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Let $S(X) = X^2 +X+1 \in \mathbb{F}_2[X]$

Prove that $\mathbb{F}_2[X]/(S) \cong \mathbb{F}_4 $

What I did:

$\{1, X \}$ is a basis of $\mathbb{F}_2[X]/(S)$ and S is irreducible in $\mathbb{F}_2$ so $ |\mathbb{F}_2[X]/(S)| = 2^2 = 4$

I need help to prove that $\mathbb{F}_2[X]/(S) \cong \mathbb{F}_4 $. Thank you.

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    $\begingroup$ Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $\mathbb{F}_{2^5}$. $\endgroup$ – Trevor Gunn Nov 21 '18 at 3:14
  • $\begingroup$ Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$ $\endgroup$ – PerelMan Nov 21 '18 at 9:58
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    $\begingroup$ What is your definition of $\Bbb{F}_4$? Many would define $\Bbb{F}_4$ as the quotient ring $\Bbb{F}_2[X]/(X^2+X+1)$ :-) $\endgroup$ – Jyrki Lahtonen Nov 21 '18 at 11:25
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What Trevor Gunn said is right. The statement you want to prove is not correct. First you need to show that $S(X) = X^5 + X^2 + 1$ is irreducible over $\mathbb{F}_2$. Then it will follow that $F[X]/\langle S \rangle$ is a field spanned by the elements $1, X, X^2, X^3, X^4$ and is thus isomorphic to $\mathbb{F}_{32}$ (remember that there is exactly one field with 32 elements).

Edit: OP has now changed the polynomial in question to $S(X) = X^2 + X + 1$. In this case the quotient ring $F[X]/\langle S(X) \rangle$ is a field spanned by the elements $1, X$ and thus has cardinality $2^2 = 4$, as mentioned in the question. The identification with $\mathbb{F}_4$ then comes from the fact that, up to isomorphism, there is only one field of cardinality $4$.

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  • $\begingroup$ I fixed my question thank you. Now $F[X]/\langle S \rangle$ is a field spawned by $1,X $ $\endgroup$ – PerelMan Nov 21 '18 at 10:00
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It is easy to see that $x^2+x+1$ is irreducible since it has degree $2$ and it has no root. Hence the quotient ring is necessarily a field. Since the quotient has $4$ elements and a finite field is uniquely determined by the number of its elements, you get $\mathbb{F}_2[x]/(x^2+x+1)\cong\mathbb{F}_4$.

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