1
$\begingroup$

This a new chapter that we are learning and the teacher is flying through it and this are also new concept that i have just learn and i was wondering if i can have some guidance in this problem.

Show that $x^3 + x + 1$ is irreducible over $F_{2}$ and let $\theta$ be a root. Compute the powers of $\theta$ in $F_{2} (\theta)$.

Proof: $F_2$ has only two element $0$ and $1$ thus to show that the polynomial is irreducible i did $$0^3+0+1 = 1 \not=0$$ and $$ 1^3+1+1 = 3 = 1 \not=0$$ since there is no root therefore the polynomial is irreducible.

Now my question is how to you go about computing the powers of $\theta$

$\endgroup$

marked as duplicate by Dietrich Burde, Brian M. Scott, user26857, user147263, Rolf Hoyer Apr 29 '15 at 22:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Hint: Field extensions behave pretty much like the well-known case of extending $\Bbb R$ by $i$, where $i$ knows nothing but that it is a root (somewhere in a bigger field) of the polynomial $x^2+1$.

So, now you have a $\theta$ which satisfies $\theta^3=\theta+1$. Can you carry on from here?

$\endgroup$
  • $\begingroup$ What i am asking is that since we have a degree 3 polynomial does that mean i have to compute $\theta ^3$, $\theta ^2$, and $\theta $ $\endgroup$ – user146269 Apr 29 '15 at 20:21
  • $\begingroup$ No, Since we have a degree $3$ polynomial, it means that $1,\theta,\theta^2$ is the basis of $F_2(\theta)$ as a vector field over $F_2$, so that all elements of the extended field, in particular, $\theta^3,\,\theta^4,\,\theta^5,\,\dots$ are expressible by a polynomial of $\theta$ of degree at most $2$. $\endgroup$ – Berci May 10 '15 at 23:06

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