Galois Field GF(4)

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}

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

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? –  Arturo Magidin Jul 19 '12 at 19:56
I wrote down some stuff about constructing the multiplication table for $\mathbf F_4$ in this answer. Feel free to ask questions there. –  Dylan Moreland Jul 19 '12 at 22:13
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. –  Dylan Moreland Jul 19 '12 at 22:54
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. –  azaz104 Jul 20 '12 at 5:52

This is the only way to define operations on a four-element set making it a field (up to a permutation of elements). I will use some general properties of fields in the following.

1. First, multiplication and addition are commutative, which saves us some guesswork (we only need to determine half the tables).
2. Furthermore, there's got to be 0 and 1. Multiplication by 1 and 0 works the same in any field, so that takes cares of two rows in the multiplication table.
3. In any field, elements distinct from zero with multiplication form an abelian group. In case of $\mathbf F_4$, it is a three-element group, and there is only one such group, so it must be the cyclic group of order three, hence $B^2=D$, $BD=1$, $D^2=B$, so we're done with multiplication.
4. The characteristic of a field is always a prime number, so it must be $2$ in case of $\mathbf F_4$, so there must be zeroes on diagonal of additive table.
5. Additive identity and inverse are unique, so $B+1\neq B,1,0$, so it must be $D$, similarly $D+1=B$
6. Knowing the above, we can easily see that $B+D=B+B+1=0+1=1$, so we're done.
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How do we construct $F_4?$ We can interpret it as a quadratic extension of $F_2$ by the roots of the polynomial $X^2 + X + 1$. If $\alpha$ denotes one root of this, then a second root is $1 + \alpha$, and from the knowledge that $1 + 1 = 0$ and $\alpha^2 = \alpha + 1$ we can work out the addition and multiplication tables of $F_4$.

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+1 Adding for OP's benefit that your $\alpha$ is his/her $B$ and his/her $D$ is your $\alpha+1$. Well, actually they can swap roles, if so desired. There's an automorphism for ya! –  Jyrki Lahtonen Jul 19 '12 at 20:12