The splitting field of $(x^3 + x^2 +1)(x^3 + x +1)(x^2 + x +1)$ over $F_2$

Question

Let f(x) = $(x^3 + x^2 +1)(x^3 + x +1)(x^2 + x +1)$ over $F_2$. I know that the factors of f are irreducible over $F_2$ since they have no root in $F_2$. How can I compute the splitting fields?

Answer 1

The splitting field of an irreducible polynomial of degree $n$ over $F_{p}$ is the field $F_{p^n}$ with $p^n$ elements.

So the splitting field must contain $F_{2^2}$ and $F_{2^{3}}$. Then it is $F_{2^6}$, as $F_{p^h} \subseteq F_{p^k}$ iff $h \mid k$.

Answer 2

I'm piggybacking off of the Andreas' response. Let $\alpha$ be a root of $x^3+x^2+1$. Since the polynomial is irreducible, we know that $[F(\alpha):F]=3$. Since $F(\alpha)$ is a vector space over $F$ with dimension $3$, we have that $F(\alpha)$ has $2^3$ elements. Now let's show that this field actually has roots for the other two polynomials. We use the fact that $1+1=0$ in $F$.

\begin{align*} (\alpha+1)^3+(\alpha+1)+1&=\alpha^3+3\alpha^2+3\alpha+1+\alpha+1+1\\ &=\alpha^3+\alpha^2+1\\ &=0 \end{align*}

You will find that the six new elements of $F(\alpha)$ are roots of the first two polynomials. To show that all the roots of the cubics are in $F(\alpha)$, it's probably easier to note that $(x^3+x^2+1)(x^3+x+1)(x+1)x=x^8-x$! In other words, your cubics do actually split in $F(\alpha)$

You still have no roots of the quadratic; it remains irreducible in $F(\alpha)$. The splitting field of your eighth degree polynomial is $F(\alpha,\beta)$, where $\beta$ is a root of $x^2+x+1$. The degree is $[F(\alpha,\beta):F]=6$.

Answer 3

It is the same as the splitting field of $(x^3+x+1)(x^2+x+1)$.

Indeed if $\omega$ satisfies the equation $x^3+x+1=0$, then $\omega^2+1$ satisfies the equation $x^3+x^2+1=0$ since \begin{align}(1+\omega^2)^3+(1+\omega^2)^2+1&=(1+\omega^2+\omega^4+\omega^6)+(1+\omega^4)+1\\ &=1+\omega^2+\omega^6=1+\omega^2+(1+\omega)^2=0 \end{align} Thus the splitting field is the compositum of the splitting fields of $x^2+x+1$ and of $x^3+x+1$.

Answer 4

Let be $\alpha$ root of polynomial $f(x)=x^3+x+1$ then $\mathbb Z_2[x]/(f(x)) \cong \mathbb Z_2(\alpha) $ then $\{ 1,\alpha,\alpha^2\} $ basis of $\mathbb Z_2(\alpha)$ over $\mathbb Z_2$, Where $$\mathbb Z_2(\alpha)=\{ 0,1,\alpha,\alpha^2,1+\alpha,1+\alpha^2, \alpha+\alpha^2, 1+ \alpha+\alpha^2 \} $$ and $f(x)=(x-\alpha)(x^2+\alpha x+1+\alpha^2 )$ where $\alpha^2, \alpha+\alpha^2 \in \mathbb Z_2(\alpha)$ are roots of $x^2+\alpha x+1+\alpha^2 $ then $\mathbb Z_2(\alpha)$ is splitting field of $x^3+x+1$ over $\mathbb Z_2$.

We note that $1+\alpha,1+\alpha^2,1+\alpha+\alpha^2 $ are roots the polynomial $x^3+x^2+1$, then $\mathbb Z_2(\alpha)$ is splitting field of $x^3+x^2+1$ over $\mathbb Z_2$.

we note that $\mathbb Z_2(\alpha)$ doesnot have root of polunomial $x^2+x+1$ then let be $\beta$ root of polynomial $g(x)=x^2+x+1$ then $\mathbb Z_2[x]/(g(x)) \cong \mathbb Z_2(\beta) $ then $\{ 1,\beta \} $ basis of $\mathbb Z_2(\beta)$ over $\mathbb Z_2$, Where $$\mathbb Z_2(\beta)=\{ 0,1,\beta,1+\beta \} $$ and $g(x)=(x-\beta)(x+\beta+1)$ where $\beta+1\in\mathbb Z_2(\beta)$ root of $(x+\beta+1) $ then $\mathbb Z_2(\beta)$ is splitting field of $x^2+x+1$ over $\mathbb Z_2$.

Finally the splitting field of $(x^3+x^2+1)(x^3+x+1)(x^2+x+1)$ over $\mathbb Z_2$ is $\mathbb Z_2(\alpha,\beta)$.