# Convert multiplication in $GF(2^{128})$ to bitwise AND?

Suppose we have $$GF(2^{128})=F_2[x]/(x^{128}+x^7+x^2+x+1)$$ and $$a,b,c \in GF(2^{128})$$ with $$a*b=c$$, where * is multiplication in $$GF(2^{128})$$. Could we convert them to $$a',b',c'$$ of length 128 bit vectors, such that $$a'\&b'=c'$$, where & is bitwise AND?

I think FFT might work, but I am confused how to do a FFT on a length 128 vector in $$F_2$$.

Could we convert them to $$a',b',c'$$ of length 128 bit vectors, such that $$a'\&b'=c'$$, where & is bitwise AND?
No. There are $$2^{128}$$ elements of $$GF(2^{128})$$, and there are $$2^{128}$$ length 128 bit vectors, and so there must be a 1:1 correspondence.
Now, consider the $$GF(2^{128})$$ element that converts to the all-1 vector; call the element $$A$$, and call the bit vector that is mapped to (namely, the all-1 vector) as $$A'$$.
Now, we know that the $$GF(2^{128})$$ equation $$B \times C = A$$ has at least $$2^{128}-1$$ solutions in $$B, C$$, however when looking that the post-mapping and function, we see that the corresponding operation $$B' \& C' = A'$$ has only one solution (namely, $$B' = C' = A'$$). Hence, we conclude that any such mapping cannot preserve the multiplication operation.
• Thank you, this makes sense. I got a new idea: if we have a*b=c^d in $GF(2^{128})$, is it possible to convert it to a'&b'=c'^d'. The arguments do not follow directly, and I am not sure if this is possible. Sep 28, 2023 at 7:27