For complex polynomials $\gcd(f,g)=1$ if and only if $f$ and $g$ have no common root

Question

Assuming the fundamental theorem of algebra, prove the following. If $f$ and $g$ are polynomials over the field of complex numbers, then $\gcd(f,g)=1$ if and only if $f$ and $g$ have no common root.

Answer 1

If $h$ divides $f$ then every root of $h$ is root of $f$. Thus if some $h$ divides $f$ and $g$ then all roots of $h$ will be common to $f$ and $g$. This yields one direction.

Conversely if there is a root $r$ of $f$, then $X-r$ divides $f$. Thus a common root directly yields a non-constant common divisor.

Answer 2

This is a rather common proof and not too complicated.

Let's say $\gcd(f, g)=1$. We will prove that $f$ and $g$ has no common root by contradiction.

Assume $f$ and $g$ have a common root $\alpha$. Then, by the Factor Theorem, $x-\alpha$ is a common root of both $f$ and $g$, which contradicts that $\gcd(f, g)=1$.

Thus, our assumption is false and if $\gcd(f, g)=1$, then $f$ and $g$ have no common root.

Now, let's say that $f$ and $g$ has no common root $\alpha$. We will prove that $\gcd(f, g)=1$ by contradiction.

Assume $f$ and $g$ have some non-constant common factor $h$. By the Fundamental Theorem of Algebra, $h$ has some zero $\alpha$. Since $h$ divides $f$ and $g$, this means that $f$ and $g$ must also have this zero, which contradicts that $f$ and $g$ have no common root.

Thus, our assumption is false and if $f$ and $g$ has no common root, then $\gcd(f, g)=1$.

By proving both of these conditionals, we have shown that $f$ and $g$ has no common root if and only if $\gcd(f, g)=1$.