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Factoring a irreducible polynomial in $\mathbb F_4\subseteq \mathbb F_{16}$.

Consider $$\mathbb F_{16}\cong \dfrac{\mathbb F_2[x]}{(x^4+x+1)}$$$$=\{a_0+a_1x+a_2x^2+a_3x^3\;|\; a_i=0\;or\;1,\quad x^4+x+1=0\}$$

Now I can find the generator of the cyclic group $$<x>=\mathbb F_{16}^*\cong \mathbb Z_{15}$$ since $x^5\neq 1, x^3\neq 1$

Then further I can find $\mathbb F_4\subseteq \mathbb F_{16}$ again using above cyclic trick that:

$$\mathbb F_{4}=\{0,1,x^5,x^{10}\}\\ =\{0,1,x^2+x,1+x+x^2\}$$

Now I want to find $Aut_{\mathbb F_{4}}(\mathbb F_{16})$ that are automorphisms, fixing $\mathbb F_{4}$

I tried to see $x^2+x=\beta$ then $\mathbb F_{4}=\{0,1,\beta,1+\beta\}$

Now I can further extend this as $$\mathbb F_{16}=\dfrac{\mathbb F_{4}[t]}{(t^2+t+1)}$$

Now to understand automorphism decribed above I need to understand the roots of the extension, which I cannot canonically see.

User not found
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    The automorphism group of $\mathbb F_{16}$ is cyclic, generated by the Frobenius automorphism $F$, which acts as $x\mapsto x^2$. The automorphisms which fix $\mathbb F_4$ are powers of $F^2$, which acts as $x\mapsto x^4$. In particular, $F^2(x^5)=x^{20}=x^5$ since $x^{15}=1$. – Andrew Hubery Oct 19 '22 at 11:40
  • I have heard bunch of these you mention, but I want a more simplier way to calculate. As the old people did, before Frobenius. – User not found Oct 19 '22 at 11:49
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    Looking at the last part of the answer to an old Q&A pair I prepared with referrals like this in mind, if $\gamma$ is the coset $\gamma=x+(x^4+x+1)\in\Bbb{F}_{16}$, and $\beta=x^2+x+(x^4+x+1)=\gamma^2+\gamma$ then, indeed, ${0,1,\beta,\beta+1}$ is the subfield with four elements. – Jyrki Lahtonen Oct 19 '22 at 18:57
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    (cont'd) But, because $\gamma$ is a zero of $t^2+t+\beta$, it follows that $$\Bbb{F}_{16}\simeq \Bbb{F}_4[t]/(t^2+t+\beta).$$ So you are moding out the wrong quadratic. Observe that $\beta$ and $\beta+1$ are the zeros of $t^2+t+1$, so in $\Bbb{F}_4[t]$ we have a factorization $$t^2+t+1=(t+\beta)(t+\beta+1).$$ Therefore $\Bbb{F}_4[t]/(t^2+t+1)$ is not a field. – Jyrki Lahtonen Oct 19 '22 at 19:00
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    And, as @AndrewHubery explained, the non-trivial $\Bbb{F}4$-automorphism of $\Bbb{F}{16}$ maps $\gamma$ to $\gamma^4=\gamma+1$. It is worth observing that $\gamma$ and $\gamma+1$ are exactly the zeros of $t^2+t+\beta$. After all $$(t+\gamma)(t+\gamma+1)=t^2+t[\gamma+(\gamma+1)]+\gamma(\gamma+1)=t^2+t+\beta.$$ – Jyrki Lahtonen Oct 19 '22 at 19:18

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