Multiplicative inverse of polynomial

Question

Question:

Determine the multiplicative inverse of $x^2 + 1$ in $GF(2^4)$ with $$m(x) = x^4 + x + 1.$$

My confusion is over the $GF (2^4)$.

Answer 1

I'm unclear what you're going for here, but it seems like you're talking about the 4-dimensional vector space over $\mathbb F_2$ the finite field of two elements, where $V=\mathbb F_2[X]/\langle m(X)\rangle$.

Working under that assumption, you need a polynomial $AX^3+BX^2+CX+D$ so that $$ (X^2+1)(AX^3+BX^2+CX+D)=(X^4+X+1)p(X)+1\equiv 1\in V, $$ where $A,B,C,D=\{0,1\}$ and $p(X)$ is some polynomial, which you can see will have degree at most $1$. You're gonna have to play around with it to come up with the answer, but this should point you in the right direction.

Answer 2

Here $\,{\rm GF}(2^4) \cong \Bbb F_2[x]/(x^4\!+x+1)\,$ denotes the finite field of size $\,2^4.\,$ We can apply the Extended Euclidean Algorithm described here over $\,\Bbb F_2\,$ to invert $\,x^2\!+1\,$ in this field

$$\begin{array}{cccc} & x^4\!+x+1 & 1 & 0\\ & x^2\!+1 & 0 & 1\\ &x^2\!+x+1 & 1 & x^2\\ & x & 1 & x^2\!+1\\ & 1 & x & x^3\!+x+1\\ \end{array}$$

where each row $\ a\ \ b\ \ c\ $ means that $\ a = b(x^4\!+x+1) + c(x^2\!+1).\ $ Hence, by the final row

$$ 1\, =\, x(x^4\!+x+1) + (x^3\!+x+1)(x^2\!+1)$$

which, finally, implies that $\ (x^3\!+x+1)(x^2\!+1) = 1\,$ in $\,\Bbb F_2[x]/(x^4\!+x+1).$