How do I prove Bézout's identity of polynomials in $F[x]$? $F$ is a field.
$a(x),b(x)\in F[X]- F$.
$\gcd(a(x),b(x)) = d(x)$ and $u(x)a(x)+v(x)b(x)=d(x)$.
I need to prove that: $\deg(u)<\deg(b)-\deg(d)$.
I use the fact that degree is replacing the absolute value, but I still stuck...
Can you give a hint please? Or what is the direction of the proof?
Thank you!!
 A: The proof is essentially the same as for integers -  descent via (euclidean) division with remainder. To compute the Bezout identity for $\,\gcd(f,g,h)\,$ note that
the set $I$ of polynomials of the form $\, a f + b g +ch$ is closed under addition and scaling so it is closed under remainder = mod, since that is a composition of such operations: $f_i\bmod g_i = f_i - q\, g_i. $ It follows that the $\rm\color{#c00}{least}$ degree $d\in I$ divides every  $e\in I\,$ (else $0\neq e\bmod d\,$ is in $\, I\,$ but has $\rm\color{#c00}{smaller}$ degree than $d).\,$
So $\,f,g,h\in I\Rightarrow d\,$ is a common divisor of $\,f,g,h\,$ necessarily a greatest (degree) common divisor by $\, d'\mid f,g,h\,\Rightarrow\, d'\mid d\!=\! \bar a f + \bar b g+\bar ch,\,$ so $\,\deg d'\le \deg d.\,$ To force the gcd $\,d\,$ to be unique the common convention is  scale it to be monic (lead coef $=1).\,$
The extended Euclidean algorithm is an efficient way to search  $I$ for a polynomial of minimal degree, while  keeping track of each element's representation as a linear combination of $\,f\,$ and $\,g.$
The same idea works for any Euclidean domain (i.e. enjoying division with (smaller) remainder).

We show there exists a Bezout identity with the sought $\rm\color{#c00}{degree\ bound}$ on the coefficient $\,\color{#c00}u\,$  of $\,a.\,$ By above there is a Bezout identity $\ u' a + v'b = d.\,$  Dividing $\,u'\,$ by $\,b/d\,$ yields $\,u' = q\, b/d + u\,$ with quotient $\,q\,$ and remainder $\, u\,$ satisyfing the sought  $\,\color{#c00}{\deg(u) <} \deg(b/d) = \deg(b)-\deg(d)$
Substituting $\,u' = q\, b/d + u\,$ into $\ d = u'a + v'b \ $ yields the sought Bezout identity, i.e.
$$ d = u'a+v'b = (q\, b/d + u)\,a + v'b = u\,a + (\color{#0a0}{v' + q\,a/d})\, b = \color{#c00}u\,a+\color{#0a0}v\,b \qquad\qquad  $$
It's clearer via mod: $\ 1 = u' a/d + v' b/d\iff u'\equiv (a/d)^{-1}\!\pmod {\!b/d}$ and any such inverse $u'$ remains an inverse when reduced $\!\bmod b/d\,$ to $\, u:= u' \bmod b/d = $ the remainder above. So the algebra above amounts to modular reducing an inverse to force its degree less than the modulus.
