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How to solve a differential equation in polynomial ring (modulo polynomial)

asked 2018-05-03 09:04:36 -0500

anonymous user

Anonymous

updated 2018-05-05 16:36:04 -0500

I'm working in an arbitrary field F=GF(p^m,'a',modulus=f) where f is a irreducible polynomial of degree m in F_p. I then took an irreducible polynomial h of degree k in PolynomialRing(F). Now i am trying to solve the differential equation for phi with degree(phi)= d with d < k fixed

phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1

I tried solving this the regular way by defining phi as a function and using desolve(), but i don't see how to implement the mod h part and how to demand phi to have degree d.

Any help would be greatly appreciated!

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Could you please provide the explicit construction of p, m, F, f, h, k, d ? Then we have something to start with.

tmonteil gravatar imagetmonteil ( 2018-05-05 05:21:32 -0500 )edit

f=x^13+x^4+x^3+x+1, p=2, m=13, F=GF(p^m, name="a", modulus=f), h is a random irreducible polynomial of degree 8 in the polynomial ring of GF, d is the degree of phi, it should be less than or equal to 4

Necoroyals gravatar imageNecoroyals ( 2018-05-05 05:56:36 -0500 )edit

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answered 2018-05-05 15:38:38 -0500

tmonteil gravatar image

updated 2018-05-05 15:42:30 -0500

What is wrong is the following argument ?

The degree of $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$ <="" p="">

Hence, $\phi-\phi' = 0 \mod h$ is equivalent to $\phi-\phi' = 0$. Hence $\phi = 0$ is the only solution to the equation.

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Sorry i checked again and it should be phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1, terribly sorry about the confusion....

Necoroyals gravatar imageNecoroyals ( 2018-05-05 16:35:30 -0500 )edit

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Asked: 2018-05-03 09:04:36 -0500

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Last updated: May 05