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I got a question about two examples in my studybook about the factorisation of a galois field. I have included a screenshot of both my examples along with some clarification as it's written in Dutch, though it should be mostly understandable.

http://postimg.org/image/fnld4txi5/

In this example I must factorise $x^7-1$ over GF(2) down to irreducable factors. To achieve this, they make $GF(8) = GF(2^3)$. The primitive polynomial is given as $x^3+x+1$ (given to us in a predetermined table). Then $\alpha$ is chosen as a root of this polynomial after which a table is made, listing the different degrees of $\alpha$ as a polynomial. The cyclotomic cosets of mod 7 over GF(2) are determined as indicated on the picture. The part where I've been stuck on for the longest is what follows next: when the minimal polynomials are determined.

I understand that the factors corresponding to the same cyclomotic coset are multiplied by each other, but I don't understand how the results came to be.

$$ \alpha^0 : x-1 = x+1 $$ This I can understand. $x-\alpha^0 = x-1$ and this becomes $x+1$ in GF(2)

$$ (x-\alpha^1)(x-\alpha^2)(x-\alpha^4) = x^3 + x + 1 $$ I understand this is the same as the original primitive polynomial, because of the factor $\alpha^1$, though I do not understand why.

$$ (x-\alpha^3)(x-\alpha^6)(x-\alpha^5) = x^3 + x^2 + 1 $$ I'm completely lost on this one. I don't know how to work these factors out. If I try to multiply the 3 factors, I will have a polynomial with a lot of terms. I can work them out mostly by substituting them to a lower degree with the help of the table in the picture, but in the end the $\alpha$ is still going to be present and it is an unknown value. So how did they find $x^3 + x^2 + 1$?

http://postimg.org/image/jvaokhitx/

This example has similar bearings to the previous one, except I cannot use the same method. Because $x^4-1$ means $n = 4$ and GF(3) means $q^k-1 = 3^1-1 = 2 or 3^2-1 = 8$ and is thus not equal to n.

Factorisation starts off simple, the root has been determined to be 1 and 2. Then from $x^2+1$, $\gamma$ is determined as a root. Cosets get listed under K, but then all of a sudden the example goes and selects $2\gamma$ as the fourth root. That seemingly came out of nowhere. Can anyone elaborate this step to me?

Babyburger
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  • Try reading this detailed answer by Jyrki Lahtonen and see if it resolves the questions you have. – Dilip Sarwate Aug 28 '14 at 02:02
  • @DilipSarwate I'm afraid not. It was a good read, but the part where the minimal polynomials are decided were not included. One thing I noted though, is that the user of that post used addition between the alphas. In my textbook it mentions that only multiplication is allowed and explicitly says it is not meant to be used for addition. It does seem to work out for him though. – Babyburger Aug 28 '14 at 09:52

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To verify that $(x-\alpha^1)(x-\alpha^2)(x-\alpha^4) = x^3 + x + 1$, just multiply out the cubic to get $$(x-\alpha^1)(x-\alpha^2)(x-\alpha^4) = x^3 - x^2(\alpha+\alpha^2+\alpha^4) + x(\alpha^3 + \alpha^5 + \alpha^6) - \alpha^7$$ and then replace $\alpha^3$ by $\alpha+1$, $\alpha^4$ by $\alpha^2+\alpha$, etc. as given in the table in your Dutch book. You will see that the coefficient of $x^2$ works out to be $0$ and the coefficient of $x$ to be $1$ while of course $\alpha^7 = 1 = -1$ and so we get $x^3+x+1$.

Repeat for $(x-\alpha^3)(x-\alpha^6)(x-\alpha^5)$, noting along the way that the coefficient of $x^2$ in this case was already computed in the previous paragraph while the coefficient of $x$ will need to use $\alpha^7 = 1$ to reduce matters to where your table can be used.

Dilip Sarwate
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  • Thank you Dilip! I had tried this before, but I somehow ended up with some alphas that I couldn't scratch away... Perhaps I made a stupid calculation mistake. It seems very convenient that all the alphas can be annihilated this way. Is that a result of the chosen primitive polynomial? – Babyburger Aug 28 '14 at 12:14
  • You began with the given primitive polynomial $x^3+x+1$ and said "Let $\alpha$ denote a root of this polynomial" from which you got the table. Note that you used this polynomial to determine that $\alpha^3 = \alpha+1$ etc. So now you are working out $(x-\alpha)(x-\alpha^2)(x-\alpha^4)$ which is a polynomial that has $\alpha$ as a root, and also has all the conjugates of $\alpha$ as roots. So you know, I hope, that this polynomial must have coefficients in $\mathbb F_2$ and also that the polynomial must be divisible by the minimal polynomial of $\alpha$. – Dilip Sarwate Aug 28 '14 at 13:19