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Well, all representations of the field $GF(2^8)$ are isomorphic. What that means is that there is a mapping between one representation of that field to another, where that mapping preserves all field properties. That is, if we had two representations $A$ and $B$, there exists a mapping $M$ from elements of $A$ to elements of $B$ such that, for any two ...


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First there are the "generic" discrete logarithm algorithms like Shanks's "baby step, giant step" and Pollard's $\rho$, which run in $O(\sqrt{L})$ and are thus of exponential complexity (in the size of $L$). Those algorithms work in virtually any group. In the special case of the multiplicative groups of finite fields, we have subexponential algorithms ...


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Also, encrytion algorithms require inverses. Some of the other possible structures do not alwyas have inverses and so decryption is an issue. They could be used in hashing functions however or where only forward encryption is used as in some block chaining modes.


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GF$(2^8)$ or $\mathbb F_{2^8}$ can also be viewed as the vector space $\mathbb F_2^8$ of $8$-bit vectors (or bytes) over GF$(2)$ or $\mathbb F_2$. Suppose $\{\beta_0, \beta_1, \cdots, \beta_7\}$ is a basis of $\mathbb F_2^8$ over $\mathbb F_2$, that is, the sum $$a_0\beta_0 \oplus a_1\beta_1 \oplus \cdots \oplus a_7\beta_7, ~ a_i \in \mathbb F_2$$ equals ...


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You can begin by enumerating all the irreducible polynomials of degree 8. This gives you all the possible fields representations. If I remember Eisenstein criterium is one of the algorithm for testing irreducibility of polynomials All these field are isomorphic to each other.


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(Note: I'm using hexadecimal numbers to denote AES field elements and decimal numbers to denote integers.) First of all, you have to fix a generator of the AES field's multiplicative group. There's quite a lot of them: 0x03 0x05 0x06 0x09 0x0b 0x0e 0x11 0x12 0x13 0x14 0x17 0x18 0x19 0x1a 0x1c 0x1e 0x1f 0x21 0x22 0x23 0x27 0x28 0x2a 0x2c 0x30 0x31 0x3c 0x3e ...


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I'll try to give additional explanation on algebraic number and the link with EC. Let $q=2^{163}$ the finite Field $F_q=GF(q)$, as selected by NIST has some features for doing cryptography. This field has been generated with the irreduccible pentanomial you gave in the table. To understand what the trace of the polynomial is, it corresponds to the ...


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Actually, you don't compute the trace of a polynomial per se, but of a finite-field element, which is expressed as a polynomial-like expresssion with $u$ acting as the indeterminate. ($u$, as you are probably aware, is a root of the generating polynomial $p(t)$.) Mathematically, the trace of $u$ is $$\mathrm{Tr}(u) = u + u^2 + u^4 + \cdots + u^{2^{162}}$$ ...



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