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Could you please provide the steps involved in calculating a multiplication table for GF(8)?

Gerry Myerson
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phoenix
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1 Answers1

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Take an irriducibile polynomial of degree $3$ over the field $\Bbb F_2$ with two elements, e.g. $P(X)=X^3+X+1$. Then you know that $$ \Bbb F_8=\Bbb F_2[X]/(P(X)) $$ and that its 8 elements are represented by the 8 polynomials of degree $\leq2$. Thus you can construct the multiplication of $\Bbb F_8$ simply by multiplying these polynomials and taking the result modulo $P(X)$.

Andrea Mori
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  • Thank you so much. What i do not understand is that in the text book i have the entry for a multiplication of 2(010) and 4(100) as 3 in GF(8). Could you explain in this context? – phoenix Oct 21 '12 at 23:55
  • I am not sure what you (or the textbook) mean with that notation. – Andrea Mori Oct 21 '12 at 23:58
  • Okay maybe i m gonna copy paste a link which shows something similar happening.Here it is : http://www.ee.unb.ca/cgi-bin/tervo/galois3.pl – phoenix Oct 22 '12 at 00:00
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    @phoenix: Looks like "010" represents $0\cdot\tau^2 + 1\cdot\tau + 0\cdot1$, and "100" represents $1\cdot\tau^2 + 0\cdot\tau + 0\cdot1$. – Niel de Beaudrap Oct 22 '12 at 00:46
  • As Andrea says, the elements of the field are (represented by) the polynomials of degree at most 2. Your source chooses a different representation; the polynomial $ax^2+bx+c$ is represented by the bit-string $abc$ (or maybe $cba$, I didn't read closely enough to tell). – Gerry Myerson Oct 22 '12 at 00:47
  • @NieldeBeaudrap...probably because if i multiply the two of them using the terms you said, i get x^3 which is not permissible here since the max degree would be 2. am i right? – phoenix Oct 22 '12 at 01:14
  • @GerryMyerson yes i think you are correct – phoenix Oct 22 '12 at 01:38