# Revision history [back]

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.

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.

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.

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.