the reader)", didn't worked out for me.

I would be more than grateful, if you suggest me a reference that helps me to tackles the math behind the algorithm.

Computing places over x in F/K(x)

2010-07-01T22:45:26Z

Let $F$ be a function field of "transcendental degree one" over its full constant field $K$. Let $x \in F \backslash K$. We know the divisor of $(x) = (x) - (1/x)$ in $K(x)$. Could you please give me an algorithm to compute the places over two above places in $F$ and the ramification degrees.

If this setting is too abstract, what if we have $F$ is the field of fraction of $K(x)[y]/f(x,y)$ where $K$ is a finite field, could you show me any algorithm to find places over zero place and infinite place of $x$.

As KConrad suggested, I'm telling you a little about how I got involved with this problem.

Once upon a time when I was a bit younger (and a bit more stupid but not much less than now) I dared to ask Noam Elkies that how I can represent a curve with an equations of different degree than the one I'm given. For example an elliptic curve of degree 5 (you see, it's not only your time that I waste, so don't take it personal). He wrote me something that time I didn't quite understand at the time but today I went back to the email and fortunately I understood almost all of it:

start from your sample curve y^2 + xy + x^3 + 1 = 0 over Z/2Z

and choose any function of degree 5, say z = x*y. Then eliminate y from the equations by computing the resultant with respect to y of y^2 + xy + x^3 + 1 with the equation satisfied by x,y,z, which is here z - x*y. This gives z^2 + x*z = x^2 + x^5 with x,z functions of degree 2 and 5 on the curve.*

Sincerely, --Noam D. Elkies

The only point which wasn't clear for me was "function of degree 5, say z = x*y". So I assumed it means that the degree of the zero divisor or the pole divisor should be 5. Although I checked it with Magma and it was the case, but I felt the need to compute the divisor for function $z$ in $K(x,y)$ myself. So I tried to compute the divisor of $x$ as the first step. Using the "Extensions = Ramified covers" rule of thumb, and looking at $x$ (the coordinate function) as the covering map to $\mathbb{P}^1$, I said that $(x)$ (the function) correspond to point $x - 0$ in $\mathbb{P}^1$ scheme so I put zero instead of $x$ in my equation and I get my two ramified points $y^2 = 1$. But for the places of over place at infinity downstairs $(1/x)$, I couldn't go that far. I changed the variable $1/\theta = x$ and put zero in $\theta$, I'll get 1=0, unless I replace $y$ with something like $\omega/\theta^2$ as well (which I don't see why) to see my ramification at infinity.

Now my question unfolded is: 1. Do you think what I'm doing makes sense and why it doesn't work for the infinite place. 2. Is there an algebraic/arithmetic way to do what I did instead of the geometric approach of covering space that I used, which I suppose would be more algorithmic friendly.

Sorry I think I gave too much of background.

Computing the function field of a curve given as a subvariety of the Jacobian of its cover or merely the degree of the covering

2011-07-19T18:06:49Z

I read following paragraph from:

G. Tamme, Teilkörper höheren Geschlechts eines algebraischen Funktionenkörpers, Arch. Math. 23 (1972), 257--259

Here $C$ is a curve of genus $\ge 2$ and $J$ is the Jacobian of $C$ and $K$ is the function field of $C$:

Es sei $\mu$ ein beliebiger $k$-Endomorphismus von $J$. Dann hat man einen $k$-Morphismus $C \underrightarrow{\hspace{2mm}\phi\hspace{2mm}} J \underrightarrow{\hspace{2mm} \mu \hspace{2mm}} J$ ; das geometrisehe Bild unter diesem $k$-Morphismus ist eine irreduzible vollstäindige Untervarietät $C^{\mu}$ von $J$. Der $k$-Morphismus $C \to C^{\mu}$ induziert einen $k$-Isomorphismus des Funktionenkörpers von $C^{\mu}$ auf einen Teilkörper von $K$, den wir mit $K^{\mu}$ bezeichnen. Es gilt $K^{\mu} = k$ d.u.n.d., wenn $\mu = 0$.

(Disclaimed translation: Let $\mu$ be an arbitrary $k$-endomorphism of $J$. Then one has a $k$-morphism $C \underrightarrow{\hspace{2mm}\phi\hspace{2mm}} J \underrightarrow{\hspace{2mm} \mu \hspace{2mm}} J$; The geometric image under this $k$-morphism is an irreducible complete subvariety $C^\mu$ of $J$. The $k$-morphism $C\to C^\mu$ induces a $k$-Isomorphism of the function field of $C^{\mu}$ to a subfield of $K$, which we denote by $K^\mu$. Note tha $K^\mu=k$ if and only if $\mu = 0$)

Now my question is that if we have $\mu$ as a function that gets a divisor and gives back a divisor in $J$, how can we compute $K^{\mu}$ as a function field over $k$. "Compute" in a sense of computing the defining equation(s) of the function field.

And if it's too much to ask, is there a way to compute the degree of $\mu \circ \phi$ map as a covering of curves?

For example if $\mu$ is simply the pullback of the Norm of a divisor (such that Norm$:K\to k(x))$ then what would be $K^{\mu}$

To be more explicit, for example, let $k := \mathbb{F}_5$, $C := y^2 -x^3-x-2$ and let $\tau^{\ast}: k(x) \hookrightarrow k(C)$, and $Norm:k(C)\to k(x)$. Let $\mu$ be the map induced by $\tau^{\ast} \circ Norm$. How can I compute $k(C)^{\mu}$ in this situation.

A non-trivial example would be $\mu := \tau^\ast \circ \tau_\ast$ such that $\tau^*: k(C') \to k(C)$

such that

$k := \mathbb{F}_5$

$C' := v^2 - u^5 + u + 1$

$C := y^4 + (2x^5 + 2x^2 + 4x+4)y^2+x^{10}+3x^7+x^4$

$\tau(u) = 2*x/(x + 3)$

$\tau(v) = 1/(x^8 + 4*x^7 + 2*x^6 + x^5 + x^4 + 3*x^3 + 3*x^2)*y^3 + $

$(3*x^3 + 2*x^2 + 4)/(x^6 + 3*x^5 + 3*x^4 + 3*x^2)*y$

Thank you very much indeed.

How to ask Magma to compute the induced morphisim on divisor group

2011-07-04T22:33:12Z

Suppose Magma has computed homomorphism $h$ between function fields $F1 \to F2$. Then we have an induced homomorphism $h$ on the divisor group. Now my question is that if there's a better way to compute this homomorphism for $D2 := h(D1)$ than this way which is basically computing the image of the two generators of each place and is very slow.

Ps, Ds := Support(D1);
D2 := Divisor(F2!1);

for i := 1 to #Ps do
  g1,g2 := TwoGenerators(Ps[i]);

  G1 := h(g1);
  G2 := h(g2);

  D2 := D2 + Ds[i]*ZeroDivisor(GCD(Divisor(G1), Divisor(G2)) );

end for;

Thank you very much indeed!

Necessary/Sufficient condition/Algorithm that tells me a function field is a kummer extension

2011-04-19T20:04:20Z

I start my question with an example. Suppose $F/K$ be the function field generated by $x^n - yx^{n-1} - 1 = 0$. It is not a cyclic over K(y), but if I set $t = yx^{n-1}$ then we have $K(x,t) \subset K(x,y)$ and because $y = tx^{p-1}$ we have $K(x,y) = K(x,t)$. So if $K$ has $n$-th root of unity then $F$ is a Kummer extension over $K(t)$.

So, my question is what are the conditions that given $F/K$ defined by equation f(x,y) = 0, of genus $\ge 2$, there exits $t \in F$ such that $F/K(t)$ is Kummer (of degree $n$ for some $n$ such that $(n,\mathrm{char}(K))=1$ ), or is there an algorithm that can answer this question.

I know already one obvious necessary condition that $\mathrm{Aut}(F/K)$ should have a cyclic subgroup of order $n$ and that the co-factor subgroup should be a finite subgroup of $PGL(2, K)$, but is there something more that I should look for?

I think the question (as Felipe said) boils down to computing the genus of $F^{\mathcal{C}}$ for cyclic $\mathcal{C}<\mathrm{Aut}(F/K)$, but is there a way to make sure that the genus of $F^{\mathcal{C}}$ is zero while avoiding the computation of the different (using only the group structure)?

For example using Riemann-Hurwitz and choosing a large $n$ (say $n > \mathrm{Genus}(F))$ we can make sure that the $Genus(F^{\mathcal{C}}) \le 2$. But how can I decide between the 3 possibilities of $0, 1, 2$?

I mean if we are going to compute the different, then my answer will be "try to see if you can build a Kummer extension then your function field" is Kummer"! I would like to look at the group structure and be able to say something about this question (without carrying out a ramification inspection).

In special case, it's to see that which function with even order automorphism group is hyper elliptic. One necessary condition in that case is that the binary subgroup be in the centre (I don't know if it's sufficient). So, I would like to know if there's a similar conditions for Kummer extensions to decide about this problem without referring to the ramification structure.

Thank you very much indeed!

Subfields of a function field

2011-02-26T00:56:20Z

Is there an algorithm for generating (some or all) subfields of a certain genus of a given function field (even a random one,I mean for example generating a random elliptic subfield of a certain given function field). I did a quick search and it seems to me that the problem is heavily treated in the case of cyclic and Hermitian function fields, but I was wondering what do we know in general case. Is there something that I can do in Magma?

On the other hand, do we have an algorithm to check if $F$ is a subfield of $E$, When $F, E$ are function fields (of one variable)? Florian Hess told me that somebody developed such an algorithm using his automorphism algorithm but I don't have much luck finding it.

In order to stick to the tradition, I give a motivation also: Subfields of function fields with a rich automorphism group are subject to cover attack in cryptography when they are not one of those few which are fixed by an automorphism of the cover.

Thank you very much indeed!

G=(16, 13) notation

2011-02-22T06:13:00Z

I hope this is an easy question but no so easy to be wiped out. I have asked few physical people (Profs) here in our department, and nobody has a clue. I'm reading this paper:

The locus of curves with prescribed automorphism group

and it uses following notation to represent a group:

G = (16, 13)

This is at the beginning of page 11. Shaska uses this notation in his other papers but I couldn't find an explanation for the kind of groups he's talking about. Could you please tell me exactly what's the meaning of this bracket notation? It could be a GAP thing but unfortunately I'm not a GAP person.

Thanks a lot.

Computing only the order of Galois group (not the group itself).

2010-06-13T03:38:47Z

My question is related to this one: http://mathoverflow.net/questions/22923/computing-the-galois-group-of-a-polynomial.

I was wondering if there is a faster algorithm just to compute the order of the group rather than the group itself.

Also, has anybody compared the performance of GAP and Magma in computing Galois groups? I just heard Magma is very good at it.

I asked this question because I encounter every so often new bug with Magma's implementation and I wanted to see if I can implement something similar. But at this time I'm just interested in the exponent at the first place. This is the last annoying error that I get for basically any deg 5 poly that has Gal group $S_5$.

k := FiniteField(2);
kx<x> := RationalFunctionField(k);
kxbyb<y> := PolynomialRing(kx);
MinP :=  y^5 + y + x^2 + x;
print GaloisGroup(MinP);

The result is:

Runtime error: too much looping

Which I don't understand what it means (Magma Ver 2.16-8).

To be more clear, my ultimate goal is to check a lot of polynomials and throw out those with $S_n$ Gal group and focus on those which are not such. As you see even an upper bound over the exponent is enough for me.

Answer by Syed for How to compute div(dx)

2010-07-21T06:31:28Z

When I saw that example in Silverman for the first, it didn't understand much out of it either. I'm still not good at it, but I'm trying to solve it using theorems and examples in Silverman. However, It is possible that I don't see some trivial stuff and try to prove them. On the bright side, in the begining, not assuming any thing, has some benefits.

I want to use Proposition II.4.3.d. which says:

$\mathrm{ord}_p(fdx) = \mathrm{ord}_p(f) + \mathrm{ord}_p(x) - 1$

To compute $\mathrm{div}(x)$ We need to check all $P \in C$ and compute $\mathrm{ord}_p(dx)$. If $P = (x_0, y_0) \neq (e_i, 0)$ then $M_P = (x - x_0, y - y_0)$. However, on can see that $x - x_0$ is a uniformizer for $P$, because $x - x_0$ can generate the whole ideal. To see that, it is enough to show that it can generate $y - y_0$. We have:

$(x-x_0)f(x) = (x - e_1)(x - e_2)(x - e_3) - (x_0 - e_1)(x_0 - e_2)(x - e_3)$

For some $f(x) \in K[E]_{(x_0,y_0)}$, because $x_0$ is a root for the right hand side. Now using the curve equation we can write the right hand side as:

$(x-x_0)f(x) = y^2 - y_0^2$

Now because $y + y_0 \not \in (y - y_0, x-x_0)$ so $1/(y+y_0) \in K[E]_{(x_0,y_0)}$, so we can multiply both side by $1/(y+y_0)$ and we get:

$(x-x_0)\frac{f(x)}{y + y_0} = y - y_0$, hence $M_{(x_0, y_0)} = (x-x_0)$ and therefore $\mathrm{ord}_{(x_0, y_0)}(dx) = \mathrm{ord}_{(x_0, y_0)}(d(x-x_0)) = \mathrm{ord}_{(x_0, y_0)}(x-x_0) - 1 = 0$

Using Proposition II.4.3.d.

So we only need to compute $\mathrm{ord}_p(dx)$ for $p = (e_i, 0)$ and $p = \infty$.

We know that $M_{(e_i, 0)} = (y, x - e_i)$ However, clearly $y^2 = (x - e_i)f(x)$ for some $f(x) \in K[E]_{(e_i,0)}^*$. So, $y$ is a unifromizer for $M_{(e_i, 0)}$ and $\mathrm{ord}_{(e_i,0)}(x - e_i) = 2$. Using the same proposition we have: $\mathrm{ord}_{(e_i, 0)}(dx) = \mathrm{ord}_{(e_i, 0)}(d(x-e_i)) = \mathrm{ord}_{(e_i, 0)}(x-e_i) - 1 = 2 -1 =1$

For the point at infinity, We can use $dx = -x^2(d(1/x))$. Now from example II.3.3, We know that $ord_{\infty}(x) = -2$. So we can easily use the proposition again:

$ord_{\infty}(dx) = ord_{\infty}(-x^2d(1/x)) =$ $2ord_{\infty}(x) + ord_{\infty}(1/x) - 1 = 2 (-2) + (2) - 1 = -3$.

Therefore $\mathrm{div}(dx) = (e_1, 0) + (e_2, 0) + (e_3, 0) - 3\infty$

Why do generic polynomials work in reality?

2010-06-17T02:39:04Z

I understand that a generic $G$-polynomial $f(t_1,...,t_n)[X]$ over field $k$ has Galois group $G$ over $k(t_1,...,t_n)$. And basically any $G$ extension of $k$ should be generated by a realization of $f$.(even a bit stronger but that is not the point here).

Now as much as I understand, our motivation for hunting these polynomials is that in real (constructive) life, we would like to plug random elements of $k$ into $t_1,...,t_n$ and get a $G$-extension. However, it's obvious that the definition doesn't guarantee it. For example as a trivial failure, we know that $X^n + t_1X^{n-1} + \cdots + t_n$ is generic for $S_n$, but not all values for $t_1, ..., t_n$ (basically all polynomials) lead to an $S_n$-extension.

So, basically, my question is this: what is the constructive value of the definition of generic polynomial. Is there any (although I know I'm saying nonsense) high probabilistic/statistic success rate in getting a $G$-extension when a random realization is chosen. Is there some kind of definition of "odd" that says those times that we don't get a $G$-extension are somehow odd and not normal?

Comment by Syed

2013-02-27T15:47:45Z

Looking at Eischler Trace Formula in Breuer's thesis, I think you are certainly right.

Comment by Syed

2013-02-13T16:31:43Z

So does it mean that the cochains cosist of these objects en.wikipedia.org/wiki/…? Is the cohomology the usual cohomology or is a cech cohomology? I'll look into his Magama code to see if this definition makes sense.

Comment by Syed

2012-01-12T07:11:05Z

Let not worry about the char p (suppose it is much larger than the automorphism order). As, in our case, we are dealing with the group generated by one automorphism, hence is always cyclic. Now how repn theory can help me to find the generators? (I don't think repns are too helpful when our group is cyclic)

Comment by Syed

2012-01-08T23:03:22Z

Part 1 and 2 of the question were actually to derive x^q -x and (x^q - x)^(q-1), resp.

Comment by Syed

2012-01-08T23:02:08Z

What if the function field is not rational? I see that if the (deg(x), deg(y)) = 1 then I can generalize it to non-rational case. But the question, now is that how to find such generators (without computing the RR spaces, because that seems to me too much for such a small thing)

Comment by Syed

2011-11-14T22:06:39Z

@Will-sawin, Me neither. I was desperate so I just tried different silly stuff, so hopefully something illuminating might happen. Later, I found out the reason of that phenomena: it's because [F:F2] = 2 and hence F is hyperelliptic, so we have that all rational sub field of F up to some degree are all in F2. then deg(v2/v1) <= |(v1)| 2g-2. and hence v2/v1 in F2. But still my question is that if there's anything special about those rational fields or we are just generating random rational function fields, in that case, the question is that what's the attraction of Felipe algorithm then?

Comment by Syed

2011-07-25T21:55:15Z

@Felipe, "something that will factor through the Jacobian of $\mathbb{P}^1$". The answer is easy in this case because, we know all zero degree divisors over $\mathbb{P}^1$ are principal. I just brought it as an example, for testing a general algorithm. I'll replace it with a less toyish example.

Comment by Syed

2011-07-25T21:49:12Z

@Felipe, the function field to be computed is the function field of $\mu \circ \phi(C)$ which is an image of a curve, hence is a curve and has function field of transcendental degree 1. My guess was that $Jac(\mu \circ \phi(C)$ is $\mu(Jac(C))$ and hence $K^\mu$ would the function field I mentioned in the comment for $\mu(Jac(C))$. I changed the title to be more appropriately represent my question.

Comment by Syed

2011-07-25T21:22:24Z

@Felipe, As much as I understood from Silverman, if you have maps between function fields, then we have map between curves. This induces maps $\phi_*$ and $\phi^*$ on the divisors.

Comment by Syed

2011-07-19T21:43:45Z

@Felipe, I made the example explicit. Could you please show me how can I use the generic point technique to compute $K^{\mu}$? Thanx. (My other problem is that I'm not sure if I know the definition of the function field "of" an Ab variety, my guess is the function field whose Jacobian is that variety :? ).

Comment by Syed

2011-05-26T18:27:55Z

@Felipe I read in this paper: arxiv.org/abs/math/0205314 at top of page 2: A curve with large automorphism group always has orbit genus 0. Large means |Aut(C)| > |4(g-1)|.

Comment by Syed

2011-04-20T19:34:31Z

I mean I need Ramification structure, is it possible to just using $Aut(F/K)$ and $G$ we can compute the genus of $F^G$?

Comment by Syed

2011-04-20T19:32:24Z

How is genus of $F^G$ computable if all I have is the automorphism group?

Comment by Syed

2011-02-26T21:18:26Z

@Inkspot I don't get "a general curve will have no morphisms to any other curve of positive genus" part. If $F$ is a subfield of $F$ then there's morphism from the defining curve of $F$ to the defining curve of $E$. Isn't in second chapter of Silverman? By random, I mean given function field $F$. Is there a way to generate a subfield of it, of given genus. As much as I understand you say there's no subfield other than rational subfields (of genus zero)?

Comment by Syed

2011-02-26T21:08:27Z

@Drar, As Felipe said, exactly the problem is that $K(t)$ is not as good as $\mathbb{Q}$ for number fields. This is why automorphism group is richer than the Galois group and why it's harder to be computed. I think the use of Hess's automorphism algorithm is to check all the (non-canonical but isomorphic)possible ways of embedding of $F$ in $E$ and check if it's work but I don't know the detail.