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Question: Why is the table of GF(4) look like the one below? I know it has to do with the fact that 4 is composite Let GF(4) = {0,1,B,D}

Addition:
$$ \begin{array}{c|cccc} + & 0& 1& B & D \\ \hline 0& 0 & 1 & B & D \\ 1 & 1 & 0 & D & B \\ B & B & D & 0 & 1 \\ D & D & B & 1 & 0 \end{array} $$

Multiplication:
$$\begin{array}{c|cccc} \cdot & 0 & 1 & B & D \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & B & D \\ B & 0 & B & D & 1 \\ D & 0 & D & 1 & B \end{array}$$

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    I don't understand your question. What do you mean "why does it look like the one below"? Do you mean, why is it not the table of $\mathbb{Z}/4\mathbb{Z}$, but something different?2012-07-19
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    I wrote down some stuff about constructing the multiplication table for $\mathbf F_4$ in [this answer](http://math.stackexchange.com/questions/168135/explicit-construction-of-a-finite-field-with-8-elements/168142#168142). Feel free to ask questions there.2012-07-19
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    I should add that my answer there and Kris' answer here both involve some abstract algebra that you may not know — it's possible to talk about finite fields without knowing anything about polynomial rings, quotient rings, etc. In that case, tomasz's answer is probably what you need.2012-07-19
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    Thank you for the explanation. I have been reading about fields and rings now and can get a basic understanding of the nomenclature. To get to a more basic question, when I deal with GF(m) where m is a prime number then no problems occur when I translate this into my understanding of Modulo(m), things got a bit stranger when I started doing it when m is composite, the case for m=4 above in the question is evident. So the essence here is to understand that I have to find a primitive polynomial of order 2 under GF(2), the lowest one is x^2+x+1 with root x and x+1 and go from there.2012-07-20
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    Why do you use B and D and not 2 and 3?2018-04-20

2 Answers 2