- #1

Bashyboy

- 1,421

- 5

## Homework Statement

Let ##f(x) = x^2 + x + 1 \in \Bbb{F}_2[x]##. Prove that ##f(x)## is irreducible and that ##f(x)## has a root ##\alpha \in \Bbb{F}_4##. Use the construction of ##\Bbb{F}_4## to display ##\alpha## explicitly

## Homework Equations

Definition: An element ##p## in a domain ##R## is irreducible if ##p## is neither ##0## nor a unit and, in every factorization ##p=uv##, either ##u## or ##v## is a unit.

Unless I am misunderstanding my book, ##\Bbb{F}_2 = \Bbb{Z}_2##, and ##\Bbb{F}_4## is the field of matrices of the form

$$\begin{bmatrix} a & b \\ b & a+b \\ \end{bmatrix}$$

where the entries come from ##\Bbb{F}_2##

## The Attempt at a Solution

For now I am focusing on the first part. Note that ##f(x)## is not the zero polynomial (zero element), nor is it a unit, since it is not a constant polynomial, which are the only units since ##\Bbb{F}_2## is a field and therefore ##\Bbb{F}_2[x]## is a domain. To see this, suppose that ##f(x)## is in fact a unit. Then there exists a ##u(x)## such ##1 = f(x) u(x)## which implies ##0 = \deg(1) = \deg(f) + deg(u)##, and therefore ##\deg(f) = 0##, which makes ##f(x)## the constant polynomial. But clearly this isn't the case, and so ##f(x)## cannot be a unit.

Now suppose that ##f(x) = u(x) v(x)## but ##v(x)## is not a unit. This is where I run into difficulties. I am trying to argue that ##u(x)## is a unit but I am having trouble. I could use a hint.