Exercise 5.1.7

Proof.

(a)

Let $f=x^3-x+1\in {𝔽}_3[x]$.

As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$, to prove the irreducibility of $f$, it is sufficient to show that $f$ has no root in ${𝔽}_3$. This is the case, since every element $\alpha$ of ${𝔽}_3$ is a root of $x^3-x$ (little Fermat’s theorem), so ${\alpha }^3-\alpha +1=1\ne 0$ : $f(0)=f(1)=f(2)=1$.

$f=x^3-x+1$ is irreducible over ${𝔽}_3$.

(b)

Let $L$ the splitting of $f$ over ${𝔽}_3$, and $\alpha$ a root of $f$ in $L$. As the characteristic of ${𝔽}_3$ is 3, $$\begin{array}{llll}\hfill f(x+1)& ={(x+1)}^3-(x+1)+1& \hfill & \\ \hfill & =(x^3+1)-(x+1)+1& \hfill & \\ \hfill & =x^3-x+1& \hfill & \\ \hfill & =f(x)& \hfill & \end{array}$$

Consequently, $\alpha ,\alpha +1,\alpha +2$ are the distinct roots of $f$, since $0,1,2$ are distinct in ${𝔽}_3$ :

$$f(x)=(x-\alpha )(x-\alpha -1)(x-\alpha -2).$$

As $\alpha +1,\alpha +2\in {𝔽}_3(\alpha )$,

$$L={𝔽}_3(\alpha ,\alpha +1,\alpha +2)={𝔽}_3(\alpha ).$$

$f=x^3-x+1$ being the minimal polynomial of $\alpha$, $[{𝔽}_3(\alpha ):{𝔽}_3]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$.

In conclusion, $L={𝔽}_3(\alpha )$ is the splitting field of $x^3-x+1$ over ${𝔽}_3$. Its degree is 3 over ${𝔽}_3$. As a vector space over ${𝔽}_3$, its dimension is 3, so $L\simeq {𝔽}_3^3$, so its cardinality is $3^3=27$.