# Bezout's identity on $F[x]$ with constraints

I have some issues with solving this exercise:

Prove: Let $$F$$ be a field. If $$f,g∈F[x]$$ are relatively prime and not both constant, then there exists $$a,b∈F[x]$$ such that $$af+bg=1$$ and $$\deg(a)<\deg⁡(g)$$, $$\deg⁡(b)<\deg⁡(f)$$.

My attempt: We know that $$F[x]$$ is a principal ideal domain, so the ideal $$I≔{af+bg∶a,b∈F[x]}$$ is equal to $$(d)$$ for some $$d\in F[x]$$ with minimal degree in the set. Let’s do the division of $$f$$ by $$d$$ $$f=dq+r$$ where $$\deg⁡(r)<\deg⁡(d)$$ and $$r=f-dq\in I$$, so $$r=0$$ and $$d\mid f$$. In the same way we get $$d\mid g$$. Now if $$c\mid f,g$$, then $$c$$ divides all the non-zero element of $$I$$ and so $$d$$ too. Therefore $$d=\gcd(f,g)$$ and $$1=d=af+bg$$ for some $$a,b\in F[x]$$.

The problem is that I'm not able to prove the fact about degrees of $$a$$ and $$b$$.

• Use your equation, apply the division algorithm to reduce the degree of the "coefficients" $a,b$. – xbh Nov 2 '18 at 15:26

If $$af+bg=1$$ and $$\deg a\geq\deg g$$ then we can write $$a=qg+r$$ with $$\deg r<\deg g$$ and so $$1=af+bg=(a-qg)f+(b+qf)g=rf+(b+qf)g.$$ Comparing degrees shows that $$\deg(b+qf)g=\deg rf$$ and hence that also $$\deg b+qf<\deg f$$.
Hint  The general solution is $$\,(\bar a,\bar b) =\!\! \overbrace{(a,b)}^{\rm particular}\!\!+\!\!\overbrace{h(g,-f)}^{\rm homogeneous}\! = (a\!+\!hg,b\!-\!hf).$$ Via division algorithm, choose $$h$$ so $$\, \deg(a\!+\!hg) < \deg(g)\$$ (this implies the other bound via $$\,\bar a f+\bar b g = 1)$$