Show the ring $R = \mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1)$ is a field 
Show that the ring $R = \mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1)$ is a field. Find the multiplicative inverse of the element [x^2+1] in that field.

From what I know,
$$\mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1) = \{ a_0 + a_1x+a_2x^2+a_3x^3 + x^4 : a _i = \{0,1\}\}$$
To show that $R$ is a field, I would need to show that it is a commutative division ring.
How can I show that formally? Especially the fact that every non-zero element of $R$ has an inverse with respect to multiplication.
 A: Why don't prove that $x^4+x+1$ is irreducible over $\Bbb Z/2\Bbb Z$? 
Denote $a=[x]$. We have $a^4=a+1$, so $a^8=a^2+1$. But $a^{15}=1$ (since $R$ is a field with $2^4=16$ elements and $a\in R^{\times}$), so $a^7$ is the inverse of $a^2+1$. Now, from $a^4=a+1$ we find $a^7=a^3+a+1$.
A: You are a little wrong.  Since the ideal is generated by $x^4+x+1$, you are saying that 
$$
x^4+x+1=0
$$
in the quotient ring.  In $\mathbb{Z}/2$ coefficients, this is the same as
$$
x^4=x+1.
$$
So, you have a generating set $1,x,x^2,x^3$.  If you don't have any theorems to show that the quotient ring is a field, you need to check that each element is invertible.  We can reduce our work a bit.  Namely, if any nonzero element is a product of two invertible elements, then it is also invertible.  First notice that
$$
x(x^3+1)=x^4+x=x+1+x=1.
$$
Thus $x$ is invertible.  Then all powers of $x$ are invertible as long as they are nonzero.  As in @user's answer, you can check, using $x^4=x+1$, that the set $\{x^n\,\,|\,\, n=0,\ldots 14\}$ generates all nonzero elements of the quotient ring.  
