Show that $\Bbb F_2[x]/(x^3+x+1)$ is a field and more... I have been going over some old papers to prepare for my exams, and this one does not have a set of solutions to compare to. 
i) Show that  $\Bbb F_2[x]/(x^3+x+1)$ is a field by considering powers of $x$
I found that $x=x, x^2=x^2, x^3=x+1, x^4=x^2+x,x^5=x^2+x+1, x^6=x^2+1, x^7=1$
So the order of $x$ is 7, but I am not sure how I can use that to show that its a field. I was tempted to use the fact that the polynomial by which we are $"quotienting"$ is irreducible over $\Bbb F_2$ and hence it is a field. But I am not sure if that is okay to do. 
Another idea is to show that $\Bbb F_2[x]/(x^3+x+1)$ is a finite (and commutative) integral domain and this would imply its a field.
ii) Show that $x$ generates the multiplicative group of $\Bbb F_2[x]/(x^3+x+1)$ and hence identify what $(x^2+1)^{-1}$ would be in this field 
Since I could not get part i) done,  I can't really proceed here either but I think I see that the multiplicative group of $\Bbb F_2[x]/(x^3+x+1)$ is {$1,x,x^2,x^3,x^4,x^5,x^6$} as we simply have to take out $0$ and form a multiplicative group from the additive group of $\Bbb F_2[x]/(x^3+x+1)$. I am not sure about finding $(x^2+1)^{-1}$ either.
 A: i) The normal way to show that ${\mathbb F}_2[x]/(f)$ is a field, is indeed by showing that $f$ is irreducible. This makes $(f)$ a maximal ideal and ${\mathbb F}_2[x]/(f)$ a field. In this case, $f = x^3 + x + 1$ is of degree $3$ and hence it is enough to show that is has no roots in ${\mathbb F}_2$.
To actually use the powers of $x$: you have shown that the powers of $x$ are all non-zero elements of ${\mathbb F}_2[x]/(f)$ and therefore all non-zero elements are invertible. This makes ${\mathbb F}_2[x]/(f)$ a field as well. 
ii) Your solution to i) in fact solves ii). You have shown that $1, x, x^2, \dots, x^6$ are all $7$ non-zero elements of ${\mathbb F}_2[x]/(f)$, i.e., $x$ generates ${\mathbb F}_2[x]/(f)^*$. Since it turns out that $x^6 = x^2 + 1$ (and because $x^6 \cdot x = x^7 = 1)$, the inverse of $x^2 + 1$ is $x$.
A: Hints: 


*

*You have already found $(x^2+1)^{-1}$ in your work with (a), since you have $x^6=x^2+1$ and $x^7=1$

*How many elements are in this ring, and how many of those elements are powers of the invertible element $x$, and thus are themselves invertible (by a similar argument to hint 1)?

