Acanti's profile - activity

"then using the vector space structure of $K$ and changing the ring of the vectors to a polynomial ring in the same variables over $\mathbb{Q}$" Didn't know I could do that, thanks for that idea!

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}:=\langle k^4\vert k\in K\setminus 0 \rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

Thank you for looking into this for me! As a new user, I'm glad to know it wasn't something ridiculous on my part. Also, thanks for the suggestion to use the polynomial ring. That will get me through what I need to do.

Thanks! This will fix my current problem, but I am still concerned for when I use larger matrices, and/or a different modulus. If anyone has ideas on a really efficient way to do this in general, I would love to hear them.

I want to conjugate a symbolic matrix, Sigma, by a matrix, garbage, over Z/9Z. If I define both matrices as symbolic matrices, I get the right answer. If I define garbage over Z/9Z, I get confusing answers. Can anyone explain my results?

Sigma=matrix(SR,2,[[1+3*A,3*B],[3*C,1+3*D]])
garbage=matrix(SR,2,[[2,1],[2,6]]);garbageinverse=matrix(SR,2,[[6,8],[7,2]])
expand(garbage*Sigma*garbageinverse);(Sigma*garbageinverse)[0,0]*garbage[1,0]+(Sigma*garbageinverse)[1,0]*garbage[1,1]
R=Integers(9)
garbage=matrix(R,2,[[2,1],[2,6]]);garbageinverse=matrix(R,2,[[6,8],[7,2]])
expand(garbage*Sigma*garbageinverse);(Sigma*garbageinverse)[0,0]*garbage[1,0]+(Sigma*garbageinverse)[1,0]*garbage[1,1]

[  36*A + 42*B + 18*C + 21*D + 19    48*A + 12*B + 24*C + 6*D + 18]
[36*A + 42*B + 108*C + 126*D + 54  48*A + 12*B + 144*C + 36*D + 28]
36*A + 42*B + 108*C + 126*D + 54
[        6*B + 3*D + 1 3*A + 3*B + 6*C + 6*D]
[            0*A + 0*D         3*A + 0*D + 1]
6*B + 0*D

I am trying to find all 2x2 matrices $S$ over $Z/9Z$ such that $S^3=I$, where I is the identity matrix. I am currently using the following procedure:

S = matrix(SR, 2, [[a,b],[c,d]]);
S3=S^3
l=solve_mod([S3[0,0]==1,S3[0,1]==0,S3[1,0]==0,S3[1,1]==1], 9);
l

The list of solutions (there are 207) I receive does not include S=[[1,3],[3,1]], for example, which does in fact satisfy $S^3=I$. I am new to Sage, is there something I am missing? How can I get a complete list of solutions?