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Consider $GF(2^8)$ with reducing polynomial $m_p = x^8+x^4+x^2+x^1+x^0$, compute multiplication between $a=x^7+x^0$ and $b=x$.

Following https://en.wikipedia.org/wiki/Finite_field_arithmetic#Multiplication, it seems that the multiplication rule $a \cdot b = (a\cdot b) \mod m_p$. However, if we use this rule, we have $(x^7+x^0)\cdot(x^1) \mod (x^8+x^4+x^2+x^1+x^0)$, this results in $x^8+x^1$, which is beyond the Galois field. Is this multiplication rule wrong? If so, what is the formulated rule for finite field multiplication?

Chong
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    You haven't calculated the remainder of $a\cdot b$ modulo $m_p$ yet. Remainders in polynomial division always have degree lower than the divisor (here $m_p$ of degree eight). Hint: subtract $m_p$ from $x^8+x^0=x^8+1$. – Jyrki Lahtonen Feb 13 '20 at 08:42
  • You may benefit from the sample calculations I did here modulo the polynomials $x^3+x+1$ and $x^4+x+1$ to produce the fields $GF(8)$ and $GF(16)$. I prefer to write the fields using a zero of $m_p$ instead of $x$. The respective zeros there are denoted $\alpha$ and $\gamma$. – Jyrki Lahtonen Feb 13 '20 at 08:48

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As $GF(2^8)$ is a field it satisfies field axioms, meaning that between others, is closed by multiplication.

You've said that multiplication is reduced modulo an irreducible polynomial over $F_2$. This is right. But seems that you failed on performing the operation as $(x^7+1)x \equiv x^4+x^2+1 \pmod{x^8+x^4+x^2+x+1}$. Moreover $x^8+x^4+x^2+x+1$ is not irreducible over $F_2$, so the structure is not $GF(2^8)$ but an univariate polynomial quotient ring $\frac{F_2[X]}{x^8+x^4+x^2+x+1}$.

kub0x
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    +1 for spotting that this octic is not irreducible. I assumed the usual Rijndael polynomial $x^8+x^4+x^3+x+1$ without checking. There was a warning sign, but I missed it. – Jyrki Lahtonen Feb 13 '20 at 08:57