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Given $\mathbb{F}_4 = \{0,1,a,a+1\}$

resulting from $a^2 + a + 1 = 0$ whereupon $a+1$ and $a$ are the roots of $x^2+x+1 \in \mathbb{F}_2[x]$

why does $(a+1)(a+1)=a$ ?

user26857
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Dylan Thomas
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1 Answers1

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Inside $\Bbb F_4$, $2=0$ since the field has characteristic two. Consequently $-1=1$. So $$(a+1)(a+1)=a^2+2a+1=a^2+1=a^1+1-(a^2+a+1)=-a=(-1)a=a.$$

Angina Seng
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    $\mathbb{F}_4$ doesn't have characteristic 4? – Clement C. May 10 '17 at 16:13
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    @ClementC. It does not; the characteristic of a finite field is always prime. – Angina Seng May 10 '17 at 16:14
  • I am even rustier than I thought... Thank you. – Clement C. May 10 '17 at 16:15
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    You should mention you are in $\mathbb{F}_2[a]/(a^2+a+1) \cong \mathbb{F}_4$ @ClementC. – reuns May 10 '17 at 16:21
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    @ClementC. $\mathbb F_4$ is not the integers mod $4$; if it were then it would not be a field since there would be no multiplicative inverse of $2$. If you start with $\mathbb F_2$ and adjoin an "imaginary" root of the equation $x^2+x+1=0,$ you get $\mathbb F_4.\qquad$ – Michael Hardy May 10 '17 at 16:43
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    @MichaelHardy Yes, I realized afterwards... I always think of $\mathbb{F}_n$ as $\mathbb{Z}/n|mathbb{Z}$, hence my confusion. – Clement C. May 10 '17 at 17:52