Let $f(x) = \sum ^{n}_{k = 0} a_nx^n,\ g(x) = \sum ^{m}_{k = 0} b_mx^m \in \mathbb{F}[x]$ be monic polynomials on some arbitrary field. Then show equivalence of the following conditions:
1). $f$ and $g$ are relatively prime
2). $\deg(lcm(f,g)) = m+n$
3). $x^if(x), x^jg(x),$ are linearly independent over $\mathbb{F}$ for $i, j \in \mathbb{N}\ ,\ 0\leq i <m, 0\leq j < n$
1) implies 2), since relatively prime, $lcm(f,g) = f g$, hence the degree.
I don't understand how 2) implies 3) and 3) implies 1).
Any input will be helpful, thanks in advance.
Edit:
2) implies 1) is pretty obvious, by considering the contrapositive. So I just need 1) iff 3).
Suppose $f$ and $g$ are not relatively prime, then $f(x) = (x-r_0)f_0(x),\ g(x) = (x-r_0)g_0(x)$ Then, $f_0, g_0$ are polynomials of degree $m-1,\ n-1$ s.t. $f(x)g_0(x) = g(x)f_0(x)$.
$f(x)g_0(x) - g(x)f_0(x) = 0$ With some argument about the degree of each term, this should conclude $3) \implies 1)$.
For $1) \implies 3)$, it is better to prove $\neg 3) \implies \neg 2)$. If $f_0, g_0$ with degree less than $m-1, n-1$ exist, then, $f(x)g_0(x)$ is a common multiple of $f$ and $g$ while its degree is less than or equal to $m+n-1$. Hence $\neg 2)$. This concludes this problem.
Can someone verify?