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 $\mathrm{\Pi}(\alpha )=1+\alpha +{\alpha}^{4}$ that is an irreducible polynomial.

In polynomial form we have:

$0111\to \alpha +{\alpha}^{2}+{\alpha}^{3}$

$1111\to 1+\alpha +{\alpha}^{2}+{\alpha}^{3}$

If I perform the polynomial division, I obtain -1 (that is the same result obtained writing $0111\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}1111)$.

How can I compute this result -1 in the right element of the field? Or perhaps this some kind of sign that the result $\notin GF(16)$?

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

In polynomial form we have:

$0111\to \alpha +{\alpha}^{2}+{\alpha}^{3}$

$1111\to 1+\alpha +{\alpha}^{2}+{\alpha}^{3}$

If I perform the polynomial division, I obtain -1 (that is the same result obtained writing $0111\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}1111)$.

How can I compute this result -1 in the right element of the field? Or perhaps this some kind of sign that the result $\notin GF(16)$?