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I'm in $\text{GF}(8) = \text{GF}(2^3)$ and have an irreducbile polynomial $p(x) = x^3 + x + 1$, then $\text{GF}(8) = \mathbb{Z}_2[x]/\langle p(x) \rangle$ .

Now I want to multiply $2$ elements of the field, namely $(x^2 + x + 1)\cdot (x^2 + x + 1)$. I get the result $x + 2$ with the remainder $2x^2 - x - 1$ if I multiply these two elements and then divide the resulting polynomial by the polynomial $p(x)$.

The result of the multiplication should be $x + 1$, but how do I get from $2x^2 - x - 1$ to $x + 1$?

mathematics2x2life
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user122299
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1 Answers1

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Think this way:

To arrive at $GF_8$ we extended the base field $\Bbb Z_2$ by a root (say $\alpha$) of $p(x)=x^3+x+1$.

Observe that $\alpha\in GF_8$ would play the role of $x$, as $x^3+x+1=0$ in the quotient $\Bbb Z_2[x]/(p(x))$ just as $\alpha^3+\alpha+1=0$ (because $\alpha$ is a root 'somewhere out in the world' of $p(x)$).

So the suggestion is to simply write $\alpha$ for $x$ and use the only equation that we know: $$\alpha^3+\alpha+1=0$$ Since we are over $\Bbb Z_2$, this is equivalent to $\alpha^3=\alpha+1$, and observe that $(A+B)^2=A^2+AB+AB+B^2=A^2+B^2$ here, because $X+X=0$ in any vector space over $\Bbb Z_2$ for any element $X$ (and a field containing $\Bbb Z_2$ is obviously a vector field over it).

Multiplying by $\alpha$, we also have $\alpha^4=\alpha^2+\alpha$, and so $$(\alpha^2+\alpha+1)^2=\alpha^4+\alpha^2+1=\alpha^2+\alpha\,+\, \alpha^2\,+\,1=\alpha+1$$

Berci
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