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Let $\Phi$ be an extension field of $\Bbb{F}_2$ of extension degree s >1. Let $a(x)$ be a non-zero polynomial with the coefficients in $\Bbb{F}_2$.

(a) Show that if $\beta$ is a root of the polynomial $a(x)$ over $\Phi$, then $\{\beta^2, \beta^4, \beta^8, \beta^{16},···\}$ are all roots of $a(x)$ over $\Phi$.

(b) Show that if $\beta$ is a primitive element in $\Phi$, and $\beta$ is a root of $a(x)$, then the degree of $a(x)$ is at least $s$.

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  • $\begingroup$ Have you covered Frobenius automorphism in class? What do you know about field extensions? In particular about the relation of the degree of a simple extension and the degree of the minimal polynomial of the generator? $\endgroup$ – Jyrki Lahtonen Apr 11 '14 at 7:55
  • $\begingroup$ And what is that $s$ in part (b)? $\endgroup$ – Jyrki Lahtonen Apr 11 '14 at 7:58
  • $\begingroup$ @JyrkiLahtonen Nope they have not taught anything about Frobenius automorphism !! About Fields , I know a bit and also if I have a generator, how to express other elements as power of that element. In part B, s is the degree ,(I edited the typo in the 1st line of the question) $\endgroup$ – kingmakerking Apr 11 '14 at 8:14
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    $\begingroup$ Ok. Probably the rule $(x+y)^2=x^2+y^2$ that holds in all such fields $\Phi$ has been explained at the very least? If $$a(x)=a_0+a_1x+\cdots a_sx^s,$$ then you know that $$0=a(\beta)=a_0+a_1\beta+\cdots a_s\beta^s.$$ What do you get if you square both sides of that equation? Keep in mind that the coefficients $a_i\in\Bbb{F}_2=\{0,1\}$. $\endgroup$ – Jyrki Lahtonen Apr 11 '14 at 8:48
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    $\begingroup$ See this answer for a detailed derivation in the more general case of what Jyrki is suggesting. $\endgroup$ – Dilip Sarwate Apr 12 '14 at 18:50

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