Question about the degree of the inverse of a polynomial $p(x)$ modulo $q(x)$

Question

I have a doubt about the modular inverse polynomial degree.

Let $p(X)$ be a polynomial in the ring $F[X]$ with $\deg(p)=\delta$, where $F$ is a finite field with characteristic 2. If $\gcd(p(X),q(X)) = 1$, what is the degree of the inverse polynomial of $p(X)$ modulo $q(X)$?

Thanks for your replies.

Answer 1

If you take a Bezout relation in $F[X]$ $$u(X)p(X)+v(X)q(X)=1$$ the inverse of $p(X)$ modulo $q(X)$ is $u(X)$. Now i don't really understand what you mean by the "degree" of the inverse, but if you want the degree of an inverse which is of degree lesser than $\deg(q)$, you should just work out this Bezout relation and see that it gives $\deg(u)+\deg(p)=\deg(v)+\deg(q)$ as long as $q$ is not trivial.