Does two's complement arithmetic produce a field isomorphic to G F ( 2 n ?From what I understand, we have these two isomorphisms:(TC,+) is isomorphic to the cyclic group Z / 2 n Z .(TC,∗) is isomorphic to the multiplicative group of polynomials.If this is correct, can we conclude that two's complement arithmetic produces a finite field isomorphic to G F ( 2 n ?If not, what algebraic structure, if any, does two's complement representation and arithmetic produce? Because there just seems to be something there.

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Does two&#039;s complement arithmetic produce a field isomorphic to   G  F  (      2          n      ?From what I understand, we have these two isomorphisms:(TC,+) is isomorphic to the cyclic group       Z        /        2    n        Z  .(TC,∗) is isomorphic to the multiplicative group of polynomials.If this is correct, can we conclude that two&#039;s complement arithmetic produces a finite field isomorphic to   G  F  (      2          n      ?If not, what algebraic structure, if any, does two&#039;s complement representation and arithmetic produce? Because there just seems to be something there.

iletsa2ym · Accepted Answer

Step 1The additive group is already a problem, since for   G  F  (      2    n    ) it is   (      Z        /    2      Z        )    n  , not       Z        /        2    n        Z  .In fact, it cannot be a field, since it has zero-divisors:       2    k    ⋅      2          n      −      k        =  0.Step 2Two&#039;s complement arithmetic is isomorphic to the quotient in your question,       Z        /        2    n        Z  , but not only as groups, but as rings, that is, with multiplication too. This structure is rarely a field, through (only if   n  =  1).

Does two's complement arithmetic produce a field isomorphic to GF(2^n)?

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