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

Question

From what I understand, we have these two isomorphisms:

• $(TC, +)$ is isomorphic to the cyclic group $\mathbb{Z}/2^n\mathbb{Z}$.
• $(TC, *)$ 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 $GF(2^{n}$)?

If not, what algebraic structure, if any, does two's complement representation and arithmetic produce? Because there just seems to be something there.

Answer 1

As Jyrki Lahtonen has already said, the additive group is already a problem, since for $GF(2^n)$ it is $(\mathbb{Z}/2\mathbb{Z})^n$, not $\mathbb{Z}/2^n \mathbb{Z}$.

In fact, it cannot be a field, since it has zero-divisors: $2^k \cdot 2^{n-k} = 0$.

Two's complement arithmetic is isomorphic to the quotient in your question, $\mathbb{Z}/2^n \mathbb{Z}$, but not only as groups, but as rings, that is, with multiplication too. This structure is rarely a field, through (only if $n=1$).