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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.

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Since $F[X]$ is a Euclidean domain, the degree of $f^{-1} \bmod q < \deg(q).$ Also, the inverse of $p$ is $u\bmod q$ for the Bézout coefficients $u, v$ in $pu + qv = 1.$ – user2468 Jul 19 '12 at 18:34
And more than @J.D. says above, can not be determined. You can’t expect to say anything reasonable about the degree of the reciprocal other than that it’s generically $1$ less than the degree of $q$. – Lubin Jul 19 '12 at 18:48
Just to complete the above comments, note that for any positive integers $m,n$, the inverse of $X^n$ modulo $X^{m+n}-1$ is $X^m$. – N. S. Jul 19 '12 at 19:03
up vote 1 down vote accepted

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.

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