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To construct $F_4$, why do we take $X^2+X+1$?

I understand that this polynomial is irreducible $F_2$, what does irreducible mean?

And why does $F_2$ come into it?

I see $F_4=\{a+b\omega | a,b \in F_2 \} = \{0,1,\omega, 1+ \omega =\omega^2\}$

Where does $a+b\omega$ come from?

Where do $0,1,\omega, \omega^2$ come from?

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Al jabra
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  • Irreducible means that $f(X) = X^2 + X+ 1$ (viewed as a polynomial with coefficients in $\mathbb{F}_2$) cannot be factored in an interesting way. The reason why $\mathbb{F}_2$ comes into play is because any field with four elements must contain $\mathbb{F}_2$ as a subfield! – D_S Feb 21 '16 at 20:48
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    Possibly the oldest incarnation of this question is this. – Jyrki Lahtonen Feb 21 '16 at 21:32

1 Answers1

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Let $K$ be the field $\mathbb{F}_2$, that is the field with two elements, $0$ and $1$. A field with four elements is the same thing as a field $F$ which contains $K$ (or an isomorphic copy of $K$) with $[F : K] = 2$.

In general, if $f$ is an irreducible polynomial in $K[X]$ with degree $d$, then the quotient ring $K[X]/(f)$ will be a field with $2^d$ elements.

So letting $f = X^2 + X + 1$, the field $F := K[X]/(f)$ is a field with $4$ elements. Here $\omega$ is the image of $X$ in $K[X]/(f)$. Any element in $F$ is by definition a coset $g + (f)$, where $g$ is a polynomial in $K[X]$. By the division algorithm, since $f$ has degree $2$, every coset $g+ (f)$ can be replaced with an equal coset $h + (f)$, where $h$ is a polynomial of degree at most $1$. So if $h = a + b X$, then the image $h+ (f)$ will be $$[a + bX] + (f)= a + b \omega$$

Since $a, b \in K = \{0,1\}$, you get your four elements as $0, 1, \omega,$ and $1+ \omega$.

D_S
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