Show that $K$ is a field

Question

Let $f = x^3+x+1 \in \mathbb{F}_2[x]$. We know that $K = \mathbb{F}_2[x] / \langle f \rangle$ is a ring. I just need to show that is also a field.

Its actually the first time that I need to work with a polynomial in $\mathbb{F}_2[x]$ and Im not sure how to show that $K$ is a field.

Answer 1

One more hint: The polynomial $f=x^3+x+1$ is irreducible because it is a cubic polynomial which has no root in $\mathbf F_2$. As $\mathbf F_2[x]$ is a P.I.D., this implies the ideal $(f)$ is prime, so that $\mathbf F_2[x]/(f)$ is an integral domain.

Answer 2

List the conditions for a ring to be a field.

• Has multiplicative identity (namely $1$)
• Every element except $0$ has a multiplicative inverse.
• Multiplication commutes (obvious)

So the core problem is:

$$f \text{ is irreducible over } \mathbb F \Leftrightarrow \text{All elements }p\in\mathbb F/\langle f\rangle \text{ except $0$ have inverses}.$$

See how this resembles proving that $\mathbb Z/p\mathbb Z$ forms a field ($p$ prime)? When we prove that every non-zero element in $\mathbb Z/p\mathbb Z$ has an inverse, we use Bezout's identity:

Suppose $(a,b)=1$. There exist $m,n\in\mathbb Z$ that $$ma+nb=1.$$

And the proof of that relies on the generalized Euclidean algorithm, which, in turn, requires division with remainder. Now we know that, in $\mathbb F_2[x]$ we can perform polynomial division analogous to division with remainder in integers. So you can generalize Bezout's identity to obtain the result.