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# Creating a matrix that has elements part of a GF

I am currently doing some implementation but I have something that I do not seem to find online and bugged me for a few hours:

e = 48;
K = GF(2^e);
KE = GF(2^(e*2));

A = matrix(KE,3,3);
E11 = 24;
E12 = 59;
E21 = 21;
E23 = 28;
E32 = 29;
E33 = 65;
A[0,0] = 2^E11;
A[0,1] = 2^E12;
A[0,2] = 0;
A[1,0] = 2^E21;
A[1,1] = 0;
A[1,2] = 2^E23;
A[2,0] = 0;
A[2,1] = 2^E32;
A[2,2] = 2^E33;

print A


When I do this, it print 0, but given that I created it in GF(2^(e*2)), I believe it shouldn't. Because of this, when I try to get the inverse of this matrix, which is invertible, I do not get anything. Please let me know if you have any thoughts.

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## 2 Answers

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The field KE = GF(2^(e*2)) has characteristic 2, hence every even number (in particular every nontrivial power of 2) will be equal to zero, which is what you got.

more Did you mean to use the ring Zmod(2^(e*2)) (the integers modulo $2^{2e}$) instead of GF(2^(e*2)) (the finite field of order $2^{2e}$)? They are different things. The matrix is also not invertible over this ring, however.

It can only be invertible in a ring where the determinant over $\mathbb{Z}$,

-1 * 2^81 * 274177 * 67280421310721


is invertible, e.g. modulo primes which do not divide this number.

more

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Asked: 2018-10-26 16:38:11 +0200

Seen: 73 times

Last updated: Oct 26 '18