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I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $\mathbb{F}_{2^{a}}$, $1 \leq a \leq 6$.

So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $\mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $\mathbb{F}_2$ which is $X^2 + X + 1$

And if $P$ were reducible over $\mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $\mathbb{F}_{2^4}$, but this is a contradiction as $\mathbb{F}_{p^k}$ is a subfield of $\mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.

Therefore, for $a = 3,5$ i.e. over $\mathbb{F}_{8}$ and $\mathbb{F}_{32}\ P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $\mathbb{F}_{16}$, since a root of this polynomial generates the field.

Now, if we let $\mathbb{F}_4 = \{0,\ 1,\ a,\ a+1\ \vert\ a^2 + a + 1=0\}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $\mathbb{F}_4$

Which leaves irreducibility of $P$ over $\mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!

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  • $\begingroup$ "...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ... $\endgroup$ – DonAntonio Nov 19 '18 at 0:53
  • $\begingroup$ Ah, a silly mistake on my part. Corrected $\endgroup$ – Naweed G. Seldon Nov 19 '18 at 0:55
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$P$ is reducible over $\mathbb F_{64}$, because reducible over the subfield $\mathbb F_4\subset \mathbb F_{64}$ ($2\mid6$).

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  • $\begingroup$ Does it also follow that $P$ has no roots over $\mathbb{F}_{64}$ $\endgroup$ – Naweed G. Seldon Nov 19 '18 at 2:05
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    $\begingroup$ Such a root would generate $\mathbb F_{16}$, right? But $4\not\mid6$. $\endgroup$ – Chris Custer Nov 19 '18 at 2:28
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$\Bbb{F}_{64}\not \supset\Bbb{F}_{16}$ which you know is the splitting field of $P$, but what about $\Bbb{F}_{4096}$? It's Galois over $\Bbb{F}_{64}$ and contains $\Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $\Bbb{F}_{64}$? What does that tell you about $P$ over $\Bbb{F}_{64}$?

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