Let $I=(x^4+x+1)\vartriangleleft\mathbb F_2 [x]$ and $R = \mathbb F_2[x] / I$. I am to:
List irreducible polynomials of $\mathbb F_2$ with degree smaller than 5.
Show that $R$ is a field and find the number of elements in $R$.
Find the inverse of $x^3+x^2+x+1$ in this field.
Show that $R^*$ is a cyclic group by explicitly naming a generator.
I approached these questions in the following way. Since a polynomial in $\mathbb K[X]$ of degree $\leq 3$ is invertible iff it has no roots, first I began by listing the irreducible polynomials of different degrees.
Degree one we have $X, X+1$ and of degree two we only have one: $X^2+X+1$. I found this by letting $f(X)=X^2+aX+c \in \mathbb F_2[X]$ be irreducible, $\implies \not \exists$ a root in $\mathbb F_2$ hence $f(0) = c = 1$ and $f(1)=1+a+c=1+a+1=a=1$.
With the same logic, the only two irreducible polynomials of degree 3 are: $X^3+X+1, X^3+X^2+1$.
Finally, the irreducible polynomials are: $$X^4+X+1, X^4+X^3+1, X^4+X^3+X^2+X+1$$ which gives us the polynomial we need to find an inverse of.
Now, since $X^4+X+1$ is irreducible in $\mathbb F_2 [x]$, $(X^4+X+1)$ is a maximal ideal, and thus R is indeed a field. Now, $[R : \mathbb F_2] = \deg(X^4+X+1)=4$, and thus $|R|=2^4 = 16$.
I also know that the multiplicative group $R^*$ has order $|R|-1=15$.I know that it is cyclic because $(X+1)^3 \not = 1 \not = (X+1)^5$.
Now my question is how do I find the inverse element of the polynomial? And how do I find the explicit generator?
''Now my question is how do I find the inverse element of the polynomial? And how do I find the explicit generator?''
Hints:
Use the extended Euclidean algorithm to divide $x^3+x^2+x+1$ into $x^4+x+1$. This gives you the inverse.
An explicit generator is the residue class of $x$ in $F = \Bbb F_2[x]/\langle x^4+x+1\rangle$, since the polynomial $x^4+x+1$ is primitive.
Note that the residue class of $x$ in $F = \Bbb F_2[x]/\langle x^4+x^3+x^2x+1\rangle$ is not a generator, since the polynomial $x^4+x+1$ is not primitive as it divides $x^5+1$.