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Considering three isomorphism finite fields of size 256 A, B and C; where A and B are $GF(2^8)/GF(2)$ constructed with two distinct irreducible polynomials of degree 8, and C is a tower field $GF(((2^2)^2)^2)$.

Having tow matrices M1 and M2, where M1 (respectively M2) maps the members of A (respectively B) to their corresponding member of B (respectively C); does $M3=M2*M1$ maps the members of A to their corresponding member of C?

Any answer would be appreciated.

Khalesi
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  • Up to isomorphism, there is only one field of size $256$. But it does allow non-trivial automorphisms, so what do you mean with "corresponding member"? Also, with respect to which bases do your matrices describe the isomorphisms? – Hagen von Eitzen Aug 29 '17 at 10:18
  • Please refer to https://math.stackexchange.com/questions/2399922/finding-matrix-for-mapping-members-of-isomorphism-finite-fields for more declaration. – Khalesi Aug 29 '17 at 10:36
  • If $T_1$ is a homomorphism (necessarily injective) from $GF(4)$ to $GF(16)$, and $T_2$ is a homomorphism from $GF(16)$ to $GF(256)$, then, yes $T_2\circ T_1$ is a homomorphism from $GF(4)$ to $GF(256)$. And as long as you use the same basis of $GF(16)$ (over $GF(4)$ or over $GF(2)$) when presenting these homomorphism (that are also linear transformations), yes, the composition becomes the matrix product. – Jyrki Lahtonen Aug 29 '17 at 12:52
  • I'm not sure exactly what you are striving to do here. Just in case it helps you, in this answer I describe a simple recursive way of constructing the tower $$GF(2)\subset GF(4)\subset GF(16)\subset GF(256)\subset GF(65536)\subset \cdots.$$ – Jyrki Lahtonen Aug 29 '17 at 12:59
  • @JyrkiLahtonen thanks for your answer. Actually, I want to compute $x^-1$ in $GF(2^8)$ constructed with a specific primitive polynomial and implement this computation in hardware. A common method to deal with the considerable area requirement of inversion computation in $GF(2^8)$ is mapping the members of $GF(2^8)$ to a tower field such as $GF(((2^2)^2)^2)$ or $GF((2^4)^2)$. We map the members into the tower field with a matrix(M2 in the above question) multiplication, compute the inversion, and then coming back to the first $GF(2^8)$ by matrix ($M2^-1$) multiplication. – Khalesi Aug 30 '17 at 05:31
  • Now, I need to know if I have matrices M1 and M2; M2 for mapping the members of $GF(2^8)$ (B in the question) with an specific irreducible polynomial into the tower field (C in the question) and M1 for mapping the members of $GF(2^8)$ (A in the question) with another specific irreducible polynomial into B; can I map the members of A to the tower field C by multiplication in $M2*M1$? – Khalesi Aug 30 '17 at 05:47
  • Ok, so $M_1$ and $M_2$ describe mappings from $GF((2^4)^2)$ to $GF(2^8)$ and from $GF(((2^2)^2)^2)$ to $GF((2^4)^2)$. Not entirely unlike here? As you see from my attempt there, I'm not well versed in this problem area. Anyway, matrix multiplication corresponds to composition of linear transformations, and inverse matrix give the inverse transformation. – Jyrki Lahtonen Aug 30 '17 at 06:13

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