Find the order of a extension field.

Question

I have the following exercice:

Let $K$ be the two element field and $P(X)=X^3+X+1\in K[X]$.

1. Show that $P$ is irreductible in $K[X]$.
2. Let $\alpha$ be a root of $P$ in an extension of $K$. Show that $\alpha^2$ is also a root of $P$.
3. Consider the extension $L=K(\alpha)$. What is the order $\vert L \vert$ of $L$.? Find $x_0,x_1,x_2\in K$ s.t. $\alpha^5=x_0+x_1\alpha+x_2\alpha^2$. Same question with $\alpha^{-2}$.
4. What is the splitting field of $P$ over $K$?

I have found the following answers but I am not sure they are correct or formal (especially point 3.):

1. Suppose there exists $Q,R\in K[X]$ s.t. $QR=P$ and $deg(Q)\geq 1$, $deg(R)\geq 1$. Then we have, w.l.o.g., $deg(R)=2$, $deg(Q)=1$. We try to find $a,b,c\in K$ such that $R=X^2+aX+b$ and $Q=X+c$. Then $X^3+X+1=P=QR=X^3+(a+c)X^2+(ac+b)X+bc$. Identifying each power of $X$ we obtain $bc=1$ (in $\mathbb{F}_2$) that force $b=c=1$. $b+ac=1\rightarrow 1+a=1\rightarrow a=0$. And $c+a=0\rightarrow c=0$, which is a contradiction. Therefore, there is no $Q,R$ and $P$ is irreducible in $K[X]$.
2. If $\alpha$ is a root of $P$ in an extension of $K$ then $P(\alpha)=\alpha^3+\alpha+1=0$. Therefore $\alpha^3=\alpha+1$. Then $P(\alpha^2)=\alpha^6+\alpha^2+1=(\alpha+1)^2+\alpha^2+1=1+\alpha^2+\underset{=0}{2\alpha}+\alpha^2+1=0$. And $\alpha^2$ is also a root of $P$.

3. To find $\vert L\vert$, we will find $l$ s.t. $\alpha^l=1$. We list the powers of $\alpha$:

   • $\alpha^3=\alpha +1$.
   • $\alpha^4=\alpha^3\alpha=(\alpha+1)\alpha=\alpha^2+\alpha$.
   • $\alpha^5=\alpha^2\alpha^3=(\alpha+1)\alpha^2=\alpha^3+\alpha^2=\alpha^2+\alpha+1$.
   • To find $x_0,x_1,x_2$ s.t. $\alpha^{-2}=x_0+x_1\alpha+x_2\alpha^2$, we solve $$\alpha+1=\alpha^3=\alpha^5\alpha^{-2}=(\alpha^2+\alpha+1)(\alpha^2x_2+\alpha x_1+x_0)=\alpha^4x_2+\alpha^3(x_1+x_2)+\alpha^2(x_0+x_1+x_2)+\alpha(x_0+x_1)+x_0=(\alpha^2+\alpha)x_2+(\alpha+1)(x_1+x_2)+\alpha^2(x_0+x_1+x_2)+\alpha(x_0+x_1)+x_0=\alpha^2(x_2+x_0+x_1+x_2)+\alpha(x_2+x_1+x_2+x_0+x_1+x_2)+x_0+x_1+x_2=\alpha^2(x_0+x_1)+\alpha(x_0+x_2)+x_0+x_1+x_2$$ We find that $x_0=x_1=0$ and $x_2=1$. $\Rightarrow \alpha^{-2}=\alpha^2$.

With this we can find that $\alpha^5=\alpha^2\alpha^3=\alpha^{-2}\alpha^3=\alpha$. That shows that $\alpha^4=1$. And $l=\vert L\vert = 4$.

4.I would say it is $L=\{ 0,1,\alpha, \alpha^2\}$.

Answer 1

1) $P$ is cubic. If it is reducible, it factors into three linear factors or a quadratic and a linear factor. Either way it has a linear factor. Since $P(0)=P(1)=1$, we find that $P$ has no linear factor and therefore is irreducible.

You forgot the three linear factors option.

2) (... and so $\alpha^2 \not \in K$, either. So we have $K(\alpha)$ is not a quadratic extension ...) (This means the Galois group is not $\mathbb{Z}/2\mathbb{Z}$ and we're about to show it's not as big as $S_3$ in part 3.)

3) You (should have) noticed that $\{1,\alpha, \alpha^2\}$ is a sufficient basis to span all the powers of $\alpha$. This implies every element of $K(\alpha)$ can be written as $k_0 \alpha^0 + k_1 \alpha^1 + k_2 \alpha^2$. Since there are $2 \times 2 \times 2$ choices for the $k_i$, there are only $8$ elements in $K(\alpha)$. I.e., $|L| = 8$.

4) What you have written is not a field. The splitting field would be the one that has eight elements: $\mathbb{F}_{2^3}$.