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2013-02-27 12:14:13 -0500 asked a question Formatting inequalities display

When solving inequalities of the type $$x^2-1>0$$ I use the code solve(x^2-1<0,x) to obtain the answer [[x > -1, x < 1]].

Question 1: Is there a way to make Sage display this solution as -1<x<1?

Question 2: Is there an already implemented way to change the display of the inequality x>1 to 1<x?

2012-09-12 10:47:28 -0500 received badge  Popular Question (source)
2012-07-11 10:11:24 -0500 answered a question elliptic curves in quartic and standard form

Unfortunately I do not know of any online source, but you can take a look into Cassel's book "Lectures on Elliptic Curves". It will tell you how to go from a quartic to a cubic model of an elliptic curve.

2012-04-29 16:26:22 -0500 asked a question Elliptic curves over function fields

Let $E$ be an elliptic curve over a function field $K=\mathbb{F}_q(t)$.

How do we compute the height pairing matrix for a set of points $P_1,\ldots,P_n\in E(K)$? or the height of a single point?

2012-02-27 16:53:35 -0500 asked a question Working with function field extensions

Let $K=k(t)$, where $k$ is a finite field. Consider a rational function $F(t)\in K$ and a simple finite extension $L=K(u)$. For instance, take $L=k(u,t)$, where $u^5=t$.

My first question is: How do we evaluate $F(u)$?

The following code I am using produces an error (... NotImplementedError)

k.<a>=GF(19^2)
Rk=k['t']
K.<t>=Frac(Rk)
Rx.<x>=PolynomialRing(K)
L.<u>=K.extension(Rx(x^5-t),'u')
F=(t^25+t^5)/(t^5+1)
F(u)
F.subs(t=u)

Notice that in my example $F(u)=(t^5+t)/(t+1)$ is a function of $t$, since $u^5=t$.

My second question is: How would I coerce $F(u)=g(t)$ back into $k(t)$?

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2012-02-11 09:46:02 -0500 marked best answer Linear transformation from polynomials

How is this?

sage: dim=4                                
sage: F = PolynomialRing(QQ, dim,'X')      
sage: I = F.ideal([x*y for x,y in tuples(F.gens(),2)])
sage: pol = [I.reduce(F.random_element()) for i in range(dim)]
sage: pol
[-3*X2, -39/2*X1 - 1/18*X2, -1/10*X0 - 3*X1 - X2 - 1/5*X3, X2 + 6*X3]
sage: matrix(dim,lambda i,j:pol[i].coefficient(F.gen(j)))
[    0     0    -3     0]
[    0 -39/2 -1/18     0]
[-1/10    -3    -1  -1/5]
[    0     0     1     6]
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2012-02-11 09:45:53 -0500 commented answer Linear transformation from polynomials

Thanks. That'll do it.

2012-02-10 19:17:30 -0500 commented question Linear transformation from polynomials

Sorry, I guess I was not too precise. I've edited and hopefully it will make sense now.

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2012-02-10 14:43:44 -0500 asked a question Linear transformation from polynomials

Suppose I have an unspecified list of degree 1 homogeneous polynomials in several variables, say [X1,X2,X3+3X4,X0]. This list will define a linear transformation [X0,X1,X2,X3,X4]|---->[X1,X2,X3+3X4,X0].

A priori I don't know how many variables or polynomials I will have, since they are found depending on some previous parameters. (The way I have done this, the variables are the generators of a polynomial ring V = PolynomialRing(QQ, dim,'X').)

My question is: How can I transform this list of polynomials into a matrix/linear transformation?

I've tried collecting the coefficients, but the .coefficients() does not work really well for multivariable polynomials since it does not "see the zero terms" (at least I don't know how to do that).