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by deadbeef57
1272 days ago
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That's flat out wrong. Working modulo 4 is not working in a finite field, because 22 = 0 when you work mod 4. When you work modulo a prime, then you are working in a finite field. Working in the finite field of 4 elements is not* the same as working mod 4. Finite fields of size p^n are not the same as (not isomorphic to!) the ring Z/(p^nZ) whenever n > 1. Just because there exists a finite field of size 2^n doesn't mean that any length-n-bit-string that is the output of a hash function lives in a finite field of size 2^n. Adding and multiplying hashes doesn't make any sense. If it does then you have a very weird hash function. Not one of the standard cryptographic hashes. So saying that a hash function maps onto a finite field sounds pretty confused to me. |
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