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We have got this nice polynomial over $\mathbb{F}_3$:

$$p(x)=x^4+x^3+x-1$$

Write it in this way: $$x^4-1=2x^3+2x$$ an so: $$(x^2+1)(x^2-1)=2x(x^2+1)$$ $$p(x)=(x^2+1)(x^2+x-1)$$ What is the splitting field of $p(x)$? The splitting fields of both the factors are $\mathbb{F}_9$ i.e. $\mathbb{F}_9=\frac{\mathbb{F}_3}{(x^2+1)}=\frac{\mathbb{F}_3}{(x^2+x-1)}$. But the splitting field of $p(x)$ is $\mathbb{F}_n$ with $n=3^2$ or maybe $n=3^{2\cdot2}$?. I have this doubt cause if I construct $\mathbb{F}_9$ respectivelly with the first factor or the second we get two different sets:

In fact computing the powers of $\alpha=\overline{x}$ for the first ideal $(x^2+1)$ using $\alpha^2=2$ we get $$\mathbb{F}_9=\{0,\alpha,2,2\alpha,1,\alpha,2,2\alpha,1\}$$ and for the second $(x^2+x-1)$ using $\alpha^2=2\alpha+1$ we get $$\mathbb{F}_9=\{0,1,\alpha,2\alpha+1,2\alpha+2,2,2\alpha,\alpha+2,\alpha+1\}$$ It seems the first one has got repetitive elements. Does this influence the choice of the splitting field? And how do you explain the strange structure of the first one? $\mathbb{F}_9$ must have 9 elements.

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Note that $\Bbb F_9$ is the splitting field of $X^9-X$, and: $$X^9-X = X(X-1)(X+1)(X^2+1)(X^2+X-1)(X^2-X-1)$$ which contains $X^2+1$ and $X^2+X-1$ as required, so $\Bbb F_9$ suffices.

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An extension field is not just the powers of $\alpha$, it is a vector space over the (base) field with basis the powers of $\alpha$. Suppose we start with the extension using $\alpha$ from $\alpha^2 = 2$. You can immediately read off from that relation that all the element of $\mathbb{F}_9$ are $$ \{0+0\alpha, 0+1\alpha, 0+2\alpha, 1+0\alpha, 1+1\alpha, 1+2\alpha, 2+0\alpha, 2+1\alpha, 2+2\alpha\} \text{,} $$ which is nine elements, as expected.

Are any of those a root of the second factor, $x^2 + x - 1$? If so, you are done extending the field. If not, you have more extending to do.

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  • $\begingroup$ Can't understand how you built $\mathbb{F}_9$ from $\alpha^2=2$ $\endgroup$ – james watt Jul 3 '18 at 16:54
  • $\begingroup$ @GiacomoTabarelli : $\alpha^2 = 2$ is the smallest positive power of $\alpha$ that is in the base field. Consequently, only $\alpha^0 = 1$ and $\alpha^1 = \alpha$ are $\mathbb{F}_3$-independent powers of $\alpha$, so the two of them form a basis of $\mathbb{F}_3(\alpha)$ over $\mathbb{F}_3$. So $\mathbb{F}_3(\alpha) \cong \mathbb{F}_9$ is $\{x \cdot 1 + y \cdot \alpha : x,y \in \mathbb{F}_3\}$. $\endgroup$ – Eric Towers Jul 4 '18 at 0:27
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The splitting field of both polynomials is $\mathbb{F}_9$; but the second polynomial $x^2+x-1$ is primitive, which means exactly that the powers of $\alpha=\overline{x}$ in $\frac{\mathbb{F}_3[x]}{(x^2+x-1)}$ give all elements of $\mathbb{F}_9^*$ (i.e. has order 8), while the first polynomial $x^2+1$ is not primitive, since $\beta=\overline{x}$ has order 4 in $\frac{\mathbb{F}_3[x]}{(x^2+1)}$. In fact it is easy to show that the two roots of the first polynomial in terms of $\alpha$ are the two elements of order 4: $\alpha^2=2\alpha+1$ and $\alpha^6=\alpha+1$.

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