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.

Calculations on GF(16) find 0111/1111
It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111\in GF\left(16\right)$ generated by $\mathrm{\Pi }\left(\alpha \right)=1+\alpha +{\alpha }^{4}$ 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 $0111\equiv -1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}1111\right)$.
How can I compute this result -1 in the right element of the field? Or perhaps this some kind of sign that the result $\notin GF\left(16\right)$?
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canhaulatlt
Step 1
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 γ the element that you refer to as α.
Step 2
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
$\begin{array}{rl}\frac{0111}{1111}& =\frac{{\alpha }^{11}}{{\alpha }^{12}}=\frac{{\alpha }^{11}}{{\alpha }^{12}}\cdot \frac{{\alpha }^{3}}{{\alpha }^{3}}\\ & =\frac{{\alpha }^{14}}{{\alpha }^{15}}=\frac{{\alpha }^{14}}{1}\\ & ={\alpha }^{14}={\alpha }^{3}+1=1001.\end{array}$
Did you like this example?
Kendrick Finley
Step 1
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 ${\alpha }^{15}=1$.
Step 2
Over the polynomial ${X}^{4}+X+1$ we have $0111={\alpha }^{11}$ and $1111={\alpha }^{12}$. Therefore:
0111/1111=α11/α12=α11+15−12=α14=1001
This is consistent with your result:
$1+\frac{1}{{\alpha }^{3}+{\alpha }^{2}+\alpha +1}=1+\frac{1}{{\alpha }^{12}}=1+\frac{{\alpha }^{15}}{{\alpha }^{12}}=1+{\alpha }^{3}$