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The problem statement is as below.

Represent the following formulas with the elements which are of the linear representations of $GF(2^2)$ where the root $\alpha$ satisfies $x^2+x+1=0$ .

$(1)$ $\alpha^4$

$(2)$ $\alpha^2-(\alpha-1)$

$(3)$ $\frac{\alpha}{\alpha^4}$

I have the following formula but I don't know how should I use it.

$GF(2^2)=\{0,1,\alpha,\alpha+1\}$ $,(\alpha^2+\alpha+1)$

Can someone let me know the solutions or some website that explain about the solutions?

user10354138
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2 Answers2

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I will do one example, that isn't on your list, and then that should show you what is meant. I choose $\alpha^2+1$.

Now the elements of the field are $0,1,\alpha,\alpha+1$, so this is equal to one of those four elements. We have the equation $\alpha^2+\alpha+1=0$, so (as we are over a field of characteristic $2$) $\alpha^2=\alpha+1$.

Then $\alpha^2+1=(\alpha+1)+1=\alpha$.

David A. Craven
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  • Can I write the formula to the middle of the formulas as $\alpha^2=-(\alpha+1)=(\alpha+1)$ ? And the answers for the problems are $\alpha,0,1$respectivelty? –  Jul 24 '20 at 09:41
  • You can put - signs anywhere you want because - and + are the same thing in this field. And yes, those are correct. – David A. Craven Jul 24 '20 at 09:54
  • I see.I'm sorry to bother you again.At last problem ,I got $\frac{\alpha}{\alpha}$ in the processes and I normally calculated it as same as an elementary way like $\alpha^1\times\alpha^-1=\alpha^0=1$ .Is is correct to do that?I have no knowledge about a division of the GF. –  Jul 24 '20 at 10:11
  • As long as your element $x$ is non-zero, $x/x=1$ in any field. Finite fields do all of the things you expect them to do, including multiplicative inverses working correctly. – David A. Craven Jul 24 '20 at 10:19
  • I see.Thank you. –  Jul 24 '20 at 10:21
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Well, by hypothesis, $\alpha^2+\alpha+1=0$ and so $\alpha^2=\alpha+1$. From here you can easily answer the questions.

First, $\alpha^4 = \alpha^2\alpha^2 = (\alpha+1)^2 = \alpha^2+1 = (\alpha+1)+1 = \alpha.$

Second, $\alpha^2 - (\alpha-1) = \alpha^2 - \alpha+1$. But $-1=+1$ in the coefficient field $GF(2)$, and so you get $\alpha^2+\alpha+1=0$.

Third, $\alpha/\alpha^4 = \alpha^{-3}$. But $\alpha^3=1$ since $\alpha$ is 3rd root of unity, and so $\alpha^{-3}=1$.

Wuestenfux
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