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.

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

  • $\begingroup$ Thank you so much! Would you mind explaining what $\theta$ (the image of $x$) represents? $\endgroup$ – Jake McNaughton May 21 at 0:45
  • $\begingroup$ You're welcome! $\endgroup$ – Bernard May 21 at 0:46
  • $\begingroup$ 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. $\endgroup$ – 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 $K[x]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.