great common divisor

**apple2009** · March 2nd 2010, 06:54 PM

Prove: let F be a field and let g(x) and h(x) be polynomials in F[x] with gcd(g(x),h(x))=1. Let a, b be elements of F with a‡b. Then gcd(h(x)-ag(x), h(x)-bg(x))=1.

**hatsoff** · March 3rd 2010, 08:40 AM

Originally Posted by apple2009

Prove: let $F$ be a field and let $g(x)$ and $h(x)$ be polynomials in $F[x]$ with $\text{gcd}(g(x),h(x))=1$ . Let $a, b$ be elements of $F$ with $a\neq b$ . Then $\text{gcd}(h(x)-ag(x), h(x)-bg(x))=1$ .

Let $p(x)$ be an irreducible monic divisor of $h(x)-ag(x)$ and $h(x)-bg(x)$ . Then $p(x)$ divides $bh(x)-abg(x)$ and $ah(x)-abg(x)$ , and therefore also $[ah(x)-abg(x)]-[bh(x)-abg(x)]=(a-b)h(x)$ and $h(x)$ . This in turn means $p(x)$ is a factor of $g(x)$ . Since $0\leq\text{deg }p(x)\leq\text{deg gcd}(h(x),g(x))=0$ , then $p(x)=1$ , which implies the conclusion.

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