Primitive polynomials of finite fields
there are 2 primitive polynomials which I can make use of to construct $GF(2^3)=GF(8)$:
$p_1(x) = x^3+x+1$
$p_2(x) = x^3+x^2+1$
$GF(8)$ developed with $p_1(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3 = \alpha + 1$
$\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$
$\alpha^5 = \alpha^4 \cdot \alpha = (\alpha^2+\alpha) \cdot \alpha=\alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1$
$\alpha^6 = \alpha^5 \cdot \alpha=(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha+1+\alpha^2+\alpha=\alpha^2+1$
$GF(8)$ developed with $p_2(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3=\alpha^2+1$
$\alpha^4=\alpha \cdot \alpha^3=\alpha \cdot (\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1$
$\alpha^5=\alpha \cdot \alpha^4=\alpha \cdot(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1$
$\alpha^6=\alpha \cdot (\alpha+1)=\alpha^2+\alpha$
So currently allow is claim I intend to add $\alpha^2 + \alpha^3$ in both areas. In area 1 I get $\alpha^2 + \alpha + 1$ and also in area 2 I get $1$. Reproduction coincides in both areas ($\alpha^i \cdot \alpha^j = \alpha^{i+j\bmod(q1)}$. So does it function so, that when some $GF(q)$ is created with various primitive polynomials after that enhancement tables will differ and also reproduction tables will coincide? Or possibly among offered polynomials ($p_1(x), p_2(x)$) is not legitimate to construct area (altough both are primitive)?
The generator $\alpha$ for your area with the first summary can not amount to the generator $\beta$ for your area with the 2nd summary. An isomorphism in between $\mathbb{F}_2(\alpha)$ and also $\mathbb{F}_2(\beta)$ is offered by taking $\alpha \mapsto \beta + 1$ ; you can examine that $\beta + 1$ pleases $p_1$ iff $\beta$ pleases $p_2$.
To extra straight address the inquiries asked:

Yes, as a whole making use of various primitive polynomials will certainly transform the procedures. If one makes use of expressions such as α ^{ i } to describe area components (as is carried out in GAP and also Magma, as an example), after that the reproduction table remains the very same, and also the enhancement table adjustments. Nonetheless, if one makes use of expressions like α ^{2}+α +1 (as is carried out in Macaulay2 and also Maple, as an example), after that enhancement table remains the very same, and also the reproduction table adjustments. Zech logarithms are made use of to successfully transform in between both depictions.

Both of your polynomials p _{1} and also p _{2} are excellent. This is confirmed in Charles Staats is solution.
The scenario is not so various in a less complex context, the area of 5 components, additionally called the integers modulo 5. Whether $\alpha$ is $2$ or $3$, the area is $0,1,\alpha,\alpha^2,\alpha^3$, yet whether $\alpha+\alpha+1=0$ relies on which $\alpha$ you pick.
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