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

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.

• I'm afraid I don't know what TC-arithmetic really is, but it sure looks like the answer is No. The finite field $GF(2^n)$ is an $n$-dimensional vector space over $\Bbb{Z}_2$. Its additive group is isomorphic to bitwise XOR of $n$-bit masks. Its multiplication is more complicated. It is a lot like polynomials with coefficients in $\Bbb{Z}_2$, but modulo a chosen irreducible polynomial. Without an irreducible polynomial you won't get a group. The arithmetic of $GF(2^n)$ most emphatically has nothing to do with arithmetic modulo $2^n$, $x+x=0$ for all $x\in GF(2^n)$ (bitwise XOR!!). – Jyrki Lahtonen Feb 22 '16 at 7:55
• It just seems like two's complement representation and arithmetic produces some kind of algebraic structure. Intuition suggests it must be something close to a finite field. If not, I was wondering what it was then. – Leo Heinsaar Feb 22 '16 at 10:10
• Leo, if your multiplication as polynomials means, among other things, that $0x0002\cdot 0x0002=0x0004$, $0x0002\cdot0x0004=0x0008$, $0x0003\cdot0x0003=0x0005$ et cetera, then it becomes a problem that this multiplication does not mesh at all well with modular integer addition. – Jyrki Lahtonen Feb 22 '16 at 10:17
• Thanks Jyrki. The answer below also makes a good point that TC arithmetic can't be a field itself because it has zero divisors. I just wanted to understand in more detail how close it gets. – Leo Heinsaar Feb 22 '16 at 10:29

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 $n=1$).