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I need to show that $\mathbb{Z}_2[X]/(X^2+X+1) = \{0, 1, X^2,X+1 \}$.

My own attempt:

Take $f \in \mathbb{Z}_2[X]$. Then dividing by $X^2+X+1$ gives $f = q(X^2+X+1)+r$ for some $q,r\in\mathbb{Z}_2[X]$ and $\deg(r)<2$. Since this implies that $r$ must be a polynomial in $\mathbb{Z}_2[X]$ of degree $-\infty$, $0$ or $1$, I'd say the only possibilities for $r$ should be $0$, $1$, $X$ and $X+1$.

Is, in this field, $X^2 = X$? If so, how do I see this?

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    $\begingroup$ I believe $X^2 = X+1$ in this field: $X^2 + X +1 = 0 \Rightarrow X^2 = -X -1 = X +1$. $\endgroup$
    – SeraPhim
    Commented Aug 6, 2020 at 11:27

1 Answer 1

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You have $X^2=X^2+X+1+X+1$ and thus $X^2\equiv X+1\pmod {X^2+X+1}$ and thus $$\mathbb{Z}_2[X]/(X^2+X+1)=\{0,1,X,X+1\}$$

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  • $\begingroup$ Ah, of course. Thanks! $\endgroup$ Commented Aug 6, 2020 at 11:31
  • $\begingroup$ You can use \pmod instead of typing mod with parenthesis. $\endgroup$
    – Ennar
    Commented Aug 6, 2020 at 11:32
  • $\begingroup$ @Ennar Thank You! I edited $\endgroup$ Commented Aug 6, 2020 at 11:37
  • $\begingroup$ @PaleBlueDot You 're welcome! $\endgroup$ Commented Aug 6, 2020 at 11:37

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