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Could somebody please show me the answers and how to get to this answer? Find the splitting field of f(x)= $x^3+x^2+1$ in $\mathbb{Z}_2$ Thanks!

gymbvghjkgkjkhgfkl
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Christ Schwabe
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  • Was this not explained to you in an answer to your previous question? – Jyrki Lahtonen Jan 11 '17 at 18:07
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    Anyway, any extension of finite fields is Galois, so if you can show that $f(x)$ is irreducible you are home free. This is because A) if $\alpha$ is one of the zeros, then $\alpha^2$ and $\alpha^4$ are the two other zeros. B) the polynomial has a zero $\alpha=x+I$ in the field $\Bbb{Z}_2[x]/I$, where $I$ is the (maximal) ideal generated by $x^3+x^2+1$. – Jyrki Lahtonen Jan 11 '17 at 18:08
  • Hmm. Are you at all familiar with the fact if $p(x)$ is an irreducible polynomial over a field $K$, then $L=K[x]/\langle p(x)\rangle$ is a field, and $\alpha=x+\langle p(x)\rangle$ is a zero of $p(x)$ in $L$? – Jyrki Lahtonen Jan 11 '17 at 18:12
  • Or are your difficulties possibly related to the fact you are used to working with number fields, where the zeros sort of exist (as complex numbers by virtue of the fundamental theorem of algebra) without you having to "construct" them first? That is not an uncommon problem students face, when getting acquainted with this stuff. – Jyrki Lahtonen Jan 11 '17 at 18:17

2 Answers2

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What do you know about quotient rings? The answer you are looking for is \begin{equation*} \mathbb{Z}_2[X]/(X^3+X^2+1)\cong\mathbb{Z}_2(\alpha) \end{equation*} where $\alpha$ is a root of $x^3+x^2+1$. In other words, the relation $\alpha^3+\alpha^2+1=0$ holds. Your field has the eight elements $0$, $1$, $\alpha$, $\alpha+1$, $\alpha^2$, $\alpha^2+1$, $\alpha^2+\alpha$, and $\alpha^2+\alpha+1$.

user404127
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$f(x)$ is irreducible over $\mathbb{F}_2$ since it is a third-degree polynomial without any root in the given field.
It follows that the splitting field is simply $\mathbb{F}_8 \simeq \mathbb{F}_2[x]/(x^3+x^2+1).$

Jack D'Aurizio
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