I want to know the method he used here to find the gcd of $x^{49}-x$ and $x^6-3$ in $\mathbb{F}_7[x]$. It seems that he is not using Euclidean algorithm but rather uses the reduction by replacing $x^6=3$ in $x^{49}-x$. I want to understand why and how it works. Thanks in advance.
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The first step of the Euclidean algorithm yields $q,r\in\Bbb{F}_7[x]$ such that $$x^{49}-x=q\cdot(x^6-3)+r,$$ with $\deg r<6$, and $\gcd(x^{49}-x,x^6-3)$ divides $r$. Reducing mod $x^6-3$ then shows that $$r\equiv x^{49}-x\pmod{x^6-3}.$$ Of course we can reduce $x^{49}-x$ mod $x^6-3$ by replacing $x^6$ with $3$, yielding $$r\equiv x^{49}-x\equiv(x^6)^8\cdot x-x\equiv x\pmod{x^6-3}.$$ As $\deg r<6$ this shows that $r=x$. It follows that $\gcd(x^{49}-x,x^6-3)$ divides $x$, from which it quickly follows that the $\gcd$ equals $1$.
J. W. Tanner
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Servaes
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In the last equality, why $(x^6)^8\cdot x-x\equiv x$? – one potato two potato Oct 08 '20 at 14:59
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@love_sodam I have omitted some intermediate steps; plug in $x^6=3$ and simplify. – Servaes Oct 08 '20 at 15:00
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Another argument is that $x^6-3$ is irreducible in $\Bbb F_7[x]$, and $x^{7^2}-x$ splits completely into linear and quadratic factors. In particular, $x^{49}-x$ has no factor irreducible of degree $6$.
Edit: See this question.
Dietrich Burde
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I thought so. Also, is there an easy way to see that $x^6-3$ is irreducible? It is clear that it has no linear factors, but what about quadratic or cubic? – Servaes Oct 08 '20 at 14:46
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