Revision history [back]
 | 1 | initial version |
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.
| | 2 | No.2 Revision |
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.things. The matrix is also not invertible over this ring, however.
| | 3 | No.3 Revision |
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 $\mathbb{Z}$,
2^81 * 274177 * 67280421310721
is invertible, e.g. modulo primes which do not divide this number.
| | 4 | No.4 Revision |
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.
