# Zeros in Splitting Factor Rings

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.

• It is helpful if you actually type out your full question, instead of posting a link for us to follow. – Morgan Rodgers May 21 at 0:35
• Welcome to Mathematics Stack Exchange. The link is apparently talking about $\Bbb F_3$ – J. W. Tanner May 21 at 0:45

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.

• Thank you so much! Would you mind explaining what $\theta$ (the image of $x$) represents? – Jake McNaughton May 21 at 0:45
• You're welcome! – Bernard May 21 at 0:46
• Well, $\theta$ represents a particular element of the extension field (=factor ring). That’s all. You see that the base field was $\Bbb F_3$ and the extension field is $\Bbb F_{3^3}=\Bbb F_{27}$, the unique (up to isomorphism) field with $27$ elements. – Lubin May 21 at 3:27

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 $$K[x]$$.