Calculations on $GF(16)$ find $0111/1111$ It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111 \in GF(16)$ generated by $\Pi(\alpha)=1+\alpha +\alpha^4$ that is an irreducible polynomial. 
In polynomial form we have:


*

*$0111 \rightarrow \alpha+\alpha^2+\alpha^3$

*$1111 \rightarrow 1+\alpha+\alpha^2+\alpha^3$
If I perform the polynomial division, I obtain $-1$ (that is the same result obtained writing $0111 \equiv -1 \pmod {1111}$).
How can I compute this result $-1$ in the right element of the field? Or perhaps this some kind of sign that the result $\not \in GF(16)$?
 A: $GF(16)$ has characteristic 2. That is, each coefficient of $\alpha$ is either $0$ or $1$. And $-1=1$.
However, the polynomial division does not just result in $-1$ or $1$. Instead we have:
$$\frac{\alpha^3+\alpha^2+\alpha}{\alpha^3+\alpha^2+\alpha+1}
= 1 + \frac{1}{\alpha^3+\alpha^2+\alpha+1}
$$
As an alternative to the answers in the comments, we can write each element (except $0$) as a power of $\alpha$. After all, $GF(16)$ has a cyclic multiplicative group and $\alpha^{15}=1$.
Over the polynomial $X^4+X+1$ we have $0111 = \alpha^{11}$ and $1111=\alpha^{12}$. Therefore:
$$0111/1111=\alpha^{11}/\alpha^{12}=\alpha^{11+15-12}=\alpha^{14}=1001$$
This is consistent with your result:
$$1 + \frac{1}{\alpha^3+\alpha^2+\alpha+1}=1+\frac{1}{\alpha^{12}}
=1+\frac{\alpha^{15}}{\alpha^{12}}=1+\alpha^3
$$
A: As it happens, the discrete logarithm table for $GF(16)$ that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by $\gamma$ the element that you refer to as $\alpha$.
Anyway, consulting that table, we see that
$$1111=1+\alpha+\alpha^2+\alpha^3=\alpha^{12},$$
and
$$
0111=\alpha+\alpha^2+\alpha^3=\alpha^{11}.
$$
Therefore
$$
\begin{aligned}
\frac{0111}{1111}&=\frac{\alpha^{11}}{\alpha^{12}}=\frac{\alpha^{11}}{\alpha^{12}}\cdot\frac{\alpha^3}{\alpha^3}\\
&=\frac{\alpha^{14}}{\alpha^{15}}=\frac{\alpha^{14}}1\\
&=\alpha^{14}=\alpha^3+1=1001.
\end{aligned}
$$
