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I have trouble finding the answer to his problem and I would really appreciate an answer to this problem.

Problem : Find the splitting fields of f(x)*g(x) for f(x) = $x^3+x+1$ and g(x) = $x^3+x^2+1$ both in $\mathbb{Z}_2[x]$

Christ Schwabe
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  • Christin, you can also try to look at the list of elements of the field $\Bbb{Z}_2[x]/\langle x^3+x+1\rangle$. I prepared one for referrals such as this. You will find that (with the notation of that list) the elements $\alpha,\alpha^2,\alpha^4$ are zeros of $f(x)$ and $\alpha^3,\alpha^5,\alpha^6$ are zeros of $g(x)$. Therefore all the zeros of $f(x)g(x)$ are in the field $\Bbb{F}_8=\Bbb{Z}_2[\alpha]$. – Jyrki Lahtonen Jan 11 '17 at 18:23

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Note that if $\alpha$ is a root of $f(x)$, so that $$\alpha^{3} + \alpha + 1 = 0,$$ then clearly $\alpha \ne 0$, so that multiplying by $\alpha^{-3}$ we get $$ 0 = \alpha^{-3} \cdot (\alpha^{3} + \alpha + 1) = 1 + \alpha^{-2} + \alpha^{-3} = (\alpha^{-1})^{3} + (\alpha^{-1})^{2} + 1 = g(\alpha^{-1}). $$

So once you have the splitting field of $f(x)$, this will also be a splitting field for $g(x)$.

There is a rather general fact behind this. If $f(x)$ is an irreducible polynomial over the finite field $F$ of order $q$, and $E/F$ is the splitting field of $f(x)$ over $F$, then $E$ has order $q^{n}$, and all polynomials in $F[x]$ of degree dividing $n$ will split in linear factors in $E[x]$.

Andreas Caranti
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  • If the extension $L$ contains a root $\beta$ of $f(x) \cdot g(x)$, so that $f(\beta) \cdot g(\beta) = 0$, then either $f(\beta) = 0$ (and then $L$ contains a splitting field of $f(x)$) or $g(\beta) = 0$ (and then $L$ contains a splitting field of $g(x)$). – Andreas Caranti Jan 10 '17 at 14:38
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Another of computing: if $\omega$ is a root of $x^3=x+1$ then $1+\omega^2$ is a root of $x^3+x^2+1$.

Indeed, we have $\omega^3=\omega+1$, hence \begin{align} (1+\omega^2)^3&=1+\omega^2+\omega^4+\omega^6=(1+\omega^2)^2+\omega^2+\omega^6\\ &=(1+\omega^2)^2+(\omega+\omega^3)^2=(1+\omega^2)^2+1. \end{align}

Bernard
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