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

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.

• 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 $\mathbb{F}_{2^5}$. – Trevor Gunn Nov 21 '18 at 3:14
• Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$ – PerelMan Nov 21 '18 at 9:58
• What is your definition of $\Bbb{F}_4$? Many would define $\Bbb{F}_4$ as the quotient ring $\Bbb{F}_2[X]/(X^2+X+1)$ :-) – Jyrki Lahtonen Nov 21 '18 at 11:25

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 $$4$$.
• I fixed my question thank you. Now $F[X]/\langle S \rangle$ is a field spawned by $1,X$ – PerelMan Nov 21 '18 at 10:00
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 $$\mathbb{F}_2[x]/(x^2+x+1)\cong\mathbb{F}_4$$.