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.

Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.