Greatest common divisor of polynomial in Finite Field(256), AES Have assigment and will use it as example, found solution computationaly, want to understand idea.
It is about SubBytes procedure in AES, particulary about finding inverse of polynomial.
Suppose we have element $A=x^5+1$ in finite field $F=\mathbb{Z}_2[x]/x^8+x^4+x^3+x+1$ and it is required to compute $A^{-1}$, so that $AA^{-1}=1$
Computationally $A^{-1}=x^6 + x^5 + x^3 + x^2 + x$, $AA^{-1}=(x^5+1)(x^6 + x^5 + x^3 + x^2 + x)=x^{11} + x^{10} + x^8 + x^7 + x^5 + x^3 + x^2 + x$
Lets name $I=x^8+x^4+x^3+x+1$ for convenience, it is ideal of $F$. Then $AA^{-1}-I*x^3-I*x^2-I=1$, $AA^{-1}$ reduces to $1$ alright.
Well, how to get $A^{-1}=x^6 + x^5 + x^3 + x^2 + x$? This is where it gets messy for me.
It is my understanding, that $A^{-1}$ is found by eucleidean algorithm, in this case $(x^5+1)*A^{-1}+I*R=1$, where $R \in F$.
What I do.. try to do:
$(x^8+x^4+x^3+x+1)=(x^5+1)(x^3)+(x^4 + x + 1)$
$(x^5+1)=(x^4 + x + 1)(x)+(x^2 + x + 1)$
.. it is kinda mess, see work at Cloud Sage[Removed]
What I am doing wrong above, when I try to find F.xgcd(A)?

Took a lot of thought, had the right box but was looking in wrong corner.
Will answer myself, for own convenience will change letters a bit.
All coefficients are modulo 2, and it is required to find polynomials $P(x)$ and $Q(x)$ so that $A(x)P(x)+B(x)Q(x)=1$, where $A(x)=x^5+1$ and $B(x)=x^8+x^4+x^3+x^1+x^0$. Actually, we need only $P(x)=A^{-1}(x)$
Anyway:
$x^8+x^4+x^3+x+1=(x^5+1)(x^3)+(x^4+x+1)$ [1]
where
$(x^5+1)=(x^4+x+1)(x)+(x^2+x+1)$ [2]
where
$(x^4+x+1)=(x^2+x+1)(x^2+x)+(1)$ [3]
where, well
$(x^2+x+1)=(x^2+x+1)(1)+0$
Lets express $1$ from [3] by what we have in [1], ie
$1=(x^4+x+1)+((x^5+1)+(x^4+x+1)(x))(x^2+x)$
$1=(x^5+1)(x^2+x)+(x^4+x+1)(x^3+x^2+x)$
excellent, now lets substitute $x^4+x+1$ from [1]
$1=(x^5+1)(x^2+x)+((x^8+x^4+x^3+x+1)+(x^5+1)(x^3))(x^3+x^2)=$
$=A(x)(x^6+x^5+x^3+x^2+x)+B(x)(x^3+x^2+1)$
Where $A^{-1}=x^6+x^5+x^3+x^2+x$, which was looked for. $\Box$
