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How many roots does $\mathbb{F_{16}}$ have in the polynomial $x^4-1$?

What I've done: $x$ is root iff $x^4=1$ iff $x$ has order $4$,$2$ or $1$. However $\mathbb{F_{16}}-\{0\}=\langle \beta \rangle$ (because of the primitive order element) and $\beta$ has order $15$. Therefore the only solution is $x=1$. How do I know the multiplicity of this root?

mathie12
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2 Answers2

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We have $$ (x-1)^4=x^4-4x^3+6x^2-4x+1=x^4-1 $$ (since $\Bbb F_{16}$ has characteristic $2$), so the root has multiplicity $4$.

Arthur
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  • Why does $\mathbb{F_{16}}$ has characteristic $2$? – mathie12 Jun 20 '18 at 22:47
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    @mathie12 Because $16=2^4$, so $\Bbb F_{16}$ must be an extension (the extension) of degree $4$ over $\Bbb F_2$. Alternately, it is finite, so it must have positive characteristic, it's a field, so the characteristic must be prime, and the additive group has order $16$, so by Lagrange's theorem the characteristic (which by definition is the size of the largest cyclic subgroup) must divide $16$. – Arthur Jun 21 '18 at 04:35
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Over $\Bbb F_2$ we have $a^2+b^2=(a+b)^2$ and $a+b=a-b$, so we have $$x^4-1=x^4+1=(x^2+1)^2=(x+1)^4=(x-1)^4$$ So $x=1$ has multiplicity $4$.

Berci
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