Zeros in Splitting Factor Rings

Question

I am looking at the following proof that $x^3-x-1$ splits in an extension field of $F_{3}[x]$.

Let us consider the field $$K = F_3[x]/\langle x^3-x-1\rangle$$ If $\theta$ is the image of $x$ in $K$, then $\theta$ is a root of $x^3-x-1$. respectively, as are $(\theta + a)$ for all a ∈ $F_3$, hence, K1 contains each root of $x^3 − x − 1$.

What I do not understand is what $\theta$ represents and why $(\theta+a)$ is also a zero.

Answer 1

This results from the particular form of the polynomials, and the Frobenius morphism: $$(x+1)^3-(x+1)-1=x^3+1^3-x-1-1=x^3-x-1,$$ so if it is $0$ for $x=a$, it is also $0$ for $x=a+1$, and similarly for the other polynomial.

Answer 2

Let $I=\langle x^3-x-1\rangle$; then you can consider $K=F_3[x]$, which is a field due to irreducibility of $x^3-x-1$. Consider the canonical map $\pi\colon F_3[x]\to K$, and set $\theta=\pi(x)=x+I$.

Then $\theta^3-\theta-(1+I)=(x+I)^3-(x+I)-(1+I)=(x^3-x-1)+I=0+I$.

Now forget that $K$ is actually a quotient ring and identify $a+I$ with $a$ (for $a\in F_3$), which is possible because the restriction of $\pi$ to $F_3$ is injective. Then we have, in $K$, $\theta^3-\theta-1=0$.

Therefore $K$ is an extension field of $F_3$ and $\theta$ is a root of $x^3-x-1\in K[x]$.

What happens to $\theta+a$? Well $$ (\theta+a)^3=\theta^3+a^3 $$ due to the field having characteristic $3$; also $a^3=a$ for $a\in F_3$. Hence $$ (\theta+a)^3-(\theta+a)-1=\theta^3+a^3-\theta-a-1=\theta^3-\theta-1=0 $$ Hence $\theta$, $\theta+1$ and $\theta+2$ are distinct roots of $x^3-x-1$ in $K$ and so $$ x^3-x-1=(x-\theta)(x-\theta-1)(x-\theta-2) $$ in $K[x]$.