Convert polynomials and fractions in a finit field?

Question

I am trying to understand how finite field works, and I am stuck on converting high power polynomials into a power of the field, also converting fractions into integers.

1. $8^{-1}\cdot44$ in $\Bbb F_{101}$ (the answer is 56 mod 101) I am not too sure how it got there
2. $x^4$ in $\Bbb F_{\large 2^3}$

any enlightenment would be much appreciated!

Answer 1

You have seen answers for the first question. For the second question:

The answer won't be unique. To construct a field of order $p^n$ for a prime $p$ you consider the polynomial ring $F_p[x]$ and a monic ireducible polynomial $f$ of degree $n$. Here $p=2$ and $n=3$.

Note that a polynomial of degree $3$ is reducible if and only if it has a zero. Hence we want a monic polynomial of degree $3$ in $F_2[x]$ with no zeros. It will be of the form

$$f(x)=x^3+ax^2+bx+c$$

where $a,b,c\in F_2$. Now $0$ shan't be a zero, so $c=1$. Then $1$ shan't be a zero as well so $(a,b)=(1,0)$ or $(a,b)=(0,1)$.

We end up with two options

$$f(x)=x^3+x^2+1$$ or $$g(x)=x^3+x+1$$.

In $F_2[x]/f$ we have $x^4=x\cdot x^3=x(x^2+1)=x^3+x=x^2+x+1$.

In $F_2[x]/g$ we have $x^4=x\cdot x^3=x(x+1)=x^2+x$.

You see that the answer is not unique.

Answer 2

$8^{-1}\cdot 44 = 2^{-1}\cdot 4^{-1}\cdot 4\cdot 11=2^{-1}\cdot 11 $. Noting that $2\cdot 51=1$, we see that this is $51\cdot 11=561=56$.

As polynomial $x^4\in F_8[X]$ is just that - $x^4$. However, interpreted as $function$ from $F_8$ to itself, the answer may differ / the expression be rewritten. Is it the latter you are spposed to do?