1. ## GCD

We have polynomials f(x),g(x) (degree>1), their GCD has degree>1.
I have to prove that there exist polynomials u(x),v(x) such that:
f(x).u(x)=g(x).v(x) and deg(u)<deg(g),deg(v)<deg(f). $a^-1(x)$

Am I right?
let d(x) be their GCD, then we have:
f(x)=a(x).d(x)
g(x)=b(x).d(x)
so $f(x).a^-1(x)=g(x).b^-1(x)$ where and $b^-1(x)$ are polynomials u(x),v(x)

is the proof of existence ok?
I need help with showing the inequality about degrees. Can anybody help me please?

2. Originally Posted by sidi
We have polynomials f(x),g(x) (degree>1), their GCD has degree>1.
I have to prove that there exist polynomials u(x),v(x) such that:
f(x).u(x)=g(x).v(x) and deg(u)<deg(g),deg(v)<deg(f). $a^{-1}(x)$
where did $a^{-1}(x)$ come from?? $\deg(v) < \deg(f).a^{-1}(x)$ has no meaning because the RHS of the inequality is not a number!!

let d(x) be their GCD, then we have:
f(x)=a(x).d(x)
g(x)=b(x).d(x)
so $f(x).a^-1(x)=g(x).b^-1(x)$ where and $b^-1(x)$ are polynomials u(x),v(x)
the inverse of a polynomial is not necessarily a polynomial. so $a^{-1}(x)$ and $b^{-1}(x)$ might not even exist in your polynomial ring!

the problem, as you posted, doesn't make any sense and you certainly won't get help if your problem has itself a problem!

3. Thank you, there was a mistake.
So the problem is:
We have polynomials f(x),g(x) (degree>1), their GCD has degree>1.
I have to prove that there exist polynomials u(x),v(x) such that:
f(x).u(x)=g(x).v(x) and deg(u)<deg(g),deg(v)<deg(f)

f(x),g(x) are from F[x], where F is a field

4. Originally Posted by sidi
Thank you, there was a mistake.
So the problem is:
We have polynomials f(x),g(x) (degree>1), their GCD has degree>1.
I have to prove that there exist polynomials u(x),v(x) such that:
f(x).u(x)=g(x).v(x) and deg(u)<deg(g),deg(v)<deg(f)

f(x),g(x) are from F[x], where F is a field

that's good now! so suppose $d(x)=\gcd(f(x),g(x)).$ we have $f(x)=d(x)f_1(x), \ g(x)=d(x)g_1(x).$ now let $u(x)=xg_1(x)$ and $v(x)=xf_1(x).$ see that
$f(x)u(x)=g(x)v(x).$ we also have: $\deg u(x)= 1 + \deg g_1(x) < \deg d(x) + \deg g_1(x) = \deg (d(x)g_1(x))=\deg g(x).$ similarly $\deg v(x) < \deg f(x).$